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ContentsPreface Page No.

1 . Newton's Laws of Motion

Exercise 001 - 025

2 . Friction

Exercise 026 - 043

3 . Gravitation

Exercise 044 - 058

4 . Work, Power & Energy

Exercise 059 - 078

5 . Circular Motion

Exercise 079 - 95

6 . Centre of Mass

Exercise 96 - 113

7. Rigid Body Dynamics

Exercise 114 - 156

8 . Unit and Dimensions

Exercise 157 - 161

Physics Sheet Solutions 2nd Dispatch

CLASS : XI

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RESONANCE SOLN_NEWTON'S LAWS OF MOTION - 1

TOPIC : NEWTON'S LAWS OF MOTION

EXERCISE-1PART - I

SECTION (A)

A-1. Gravitational, Electromagnetic, Nuclear.

A-4. Newton's IIIrd Law

A-6. Vertical wall does not exert force on sphere (N' = 0).

A-8.

action reaction pairs (1) and (2)(3) and (4)(5) and (6)(7) and (8)

SECTION (B)

B-1. N = F + mg [equilibrium] N = mg + mg N = 2mg

B-3.

If is obvious that block can`t accelerate in y direction N � mg cos = 0 N = mg cos


B-5.

Due to symmetry normal reactions due to left and right wall are same in magnitudeW � N cos 60 � N cos 60 = 0 [Equilibrium in vertical direction]

W � 2N

� 2N

= 0 N = W

B-7. N1 cos300 = 50 +

2

N2

N1

23

� 2

N2 = 50 ................ (1)

N1 sin300 =

2

N2

N1 = 2 N

2 ..............................(2)

Solving equation (1) & (2)N

1 = 136.57 N

N2 = 96.58 N

SECTION (C)

C-1. Since string 2 is massless net force on it must be zero. T

2 = F = 10 N

T1 + mg = T

2[Equilibrium of block]

T1 + 1 × 10 = 10

T1 = 0

C-3.


TC � 10 = 0 [Equilibrium of block]

TC = 10 N

TB � T

C � 5 = 0 [Equilibrium of 2]

TB � 10 � 5 = 0

TB = 15 N

TA � T

B � 5 = 0 [Equilibrium of 1]

TA = 20 N

C-5. For finding distance travelled we need to knowthe acceleration and initial velocity of block.m

2g � T = m

2a [Newton�s second law for m

2]

T � m1g = m

1a [Newton�s second law for m

1]

m2g � m

1g = (m

2 + m

1)a [adding both the equation]

a = 12

12

mmg)m�m(

a = 363�6

× g

a = 3g

= 3

10 m/s2

s = u t + 21

at2 = 0 × 2 + 21

× 3

10 × 22

S = 3

20 m

T � m1g = m, a

T = m1

3g

g = 3 × 340

T = 40 N

Force exerted by clamp on pulley is 2T 2 × 40 = 80 N

C-7. VA = V

P2 = 10m/s

For pulley

1 2

VP

V2V1

P P

A

B

C

VPP1

P2

VA.

VB.

V = 10m/s p2 .

2m/s = VC

VP =

2VV 21

VP2

= 2

V�V CB

10 = 2

2VB VB = 22 m/s

and2

VVV 2

1

PAP

0 = 2

10VA V

A = 10 m/s

SECTION (D)

D-1. Since string is inextensible length of string can�t change

rate of decreases of length of left string= rate of increase of length of right string

V1 cos

1 = V

2 cos

2

2

1

VV

= 1

2

coscos


D-3. Since rod is rigid, its length can�t increase.

velocity of approach of A and B point of rod is zero. u sin � v cos = 0 v = u tan at any angle x and y coordinates of center of mass are

cos2

X

...............(i)

sin2

Y

...............(ii)

from (i) and (ii)

4YX

222

equation of circle.

D-5. V1 =

220�10

[constrained relation of P1 ]

V1 = � 5 m/s

10 = 2

V5� 2

V2 = 25 m/s

VC = V

2 = 25 m/s upwards

1PV = V1 = � 5 m/s V

P = 5 m/s downward

[because we have assumed upward direction as +ve for V1]

SECTION (E)

E-1. Since point A is massless net force on it must bezero other wise it will have acceleration. F

1 � 60 cos 45 = 0

F1 = 30 2 N

F2 � 60 cos 45 = 0

F2 = 30 2 N

W � 60 sin 45 = 0

W = 30 2 N

E-3.

F =

am

a = i�ax + j�ay = 2

2

dt

xd i� + 2

2

dt

yd i� = (10) i� + (18 t) j�

at t = 2 sec t = 2 sec

a = 10 i� + j�36

F = 3 )j�36i�10(

= j�108i�30

F = 22 10830 = 112.08 N


E-5. R4 � mg = ma

R4 � 1 = 0.1 × 2

R4 = 1.2 N

R3 � mg � R

4 = ma

R3 � 1 � 1.2 = 0.1 × 2

R3 = 2.4 N

SimilarlyR

2 = 3.6 N

R1 = 4.8 N

F = 6 N F

net = ma

= 0.1 × 2

= 0.2 N

E-7. (a) When the block m is pulled 2x towards leftthe pully rises vertically up by x amount. a

B = 2a

A

F.B.D. of blocks mB

2m

2x

x

a

T

A

2a

>

> >>2T

T

m T

2aB

T = m2a ............. (1)

F.B.D.

FBD of A FBD

2T

2m a

2mg

A2mg � 2 T = 2ma

mg � T = ma ................(2)

(1) + (2) mg = 3maa = g/3 a

B = 2g/3


(b) = xB + 3x

A

0 = 2B

2

dt

xd+3 2

A2

dt

xdmB

T

T

2T

T

aB

3m^

^^^

^ xB

A

3mg

aA

0 = � aB + 3a

A

aB = 3a

A ........... (1)

For B T = maB ....................... (2)

For A 3mg � 3T = 3maA ............... (3)

mg � T = maA

By (1) , (2) & (3)

aB = 3g/4 Ans.

SECTION (F)F-1. Reading of weighing machine is equal to the normal reaction

Normal reaction is not affected byvelocity of lift, it is only affected by acceleration of lift.For I, II and III a = 0

N � mg = 0 [Equilibrium of man]N = mg = 600 N

For IV, VI and VII a = +2 m/s2

N � mg = ma [Newtons II law]N = 60 × 2 + 60 × 10 = 720 N

For V and VIII a = � 2 m/s2

N � mg = ma [Newtons II law]N = 60 × (�2) + 60 × 10 = 480 N

F-3. (a) TAB

= 2mg, TBC

= mgFor A 2mg + mg = ma

A a

A = 3g

For B TAB

� mg � TBC

= maB

A m

TAB

mg

B m

TAB

mgTBC

C m

mg

TBC

2mg � mg � mg = maB ma

B = a

B = 0

TBC � mg = ma

c a

c = 0.

(b) TAB

= 2mgT

AB � mg = ma

B

2mg � mg = maB

B m

mg

TAB

m aB

aB = g ()

aA = 0 & a

C = g().

SECTION (G)

G-1. If we take both A and B as a system then there is no external force on system. m

Aa

A + m

Ba

B = 0 [Newton�s II law for system ]

60 aA + 75 × 3 = 0

aA =

415�

m/s2

�ve sign means that acceleration is in direction opposite to the assumed direction.


G-3. 4F � (M + m)g = (M + m)a

a =mM

g)mM(�F4

=

mMF4

� g

G-5. tan 45º = B

A

aa

(wedge constrained relation)

N sin 45 = ma ...........(i)For Rod A mg � N cos 45 = ma ...........(ii)

From equation (1) & (2) a = 2g

SECTION (H)

H-1. Pseudo force depends on mass of object and acceleration of observer (frame) which is zero in thisproblem. Pseudo force is zero.

H-3.

F.B.D. in frame of liftIt is obevious that block can accelerate only in x direction. ma is Pseudo force. mg sin + ma sin = ma

x[Newton`s II law for block in x direction]

ax = (g + a) sin

PART - IISECTION (A)A-1. Experimental fact.

A-3. Force exerted by string is always along the string and of pull type.When there is a contact between a point and a surface the normal reaction is perpendicular to thesurface and of push type.

SECTION (B)

B-1.

F � N = 2 ma, [Newton`s II law for block A]N = ma

1[Newton`s II law for block B]

N = 3F

N = 2 ma2

[Newtons II law for block A]F � N = m

2a [Newtons II law for block B]

N = 2F/3 so the ratio is 1 : 2


B-3.

F � N = Ma [Newtons law for block of mass M]N � N� = ma [Newtons law for block of mass m]N� = M�a [Newtons law for block of mass M�]

N� = M� 'MmM

F

N = (m + M�) 'MmM

F

N > N�

SECTION (C)C-1. Point A is mass less so net force on it most

be zero otherwise it will have acceleration. F � Tsin = 0[Equilibrium of A in horizontal direction]

T = sin

F

C-3. 10 � T2 = 1 a

[ Newton�s II law for A ]

T2 + 30 � T

1 = 3 a

[ Newton�s II law for B ]

T1 � 30 = 3a

[ Newton�s II law for C ]

a = 7g

T2 =

7g6

C-5. Mg � T = Ma

[ Newton�s II law for M]

T � mg = ma

[ Newton�s II law for m]

T = MmMgm2

If m << M than m + M M

T = M

Mgm2

T = 2 mgTotal downward force on pulley is 2T = 4 mg.


SECTION (D)D-1. The length of string AB is constant.

speed A and B along the string are same u sin = V

u sin = V u = sin

V

D-3. By symmetry we can conclude that block will move

only in vertical direction.

Length of string AB remains constant

Velocity of point A and B along the string is same.

V cos = u V = cos

u

D-5. Let AB = , B = (x , y)

Bv

= vx i� + v

y j�

Bv

= i�3 + j�v y (i)

x2 + y2 = 2

2x vx + 2y v

y = 0 3 +

xy

vy = 0

3 + (tan600) vy = 0

vy = � 1

Hence from (i)

Bv

= i�3 � j�

Hencev

B = 2 m/s

D-7. V = (velcoity of B w.r.t ground)

24�V

= 2 V = 8 m/s (velcoity of B w.r.t ground)

V' = 6 m/s (velcoity of B w.r.t lift )

D-8. u cos 45° = v cos 60°

or v = u2

SECTION (E)

E-1. amF

dt

vda

E-4. In free fall gravitation force acts.


E-5. M2 g sin � T = M

2a [Newton�s II law for M

2]

T � M1g sin = M

1a [Newton�s II law for M

1]

By adding both equations

a =

21

12

MMsinM�sinM

g

E-6. Case 1

T1 � mg = ma

1[Newton�s II law for m]

2 mg � T1 = 2 ma

1[Newton�s II law for 2m]

a1 =

3g

Case 2

F � mg = ma2

[Newton�s II law for m]

2 mg � mg = ma2

a2 = g a

2 > a

1

E-7.

F = m1 4 [Newton�s II law for m

1]

F = m26 [Newton�s II law for m

2]

F = (m1 + m

2)a [Newton�s II law for (m

1 + m

2)]

F = a6F

4F

1 = a

61

41

a = 2.4 m/s2.

E-10. k�10j�8�i�6F

amF

amF 222 1086 = m 1 m = 10 2 kg.

E-11. as222 xm

F210 22

x = F2m�

as222

O2 = 32 + xmF2 1

0 = 9 +

F2

m�

m

F2 1

F1 = 9F


E-12.

Mg sin � T = Ma [Newton�s II law for block 1]

T = Ma [Newton�s II law for block 2]

By dividing both equations

2 T = Mg sin T = 2sinMg

SECTION (F)F-1. T � mg = 0 [ Equilibrium of block]

T � 10 = 0

T = 10Reading of spring balance is same as tension in spring balance.

F-2.

F � k x = m1 a

1[Newton�s II law for M

1]

kx = m2a

2[Newton�s II law for M

2]

By adding both equations.

F = m1a

1 + m

2a

2 a

2 =

2

11

mam�F

F-4. Weight of man in stationary lift is mg.mg � n = ma [Newton�s II law for man]

N =m (g � a)

Weight of man in moving lift is equal to N.

23

)a�g(mgm

a = 3g

SECTION (G)

G-2.

F = m1a

1[Newton�s II law for m

1]

180 = 20 a1

a1 = 9 m/s2

Net force on m2 is 0 therefore acceleration of m

2 is 0.


G-3.

30 � T2 = 3 a [Newton�s II law for 3 kg block]

T2 � T

1 = 6 a [Newton�s II law for 6 kg block]

T1 � 10 = 1 a [Newton�s II law for 1 kg block]

By adding three equations30 � 10 = 10 a a = 2 m/s2.

SECTION (H)H-1.

FBD of block is shown w.r.t. wedge and FBD of wedge is shown w.r.t. ground. FP is pseudo force.

mg sin 37 � ma cos 37 = mab

ab = g sin 37 � a cos 37 = 10 ×

53

� 5 × 54

= 2 m.s2 w.r.t. wedge

block is not stationary w.r.t. wedgeN � ma sin 37 � mg cos 37 = 0 [Newton�s II law for block]

N = 1 × 10 × 54

+ 1 × 5 × 53

N = 11 N.Net force acting on block w.r.t. ground.

F = 22 )N�37cosmg()37sinmg(

= 22

11�54

1053

10

= 22 36

F = 53 N.


H-3.

F.B.D. of wedge is w.r.t. ground andF.B.D. of block is w.r.t. wedge.Let a is the acceleration of wedge due to force F.F

P is pseudo force on block

mg sin 30º � ma cos 30º = 0 [Equilibrium of block in x direction w.r.t. wedge]a = g tan 30º

F = ( M + m)a [Newtons II law for the system of block and wedge in horizontal direction] F = (M + m) g tan 30º.

H-4.

acceleration of point A and B must be some along the line to the surface a sin = g cos

a = g cot

H-5.

F.B.D. of block is w.r.t. wedge andF.B.D. of wedge is w.r.t ground.F

P is pseudo force on block .

mg sin � ma cos = 0 [ Equilibrium of block w.r.t. wedge along x direction ] a = g tan


EXERCISE-2PART - I

1.

mg � NAB

= maA

[Newton�s II law for block A in vertical direction]

mg sin + NAB

sin = maB

[Newton�s II law for block B in x direction]

aA = a

B sin [Constrained relation for contact surface between block A and B]

Solving above three equations we get

NAB

=

2

2

sin1

cosmg

mg cos + NAB

cos � nBC

= 0 [Equilibrium of block B in y direction]

NBC

= mg cos +

2

2

sin1

coscosmg N

BC =

2sin1

cosmg2

NBC

sin � NWC

= 0 [Equilibrium of block in horizontal direction]

NWC

=

2sin1

cossinmg2

NBC

cos + mg � NFC

= 0 [Equilibrium of block C in vertical direction ]

NFC

= mgsin1

cosmg22

2

NFC

=

2

2

sin1

)cos2(mg

3. mg � Ncos 37 = maB

[Newton�s II law for block B in vertical direction]

N sin 37 = maA

[Newton�s II law for block A in horizontal direction]

aB cos 37 = a

A sin 37

[constrained relation for contact surfacebetween block A and B]By solving above three equations we get

aA =

25g12

aB =

25g9

N = 5mg4

NBW

= N sin 37[Equilibrium of block B in horizontal direction]

NBC

= 25mg12


6.

aB + a = 2a

A[constrained relation for pulley 1]

O + a = 2aB

[contrained relation for pulley 2]From above two equations3a

B = 2a

A

aA =

23

aB

..........I

F � 2T = 2maA

[Netwon's II law for block A] ..........II3T = 4 m a

B [Netwon's II law for block B] ...........III

From equation I, II and III

aB =

m17F3

.

8. mAg � 2T = m

Aa

A [Newton's II law for block A]

T � mBg = m

Ba

B [Newton's II law for block B]

aB + O = 2a

A [constrained relation for pulley P1]

mA = 4m

B [Given in question]

From above four equations

aA =

4g

= 2.5 m/s2

aB =

2g

= 5 m/s2

h = 21

aAt2 [Equation of motion for block A]

t = 52

sec.

H is the distance travelled by block

B in vertical direction till 52

second

H = 21

aBt2 [Equation of motion for block B]

21

52

5

2

H = 0.4 mH´ is the distance travelled by block B due to gained velocity.

v1 = a

Bt

= 5 × 0.4

v1 = 2 m/s

v2

2 = v1

2 + 2a H´

02 = 22 + 2 (�10) H´

H´ = 102

= 0.2 m

Total distance = H + H´

= 0.6 m = 60 cm


9. 4F1 � F

2 = ma [Newtons II law for block]

a = m

F�F4 21

For t = 0 to 2 sec.F

1 = 30N

F2 = 10N

a = 40

10�304 = 2.75 m/s2

For t = 2 to 4 secF

1 = 30N

F2 = 20N a =

4020�304

= 2.5 m/s2

For t = 4 to 6 sec.F

1 = 10N

F2 = 40N a =

4040�104

= 0 m/s2

For t = 6 to 12 secF

1 = 0 , F

2 = 0 a = 0 m/s2

V12

� V0 = a

0�2(2 � 0) + a

2�4(4 � 2) + a

4�6(6 � 4) + a

6�12(12�6)

V12

� 1.5 = 2.75 × 2 + 2.5 × 2 + 0 × 2 + 0 × 6

V12

= 12 m/s

11.

All the forces shown are in ground frame. aw is the acceleration of wedge w.r.t ground and a is the accelera-

tion of blocks w.r.t wedge.m

Ag sin45º � T = m

A (a � a

w cos 45º) [Newton's II law for block A along the wedge in ground frame]

mAgcos � N = m

A a

wsin45º [Newton's Ii law for block A in direction to the wedge in ground

frame.]T � m

Bg sin 45 = m

B (a � a

wcos 45) [Newton's II law for block B along the wedge in ground frame.]

NB � m

Bg cos 45º = m

B (a

wsin45) [Newton's II law for block B in direction to the wedge in ground frame]

NAsin45 + T cos 45 � N

B sin 45 � T cos45 = m

wa

w

[Newton's II law for wedge in horizontal direction in ground frame].After solving above five equations we will get

aw =

52

m/s2

acceleration of B w.r.t ground in 1352

m/s2.


13. 2aA = a + a

B

2aA = 3 + a

B

2T � 100 = 10aA

50 � T = 5aB

aB + a

A = 0

2aA � 3 + a

A = 0

aA = 1m/s2

aB = � 1m/s2 .

16. Fnet

= mg � 2F cos

anet

= g �mk2

L�xL 2222 xL

x

17. = 2

Fs = K

< 2

mg2

Fs < mg

T + Fs

= mg

T = mg � 2

K

If it is So

Fs > mg

i.e. < 2

string unstretched & T = 0.

19. N sin = mbN sin = m(a cos � b)

2mg � N cos = ma sin

a =

2sin1

sing4

h = 21

a sin t2 t =

2

2

sing2

)sin1(h .

20. acceleration of bead along rod is

mcosma

= a cos

21

a cos t2 =

t = cosa

2

N = 22 )mg()sinma( .


22. By newtons law on system of (A, B, C) along the string.(a) (M + m � M) g = (2M + m) a

a = mM2

mg

(b) free body diagram �C� block FBD

mg � N = ma

N = m

mM2

gmg

N = mM2

M2

gm

(c) T � Mg = M mM2

mg

for A block

T = Mg + mM2

Mmg

for pulleyP = 2T + Mg

= 2Mg + mM2

Mmg2

+ Mg = mM2m2m3M6

Mg

P =

mM2m5M6

Mg

23.

mg � T = ma 2T � 1.8 mg = 1.8 m 2a

0.2 mg = 2.9 ma a = 29

g2

arel

= a + 2a

= 2a3

= 58

g6S =

21

arel

t2

1 = 21

58

g6 t2 t = g3

58t = 1.4 sec.


PART - II1. Slope of v

rel � t curve is Constant.

arel

= Const. = a1 � a

2 0

Inference that at least one reference frame is accelerating both can�t be non - accelerating simultaneously.

3. T1cos45º = T

2cos45º T

1 = T

2

(T1 + T

2) sin45º = mg

2 T1 = mg

T1 =

2

mg. T sin = mg +

2

T1

T sin = mg + 2

mg.........(i)

T cos = 2

T1 =

2mg

.........(ii)

dividing (i) and (ii)

tan = 2/m

2/mM = 1 +

mM2

Ans.

6. w � f = ma w � ma = g

w

awm

�1 = f w

amgm

�1 = f w

ga

�1 = f

8.

a1 =

mmg�mg2

a2 =

mm2

gm�m2

a3 = 0

a1 = g a

2 =

3g

So, a1 > a

2 > a

3

10. By setting string length constant

L = 31 + 2

2

3v0 = 2v

A

vA =

23

v0

vAB

= vA� v

B

= 2

v0 towards right.


12.2T

= 32

32

mm

mm2

g

2gm1 =

32

32

mm

mm2

g

m1

= 32

32

mm

mm4

2m1

+ 3m

1 =

1m4

13. T sin = m (g sin + a0)

T cos = mg cos

tan =

cosg

asing 0

= tan�1

cosg

asing 0

16. T1 + T

2 = mg

If upper spring is cut

mg � T2 = m × 6 .....(i)

If lower spring is cut :

mg � T1 = ma ......(ii)

adding (i) and (ii)2mg � {T

1 + T

2} = m (a + 6)

2mg � mg = m (a + 6)

mg = m (a + 6)g = a + 6 a = 4m/s2.

18. A + B + C + D + E = 300 i ..........(1)B + C + D + E = � 100 i ........... (2)A + C + D + E = 2400 j ........... (3)

equation (1) - equation (3) giveB = 300 i � 2400 j .............(4)

equation (1) - equation (2) giveA = 400 i ............. (5)

Adding equation (4) and (5) A + B = 700 i � 2400 j

a(A + B)

= 100

BA

= 7 i � 24 j = 25 m/s2


20. For first case tension in spring will beT

s = 2mg just after 'A' is released.

2mg � mg = ma a = gIn second case T

s = mg

2mg � mg = 2mb

b = g/2a/b = 2

22. (Force diagram in the frame of the car)Applying Newton�s law perpendicular to string

mg sin = ma cos

tan = ga

Applying Newton�s law along string

T � m 22 ag = ma T = m 22 ag + ma Ans.

24. F.B.D. of mass m is w.r.t. trolley T sin ( � ) + mg sin � F

P = 0

[Equilibrium of mass in x direction w.r.t. trolley] T sin ( � ) + mg sin � mg sin = 0 T sin (� ) = 0since T cant be zero , sin (� ) must be zero =

25. Maximum acceleration of block is 10 m/s2 .

S = 21

at2

=21

× 10 × 0.22 = 0.2 m = 20 cm.

26.* T = m1g

when thred is burnt, tension in spring remains same = m1g.

m1g � m

2g = m

2a

2

21m

)m�m( g = a = upwards

for m1

a = o

30.* By string constraint

aA = 2a

B ................................(1)

equation for block A.

10 × 10 × 2

1 � T = 10 a

A ......(2)

equation for block B.

2T � 2

400 = 40 a

B .........................(3)

Solving equation (1) , (2) & (3) we get

aA =

2

5 m/s2 a

B =

22

5m/s2 T =

2

150 N


31.* Apply NLM on the system200 = 20 a + 12 × 10

2080

= a = 4 m/s2

spring Force = 10 × 12 = 120 N

32.* There is no horizontal force on block A, therefore it does not move in x-directing, whereas there is netdownward force (mg � N) is acting on it, making its acceleration along negative y-direction.

Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equaldue to constrain, thus w.r.t. B, A moves in positive x-direction.

Due to the component of normal exerted by C on B, it moves in negative x-direction.

EXERCISE-31. (A) Let the horizontal component of velocity be u

x. Then between the two instants (time interval T) the

projectile is at same height, the net displacement (uxT) is horizontal

average velocity = TTux = u

x (A) p, r

(B) Let j�andi� be unit vectors in direction of east and north respectively..

j�20VDC , i�20VBC and j�20VBA

BACBDCAD VVVV = j�20i�20j�20 = i�20

i�20VAD Hence BCAD VV (B) p, r

(C) Net force exerted by earth on block of mass 8 kg is shown in FBD and normal reaction exerted by 8 kgblock on earth is 120 N downwards.

Hence both forces in the statement are different in magnitude and opposite in direction. (C) q, s(D) For magnitude of displacement to be less than distance, the particle should turn back. Since the magni-tude of final velocity (v) is less than magnitude of initial velocity (u), the nature of motion is as shown.

Average velocity is in direction of initial velocity and magnitude of average velocity = 2

vu is

less than u because v < u. (C) q, r

2. Let a be acceleration of two block system towards right

a = 21

12

mmFF

The F.B.D. of m2 is

F2 � T = m

2 a

Solving T =

1

1

2

2

21

21

m

F

m

F

mm

mm

(B) Replace F1 by � F

1 is result of A

T =

1

1

2

2

21

21

m

F

m

F

mm

mm


(C) Let a be acceleration of two block system towards left

a = 21

12

mmFF

The FBD of m2 is

F2 � N

2 = m

2a

Solving N =

2

2

1

1

21

21

m

F

m

F

mm

mm

(D) Replace F1 by �F

1 in result of C

N =

1

1

2

2

21

21

m

F

m

F

mm

mm

3. FBD of Block in ground frame :applying N.L. 150 + 450 � 10 M = 5M

15 M = 600 M = 15600

Mg = 10 M

5 m/s2

150 N

450 N

M = 40 Kg Ans.Normal on block is the reading of weighing machine i.e. 150 N.

4. If lift is stopped & equilibrium is reached then

Mg = 400 M

T = 450 N

N450 + N = 400

N = � 50

So block will lose the contact with weighing machine thus reading ofweighing machine will be zero.

40 g

T

T = 40 g So reading of spring balance will be 40 Kg.

5.

Mg = 400 N

T = 450 N

N = 400 N

40 Kga

a = 40

400950

a = 40

450 =

445

m/s2 Ans.

6. ap =

10t10 = t

dtdv

= t

t

0

v

0

dttdv v = 2t2

Putting v = 2 we have t = 2 sec.

Now 2t

dtdx 2

xp =

2

0

3

6t

=

34

xB = 2 × 2 = 4 m

Hence relative displacement = 4 � 34

= 38

m


7. From above

2t = 6t3

t2 = 12 t = 2 3 sec.

8. a = t = 4 after 4 seconds VB =2 m/s

Vp =

242

= 8 m/s Vrel

= 8 � 2 = 6 m/s.

9. Inertia is the propety to resist change in state of motion or rest.

10. The FBD of block A isThe force exerted by B on A is N (normal reaction).The forces acting on A are N (horizontal) and mg (weight downwards).Hence statemt I is false.

11. If the lift is retarding while it moves upward, the man shall feel lesser weight as compared to when lift was atrest. Hence statement1 is false and statement 2 is true.

12. Newton's third law of motion is valid in all reference frames. Hence statement-1 is incorrect.

13. (i) (True)

(ii) (True)

14. (i) Earth (ii) 4N (iii) No (iv) 4N , Earth, book , upward(v) 4N , hand , book , downward (vi) nd (vii) rd

EXERCISE-4PART - I

1. 2mg cos = 2 mg

cos = 2

1= cos 45° = 45°

2. After string is cut, FBD of m

a = mmg

= g

FBD of 2m (when string is cut tension in the spring takesfinite time to become zero. How ever tension in the stringimmediately become zero.) 2m

2mg

3mg

a = m2

mg2mg3 =

2g

3. F = 2T sin

a = m

cosT

a =

sinm2cosF

= m2F

22 xa

x

4. ma cos = mg cos (90 � )

tanga

ga

= dxdy

dxd

(kx2) = ga

x = gk2a

= D


PART�II

1.

asystem

= mM

F

FBD of m

T = masystem

= mM

mF

2. V1 = 12

2

dt

dx

1

V2 = 1

dt

dx

2

Impulse = |P| = |m(V2 � V1)| = |0.4 (�1 �1)| = 0.8 Ns

3. Vertical component of acceleration of Aa1 = (g sin ). sin

= g sin 60º . sin 60º = g . 43

That for B

a2 = g sin 30º . sin 30º = g 41

(aAB)= 4g3

� 4g

= 2g

= 4.9 m/s2

4. A =

52

, B =

53

K = KA

A = K

B

B

K =

52

K A

KA =

2K5

KB =

3K5

.

5. F = ma = F0 e�bt

bt0 em

F

dtdv

t

0

bt0v

0

dtem

Fdv v =

t

0

bt0

be

m

F

v = bt0 e1mb

F

RESONANCE SOLN_FRICTION - 26

TOPIC : FRICTION

EXERCISE-1PART - I

SECTION (A)

A-1.

A-4. Friction is kinetic because their is relative motion. Direction of friction is such that it opposes the relativevelocity.

A-5. a = � µmg/m = � µg = � 1 m/s2

Vf

2 � Vi2 = 2as (V

f = 0 V

i = 5 m/s)

s = 12

25

= 12.5m.

SECTION (B)

B-4.

R = mg + 60 = 160 N

f = 80 N ( No sliding )

angle of friction = tan�1 Rf

= tan�1 16080

= tan�1 21

Ans.

B-5. ablock = mmg

= g = 0.15 × 10 = 1.5

aT = 2ST � Sb = 5

21

aT t2 � 21

aB t2 = 5

21

t2 [2 � 1.5] = 5

t2 = 20

ST = 21

aTt2

= 21

× 2 × 20 = 20 m.


B-6. N = mg � F sin

F cos = N = [mg � F sin ]

F =

sincosmg

F is minimum when cos + sin is max

d

d (cos + sin ) = 0

� sin + cos = 0 = tan or ;k = tan�1

also vr% cos + sin

= 21 for = tan-1 ds fy,

thus Fmin = 21

mg

SECTION (C)

C-1.

30 = smg 30 = s × 5 × 10

s = 0.6.Again,

kNkmg

S = 21

at2

a = 2t

S2 =

25102

= 0.8.

30 � kmg = m × 0.8 k = mg8.0m30

= 0.52.

C-2. (i) aA =mF

= 5

15 = 3

aB = 100

= 0

fAB = 0, fBG = 0.

(ii)

fBG 75Since fAB can�t be greater than fBG therefore acceleration of B will be zero.

and aA = 5

2530 = 1m/sec2

fAB = 25 N, fBG = 25 N.


(iii)

fAB 25 aA 5

25or aA 5

Let there is no sliding between A and B then common acceleration of A and B.

=15

75200 = 8.33

Since aA 5 Hence, there will be sliding between A and B in that case.

aA = 5 m/sec2, aB = 10

100200 = 10 m/sec2

fAB = 25 N, fBG = 75 N.

(iv)

aA 5Let A and B move together then common acceleration.

=15

7590 = 1m/sec2

As common acceleration is less than aA hence A and B will move together aA = 1m/sec2, aB = 1m/sec2

fAB = mA × 1 = 5N, fBG = 75 N.

PART - II

SECTION (A)

A-1.

Let acceleration in Ist case is a1 and that in second case is a2

Now ,21

a1t2 =

21

a2(2t)2 a2 = 4a1 ............(i)

Clearly a1 = msinmg

= g sin ............(ii)

and rFkk a2 = m

cosmgsinmg = g sin � g cos ............(iii)

From (i), (ii) and (iii),

we get = 0.75.

A-2. The normal reaction on the block isN = mg � F sin

Net force on block isFcos � µN = Fcos � µ mg + µFsin

or acceleration of the block is

a = m

µmg)sinµ(cosF =

mF

(cos + µsin) � µg


A-3.

N = 50 � 40 sin30 = 30

a = 5

302.0º30cos40 = 5.73 m/sec2

A-4.

Apply Newton�s law for system of m1 and m2

a = 21

2121

mm]º37cosgmº37cosgm[º37sing)mm(

= g[sin37º � cos37º]

Now apply Newton�s law for M1

m1g sin37º + T � m1gcos37º = m1a = m1g[sin37º � cos37º]

T = 0 and a = 4m/sec2

SECTION (B)

B-3.

Solving from the frame of cart , we getN = ma, mg = N

mg = ma a =

g

B-4. Solving from the frame of truck F= 5×1

= 5N

pseudo

Restf mg = 6 f = 5N.

B-7. Apply Newton�s law for system along the string

mB g = (mA + mC) × g

mC =

Bm � mA =

2.05

� 10 = 15 kg


SECTION (C) :

C-2.

a = m

ff ks = m

mg)( ks = (S � k) g

= (0.5 � 0.4)10 = 1 m/sec2

C-3. When F is less than µsmg then tension in the string is zero.

When µsmg F < µ

s2mg then friction on block B is static.

If F is further increase friction on block B is kinetic.

C-4. Solving from the frame of elevator

geff. = g + 4g

= 12.5

f = 51

× 2 × 12.5 = 5N

a2 = 25

= 2.5 m/s2

a1 = 8

530 =

825

m/sec2

EXERCISE-2PART - I

SECTION (A)

2. a1 = m2

mgmg k = 2g

(1 � k)

a2 = m4mgk =

4gk

s1 = 21

a1 t2

s2 = 21

a2 t2

s1 � s2 = 87

21

2g

(1 � k)t2 �

21

4g

k t2 =

87

t2 = kk g)1(g2

7

= )32(g7

k

s2 = 21

a2 t2 =

21

× 4gk × )32(g

7

k

= )3�2(87

k

k


4. Solving from the frame of rod.

ab = m

sinmacosma = a [cos � sin ]

Now, = 21

abt2

ba2

= ]sin[cosa2

5. f1 = 3 × 0.25 × 10 = 7.5

F = 17.5 + 25 + 37.5 = 80 NIf F = 200 then aB = aC T � f1 � f2 = mBa .........(1)

F � T � f2 � f3 = mCa .........(2)from equation (1) and (2)F � f1 � 2f2 � f3 = (mB + mC)a

CB

321

mm

ff2fFa

=

125.37355.7200

= 10 m/sec2

6. The F.B.D. of A and B are

(force of friction)

For sliding to start between A and B, the frictional f = µ N = 41

× 2 × 10 = 5 N = fmax

Applying Newton�s second law to system of A + B

F = (mA + m

B) a = 6a .....................(1)

Applying Newton�s second law to A

f = mA a a

max =

A

max

m

f =

25

= 2.5 m/s2 .......................(2)

from (1) and (2) Fmin

= (mA + m

B) 2.5 m/s2 = 6 × 2.5 = 15 N

7. (i) The F.B.D. of A and B are

For A to be in equilibrium ; F = N sin .....................(1)For B to just lift off ; N cos = mg + µ

s N .....................(2)

For horizontal equilibrium of B ; N = N sin ................(3)From (2) and (3)

N (cos � µs sin ) = mg or N

5

3

3

2

5

4= mg or N =

25

mg ...............(4)

From equation (1) F = N × 53

F = 23

mg


(ii) The acceleration of the block A be a and B be bF � N sin = 2ma ...............(1)N cos � mg � µ

kN = mb ...............(2)

N = N sin ................(3)From constraint =

a sin = b cos ................(4)

Solving (1), (2) , (3) and (4) we get b = 22g3

10. Considering the forces on the chain for the given situation we have

F � k ( � x)g = a

F �

g)x(k =

dxdv

.v..

0

dxF

�

0

k dxg)x(

= v

0

vdv

0

xF

�

v

0

2

0

2

k 2

v

2

xxg

F �

2g k

= 2v2

gF2

k

= v..

PART - II3. On smooth surface a1 = g sin

v2 = u2 + 2a1s1= 0 + 2 g sin .mOn rough surface

a2 = g sin � g cos

v´2 = v2 + 2a2s2

O = 2mg sin + 2g (sin � µ cos )n

=

nnm

tan

6.

If acceleration of the car is a0, acceleration of the block 2a0 = 2 × 2 = 4 m/s2 ()

F = N = 0.3 × 50 × 10 = 150

T � F = ma

T � 150 = 50 × 4

T = 350 N.


8. FBD of A

N

N

a

8mg

8m

T

a

mg

C

If the acceleration of �C� is a

For block �A� N = 8 ma .... (1)8 mg � N = 0 .... (2)

and acceleration a can be written by the equation of system (A + B + C)m1 g = (10 m + m1) a .... (3)

8 mg =

1

1

mm10

gmm8

10 m + m1 = m1

10 m = ( � 1) m1 m1 = 1

m10

Ans.

10.

(i) Let the blocks does not movethen T1 = 20 � 4T2 = T1 � 8 = 20 � 4 � 8 = 8

Since T2 < max possible friction force for 6 kg blockhence it will be at rest and this assumption is right. Therefore tension in the string connecting 4kg and 6 kgblock = 8N

(ii) friction of 4 kg block = N = 0.2 × 4 × 10 = 8N

(iii) friction force on 6 kg block = 8N

11. So block �Q� is moving due to force while block �P� due to friction.

Friction direction on both +Q blocks as shown.

P4

5

f =8max

Ff =8max

f =9max

Q

First block �Q� will move and P will move with �Q� so by FBD taking �P� and �Q� as system

F � 9 = 0 F = 9 NWhen applied force is 4 N then FBD

Q

P

0

44

0

4 kg block is moving due to friction and maximum friction force is 8 N.

So acceleration = 48

= 2 m/s2 = amax.

Slipping will start at when Q has +ve acceleration equal to maximum acceleration of P i.e. 2 m/s2.F � 17 = 5 × 2 F = 27 N.


13.

Applying Newton�s law for the system of m and 3m along the length of the string

we get3mg sin45 � 3mg cos 45 � mg sin45 = (3m + m)a

= 5

2as a =

25

g

now making the F.B.D. of m we get �T � mg sin 45 = m a

T = 25

mg +

2

mg

T = 25

mg 6

Now from F.B.D. of pulley we get

Force exerted by string on pulley

= T 2 = 25

mg6 × 2 =

5mg6

(downward)

14.

F.B.D. for A block

F.B.D. for B block


for block Amg sin � f

1 = ma .........(1)

for motion w.r.t. block Bmgsin � µmg cos = ma .........(2)for limiting case

a = 0and a = b = 0 mg sin = µmg cos

µ = tan

= tan�1 µfor block Bmgsin + f

1 � f

2 = mb

for motion w.r.t wedgef2 = 2µmg cos

mgsin + f1 2µmg cos = mb ..........(3)

for no relative motion between A and B block from equation (1) & (3) : a = b2mg sin � 2 µmg cos = 2ma

for limiting case a = 0 = tan�1 (µ)

for motion tan�1 (µ)

when block B is moving w.r.t wedgemgsin + f

1 � 2µ mgcos = mb

But f1 = µmg cos mg sin � µmg cos = mb

for block Amg sin � µmgcos = ma a = b.

16. * The free body diagram of the block isN is the normal reaction exerted by inclined plane on the block.

Applying Newton�s second law to the block along and normal to the incline.

mg sin 45° = T cos 45° + N ............... (1)N = mg cos 45° + T sin 45° ............... (2)

Solving we get = 1/2

so any value of greater than 0.5 is answer

18.* Applying NLM on the part that moves through slit.T

2 � f � T

1 =0

For 4 kg mass 40 � T2 = 4a

For 2 kg mass T1 � 20 = 2a

T1

T2

f

T2 T1

m1 m1

40 20

On solving 10 = 6a

a = 35

m/s2

Force exerted on 2kg mass by string = T1 =

370

N.

Tension in the string will not be same throughout, due to the friction force exerted by the slit.


19.* The breaking force is insufficient, so the block will not slide. So friction force = 100 Nand acceleration will be 20 m/sec2 only

Net contact force on the block = 22 )100()200( = 100 5 N

All mechanical interactions are electromagnetic at microscopic level.

20.*There are two possibilities(i) 100 kg block slides down the incline(ii) 100 kg block slides up the incline

case-(i)

we get, 100 g sin 37 � ×(100 g) cos 37 � mg = 0

m = 5

3100 � 0.3 × 100 ×

54

= 36 kg

case (ii) mg = 100 g sin 37 + g cos 37 × 100

m = 5

3100 +

54100

× 0.3

= 84 kg

To remain in equilibrium, m [36, 84] kgtherefore, m can be 37 and 83 kg.

21.*_

N 20 N

mg = 50 N

8 + 62 2

N = 50 � 20 = 30 N

Limiting friction force = µN = 12 N and applied force in horizontal direction is less than the limiting

friction force, therefore the block will not slide.For equilibrium in horizontal direction, friction force must be equal to 10 N.

53°

f

6i + 8j

From the top view, it is clear that = 37° i.e. 127° from the x-axis that is the direction of the friction

force. It is opposite to the applied force.

Contact force = 22 fN = 1010 N


22.*

F1 = mgsin + mg cos .F2 = mgsin � mg cos .But q mg = w = tan

F1 = w (sin +

cossin

cos) F1 = w sin( + )sec

Now F1 = 2 F2mg sin+ mg cos = 2 (mg sin � mg cos)sin+ cos = 2 sin � 2 cos 3cos = sin tan = 3tan = 3tan.

23.*_ mgsin + mg cos = maa = g sin + g cos

= 10

54

53

= 14 m/sec2.

If fr = mg sin = mg ×

53

= 5mg3

< fr max

fr < f

r max

= 5mg3

< 5mg4

hence insect can

move with constant velocity. mg sin

f r

v = constant

EXERCISE-3PART - I

1. (i) FBD in (case (i)) {1 = 0, 2 = 0.1}

A

mg

1 kg

O 2N

B

mg

1 kg

2N

N = 10 N = 10

While friction�s work is to oppose the relative motion and here if friction comes then relative motion will start

and without friction there is no relative motion so both the block move together with same acceleration andfriction will not come.

mg

A

mg

B

aA = aB = 10 m/s2


(ii)

A

10

1 kg

1

10

0

B

10

1 kg

0

10

1

Friction between wall and block A oppose relative motion since wall is stationary so friction wants to stopblock A also and maximum friction will act between wall and block while there is no friction between block.

Note : Friction between wall and block will oppose relative motion between wall and block only it will not doanything for two block motion.

10

A B

1

10aA = 9 m/s2 ; aB = 10 m/s2

(iii)

10 10

A B

1 f

f10

Friction between wall and block will be applied maximum equal to 1N but maximum friction available betweenblock A and B is 10 N but if this will be there then relative motion will increase while friction is to opposerelative motion. So friction will come less than 10 so friction will be f that will be static.

10 10

A B

1 f

f

by system (20 � 1) = 2 × a a = 2

19 = 9.5 m/s2

(iv)

10 10

A B

10 1

1

10

aA = 1

1011 = 1 m/s2

aB = 1

110 = 9 m/s2


2. The acceleration of two block system for all cases is a = 2 m/s2

In option (p) the net force on 2 kg block is frictional force Frictional force on 2 kg block is

f = 2 × 2 = 4N towards right

In option (q) the net force on 4 kg block is frictional force Frictional force on 4 kg block is

f = 4 × 2 = 8N towards right

In option (r) the net force on 2 kg block is 2 × 2 = 4N

Friction force f on 2 kg block is towards left. 6 � f = 2 × 2 or f = 2N

In option (s) the net force on 2 kg block is ma= 2 × 2 = 4N towards right.

Friction force on 2 kg block is 12N towards right.

(A) 4.2 m/s2 (B*) 3.2 m/s2 (C) 16/3 m/s2 (D) 2.0 m/s2

3. & 4.First, let us check upto what value of F, both blocks move together. Till friction becomes limiting, they willbe moving together. Using the FBDs

F

a2

15 kgFF

F

f

a1

F

f

10 kg block will not slip over the 15 kg block till acceleration of 15 kg block becomes maximum as it iscreated only by friction force exerted by 10 kg block on it

a1 > a

2(max)

10fF

= 15f

for limiting condition as f maximum is 60 N.

F = 100 N.Therefore for F = 80 N, both will move together.Their combined acceleration, by applying NLM using both as system F = 25a

a = 2580

= 3.2 m/s2

5. If F = 120 N, then there will be slipping, so using FBDs of both (friction will be 60 N)For 10 kg block120 � 60 = 10 a a = 6 m/s2

For 15 kg block60 = 15a a = 4 m/s2

6. & 7.In case 80 N force is applied vertically, then

FF

F

F

f

80

f

For 10 kg block 80 � 60 = 10a

a = 2 m/s2

For 15 kg block in horizontal direction.F � f = 15a

a = 4/3 m/s2, towards left.


8. F sin + f = mgand Fcos

= N

for minimum ; f = N = Fcos

Fmin.

= cossin

mg

9. As f = 0 F sin = mg

F = sin

mg

10. If F < Fmin.

; block slides down due to mg

11. Friction always opposes relative motion.

12. Due to pseudo force, the person observes the block to move back. Also the accelerating person doesnot observe any relative motion between body and the rough surface.

13._ The minimum force required to pull the block of mass m lying on rough horizontal surface is

F = 1

mg2

= 60

N, inclined at an angle tan�1 with horizontal (where is the coefficient of friction). Hence

statement 1 is true and statement 2 is false.

14._ There is no tendency of relative motion between the blocks. Hence Statement-1 is True, Statement-2 is True;Statement-2 is a correct explanation for Statement-1

15._ R = 22 Nf = mg if body does not move.But if it moves then f < mg sin

Bf

Fext

fA

R = 22 )cosmg(f < mg

16. (i) Since the initial velocity of block is along positive x-axis. So the direction of frictional force will

be in � i� at that moment � i� ... Ans.

(iii) The block begins to slide ifF cos 37° = µ (mg � F sin 37°)

5t [cos 37° + µ sin 37°] = µ mg

5t

53

54

= 70 or ;k t = 10 second

EXERCISE-4PART - I

1.

N = mg + F sin 60 = 3 × 10 + 2

3F ...... (i)

F cos 60 = N ................. (ii)

2

F=

32

1 × (10 3 +

23F

)

2F

= 5 + 4F

4F

= 5 F = 20 N


2. aA = g [sin 45 �

A cos 45] =

2

8 , a

B = g [sin 45 �

B cos 45] =

2

7

aAB

= aA � a

B = g (

B �

A) cos 45 =

2

1 , s

AB = 2

Now sAB

= 2

1 a

AB t2 2 =

2

1 ×

2

1 t2 t = 2 sec.

Again sA =

2

1 a

A t2 =

2

1 (

2

8 ) 4 s

A = 8 2 m

3.

Solving from the frame of disc

Let accleration of the block relative to the disc is a then25 m cos � N = m a ..........(i)Now, there will be two normal as there are two contacts (i) Horizontal and (ii) vertical

NH = 25 m×sin = 25 × m ×

53

= 15 m

NV = mg = 10 m

f = NH + N

V=

52

(15 m) + 52

(10 m) = 10 m

from (i) we get a = m

10)� cos (25 m = 10 m/sec2

4. Statement-1 is also practical experience based; so it is true.Statement-2 is also true but is not the correct explanation of statement-1. Correct explanation is ''thereis increase in normal reaction when the object is pushed and there is decrease in normal reaction whenobject is pulled".

5.

P1 = mgsin � mgcosP2 = mgsin + mgcosInitially block has tendency to slide down and as tan > , maximum friction mgcos will act in positivedirection. When magnitude P is increased from P1 to P2, friction reverse its direction from positive to negativeand becomes maximum i.e.mgcos in opposite direction.


6.

F1 =

2

mg

2

mg

F2 =

2

mg

2

mg

F1 = 3F

2

1 + = 3 � 3

4 = 2

= 21

N = 10N = 5 Ans.

7.

f = 0, If sin = cos = 45°

f towards Q, sin > cos > 45°

f towards P, sin < cos < 45°


PART - II

1. Force, F = R = Mgweight of block = R = 0.2 ×10 = 2N

2. F = ma mg = ma = ga

Now , v = u + at or 0 = 6 + 10a

or10

6.0= a = � 0.6 so = 06.0

106.0

ga

3. Let the mass of block be m.Frictional force in rest positionF = mg sin 30º

10 = m × 10 ×21

mg30º

mg cos30ºmg sin 30º

R F

m = 10

102= 2 kg

4. When fiction is absent

a1 = g sin s

1 = 2

11ta21

........ (i)

When friction is present

a2 = g sin � g cos s

2 =

21

a2t2

2 ........ (ii)

From Eq. (i) and (ii)

222

211 ta

21

ta21

or 211ta = a

2 (nt

1)2 ( t

2 = nt

1) or a

1 = n2a

2

or 21

2

n1

singcosgsing

aa

or 2n

1º45sing

º45cosgº45sing

or 1 � k = 2n

1or

k = 1 � 2n

1

5. According to work-energy theorem,W - K = 0

(Initial and final speed are zero)Work done by friction + work done by gravity = 0

� (mg cos)2

+ mgl sin = 0

cos = 2 sin = 2 tan

6. s = 105.02100100

g2v

k

2

=

25100100

= 1000 m

7. F1 = mg sin + mg cos

F2 = mg sin � mg cos

2

1

FF

=

cossincossin

tantan

=

22

=

3 = 3.

RESONANCE SOLN_GRAVITATION - 44

TOPIC : GRAVITATION

EXERCISE-1PART - I

SECTION (A)

A 2. mass of each spherem = Volume ×

= 34r3

F = G 2)r2(

m.m

= 2

23

r4

r34

G

= 94

G22r4 N. Ans.

A 3. tan = 68

= 34

= 53°

F = 2r

GmM

= 2)1.0(

01.0260.0G

a = mcosF2

= 2G 2)1.0(

260.0

5

3

= 31.2 G m/s2

SECTION (B)

B 1. Ex = �

xv

= �

x

(20x + 40y) = � 20

Ey = �

xv

= � y

(20x + 40y) = � 40

E

= Ex i� + E

yj� = � 20 i� � 40 j� Ans.

It is independent of co ordinates

Force = F

= m E

= 0.25 {� 20 i� � 40 j� } = � 5 i� � 10 j�

magnitude of F

= 52 105 = 55 N


SECTION (C)

C 1. Potential energy at ground surface

potential energy = R

GMm�

potential energy at a height of R is

potential energy = R2

GMm�

When a body comes to groundLoss in potential energy = Gain in kinetic energy

R2

GMm� �

R

GMm� =

21

mv2 R2

GMm =

21

mv2

gR = v2

g

R

GM2

v = gR

C 2. Initial kinetic energy = 21

MSV2

Initial potential energy = �2/d

MGM SA �

2/d

MGM SB = �

d

GM2 S (M

A + M

B)

Total initial energy = 21

MSV2 �

d

GM2 S (M

A + M

B)

Finally, Potential energy = 0Kinetic energy = 0 Limiting caseApplying energy cnservation

21

MSV2 �

d

GM2 S (M

A + M

B) = 0 V = 2

d

)MM(G BA Ans.

SECTION (D)

D 2. T1 = 2

e

3

GMr

, T2 = 2

e

3

GM)r01.1(

1

2

TT

=

2/3

rr01.1

1

2

TT

= [1 + 0.01]3/2 = 1 + 23

× 0.01

1

2

TT

� 1 = 0.005 × 3 1

12

T)TT(

× 100 = 0.015 × 100 = 1.5%.

D 3. (a) F = 2)R2(

GMm = 2

2

R4

GM

(b)R

Mv2

= 2

2

R4

GM

v = R4

GM

T = vR2

=

R4GM

R2 = 4

GMR3


(c) Angular speed

= T2

=

GMR

4

2

3 = 3R4

GM

(d) Energy required to separate = � { total energy }

= � { Kinetic energy + Potential energy }

= �

R2

GM�Mv

21

Mv21 2

22 = �

R2GM

�Mv2

2

= �

R2GM

�R4

GMM

2

= �

R4GM

�2

= R4

GM2

Ans.

(e) Let its velocity = �v�

Kinetic energy = 21

mv2

Potential at centre of mass = � R

GM �

RGM

= � RGM2

Potential energy at centre of mass = � R

GMm2

For particle to reach infinityKinetic energy + Potential energy = 0

21

mv2 × R

GMm2 = 0

v = RGM4

Ans.

D 4. (a)B

A

UU

=

B

B

A

A

rGMm

rGMm

= B

A

mm

A

B

rr

rB = 19200 + 6400 = 25600 Km

rA = 6400 + 6400 = 12800 km, m

A = m

B

B

A

UU

= 1280025600

= 2

(b)B

A

KK

=

B

B

A

A

r2GMm

r2GMm

= B

A

mm

A

B

rr

= 2

(c) As T.E. = r2

GMm� ,

Clearly farther the satellite from the earth, the greater is its total energy. Thus B is having more energy.


SECTION (E) :

E 2. Period of pendulum = g

2

Let T1 be the time period at pole and T

2 is time period at equator.

2

1

TT

= 1

2

g

g

21

2e

2e

1

gR

�1g

gR

�1g

1T

T1 = 1 � g2

R 2e

. Since gR 2

e << 1

So , T1 = 1 �

21

gR 2

e = 1 � 21

2

2

)86400(

)2( ×

8.9

106400 3

= 0.998 second Ans.

PART - IISECTION (A)

A-2. Net torque = F2 .

2 � F

1 .

2

= (F2 � F

1)

2

F2 = mg

H2 = mg

R

H2�1 2

F1 = mg

H1 = mg

R

H2�1 1

= (F2 � F

1)

2

= R

)H�H(mg 21

Ans.

A-4. 2 Fg cos 30 =

RMV 2

2

2

2

L

GM

23

= 3L/

MV2

V = L

GM

SECTION (B)B 2. dE

net = 2dE sin

= 2 2r

Gdm sin

= 2G . 2r

rdsin

= rG2

sind


Enet

= netdE =

2/

0

sinrG2

d = rG2

=

m and r =

Enet

= 2

Gm2

Along + y axis Ans.

B 4. For point �A� :For any point outside, the shells acts as point situated at centre.

So, FA = 2

21

p

)MM(G m

For point �B� :There will be no force by shell B.

So, FB = 2

1

q

mGM

,, For point �C� :There will be no gravitational field.So, F

C = 0

B-6. Let the possible direction of gravitational field at point B be shown by 1, 2, 3 and 4(Figure 1). Rotate the figureupside down. It will be as shown in figure 2.

Figure 1 B1

2

3

4

= B Figure 3

Figure 2 B1

2

3

4

Now on placing upper half of figure 1 on the lower half of figure 2 we get complete sphere. Gravitationalfield at point B must be zero, which is only possible if the gravitational field is along direction 3. Hencegravitational field at all points on circular base of hemisphere is normal to plane of circular base. Circular base of hemisphere is an equipotential surface.Aliter : Consider a shaded circle which divides a uniformly thin spherical shell into two equal halves.Thepotential at points A,B and C lying on the shaded circle is same. The potential at all these points dueto upper hemisphere is half that due to complete sphere.Hence potential at points A,B and Cis alsosame due to upper hemispehre


SECTION (C)

C 2. (a)

120°

Fm

F

F

m

m

120°

120°

Due to geometry net force is zero.

(b) By geometry , x2 + 4a2

= a2 and F1 = F

2

F

F2

F1 x

a

x2 = 4a3 2

x = 2

a3

Fnet

= F = 2

2

x

Gm =

34

2

2

a

Gm

(c) Initial potential energy = �

a

Gma

Gma

Gm 222

= � 23

aGm2

Work done on system = Final potential energy � intial potential energy

= � 23

aGm2

�

aGm3

�2

= 23

aGm2

Ans.

(d) Initial kinetic energy = 0

Initial potential energy = � a

Gm2 �

aGm2

= � a

Gm2 2

Total initial energy = � a

Gm2 2

Now, kinetic energy = 21

mv2

Potential energy = � 2/a

Gm2 2 �

2/aGm2

= � a

Gm4 2

Total energy = 21

mv2 � a

Gm4 2

a

Gm2 2

= 21

mv2

aGm4

= v

v = a

Gm2 Ans.


SECTION : (D)

D-1. 2r

GMm =

rmv2

v = 2r

GM M

r

Vm

T = v

r2 =

GM

r2 23

=

3

23

r34

G

r2

T

1

D-5*. PE = �G m1 m

2/r, ME = � G m

1 m

2 / 2r

On decreasing the radius of orbit PE and ME decreases

D-6. According to kepler's law applying angular momentum conservation m1v

1r

1 = m

2v

2r

2 V

max is (a) ans.

SECTION (E)E 1. w

e = 50 × 10 = 500 N

wp = 50 × 5 = 250 N

Hence option A is correct

E 2*. In case of earth the gravitational field is zero at infinity as well as the the centre and the potential isminimum at the centre .

EXERCISE-2PART - I

2. (a) r < y < 2r y

rx

y

Field due to outershell = 0Distance from centre of solid spere = (y � r)

Gravitation field intensity

= � 3)radius(

GM × distance from centre

= � 3r

GM (y � r) in y - direction

= � 3r

GM (y � r) j� = 3r

GM (y � r) (� j� )

(b) Field due to outshell = 0Distance from centre of solid spere = (y � r)

y

rx

y

4rE = 0 � 2)r�y(

GM j� =

2)r�y(

GM (� j� )

(c) y > 8r

For any point outside, the shells acts as point situated at centre.Distance from centre of hollow shell = (y � 4r)

Field due to hollow shell = � )r4�y(

GM4

Distance from centre of solid spere = (y � r)

Field due to solid spere = � 2)r�y(

GM

Total field =

2)r�y(

GMr4�y

GM4(� j� )


3. (a) Force will be due to the mass of the sphere upto the radius r

In case (i) 0 < r < b ; Mass M = 0, therefore F(r) = 0

In case (ii) b < r < a ; Mass M = 34

(r3 � b3), therefore F(r) =

2

3

r

brGpm

34

(iii) a < r < ; Mass M = 34

(a3 � b3), therefore F(r) =

2

33

r

baGpm

34

(b) Uf � Ui = � 2

1

r

r

c dr.F

(i) 0 < r < b ; u(r) = - 2 Gm(a2 � b2)

(ii) b < r < a ; u(r) = r3

mG2 (3ra2 - 2b3 - r3)

(iii) a < r < ; u (r) = )ba(r3

mG4 33

5. (a) The gravitation field is uniform inside the cavity and is directed along ´OO . Hence the particle will

strike at A.

(b) The gravitational field at any point P inside cavity.

|E

| = 3

4G �

34

G = 3

4Gy ´OO =

32

GR

Total workdone = m |E

| . S

= m . 32 GR .

2R

Applying work - energy theoremWorkdone by all force = Change in kinetic energy

m . 32 GR .

2R =

21 mv2

v = 3

RG2 2

Ans.

6. (a) 21

mv2 = R

mGMs or V = R

G2 S

(b) 21

mve

2 � R2mG

= 0 or Ve =

RG

21

m (V + Ve)2 =

R

GmMs or V + Ve = R

GM2 s

V = 1�2 R

GMs


7. Applying angular momentum conservation :mv

0 = mvd

v0= vd .......... (i)

Intial energy = 21

mv0

2 + 0

Final energy = 21

mv2 � d

GMs

Applying energy conservation ,

21

mv0

2 = 21

mv2 � d

mGMs

v0

2 = v2 � d

GM2 s .......... (ii)

From equation (i) and (ii) :

v0

2 = 2

220

d

v

� d

GM2 s

d2 + 20

s

v

GM2d � 2 = 0

Solving this quadratic

d = � 20

s

v

GM +

2

2

20

s

v

GM

= 20

s

v

GM

1�

GM

v1

220

Ans.

PART - II1. Gravitational field at �m� due to hollowed - out lead sphere

= { Field due to solid spere } � { Field due to mass that was removed }

Field due to solid sphere = 2d

GM = E

1 = 2R4

GM

Field due to removed mass = 2x

'GM = E

2

M� = 3R

34

M

×

3

2R

34

=

8M

And x = d � 2R

So , E2 = 2

2R

�d8

GM

= 2

2R3

8

GM

= 2R18

GM

Enet

= E1 � E

2

= 2R

GM

18

1�

4

1 = 2R36

GM7

Fnet

= mEnet

= 2R36

GMm7Ans.


4.

2

2

1

1

rGm

rGm

= 43

21

1

r4

m

= 2

2

2

r4

m

m1 + m

2 = m 2R4

m

= 2

1

1

r4

m

or =

1

1

rGm

R

Gm

= 35

Ans.

5. Ve =

RGM2

V = KVe =

RGM2

K

Initial total energy = 21

mv2 � R

GMm2 =

21

m.K2 RGM2

� R

GMm2

Final total energy = 21

m02 � x

GMm2

Applying energy conservation

21

mx2. RGM2

� R

GMm2 = 0 �

xGMm2

x1

= R1

� Rx2

x = 2k�1

RAns.

9. Fg = 3R

GMmr

pressing force = Fg cos = 3R

cosGMmr

= 2R2

GMm = constant

a = msinFg

= 3R

sinGMr

a = 3R

GMy

10.* In elliptical orbit sun is at one of the foci hence the distance between the planet and sun changes as planetrevolves hence linear speed, kinetic energy and potential energy of planet donot remain constant

11.* S

= 5.1

2,

E =

242

west to east

= 2

24

1�

5.1

1

Twest to east

= easttowest

2

= 1.6 hours

Similarly

east to west

= 2

241

5.11

Teast to west

= 1724

hours


EXERCISE-3

1. P.E. = �r

GMm K.E. =

21

mV2

Total energy = r

GMm +

21

mV2

T.E. = 0 if 0mV21

rGMm 2

v = r

GM2

For v < r

GM2 T.E. is � ve for v >

rGM2

, T.E. is + ve

If V is r

GM i.e. equal to orbital velocity, path is circular..

If T.E. is negative, path is elliptical.If T.E. is zero, path is parabolic.If T.E. is positive, path is hyperbolic.

2. (A) At centre of thin spherical shell V 0, E = 0.(B) At centre of solid sphere V 0 , E = 0.(C) At centre of spherical cavity inside solid sphere V 0, E 0.(D) At centre of two point masses V 0, E=0.

5. T2 = 3

2

RGM4

R =

3/1

24

GM

T2 / 3 log R =

32

log T + 31

log

24

GM

y = mx + c

(3) Slope = m = 32

intercept c = 31

log

24

GM = 6 log

104

M10320 11

= 18

(4) M = 6 × 1029 Kg(5) T2 R3

3

B

A

R

R

=

2

B

A

R

R

=

2

A

B

3

R4R

=

2

A

B

A

B =

81

rel

= 80 �

0 = 7

0

rel = (

rel) t 2 = (T

0) t

t = 0T

2

6. Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then escape

velocity from earth e = )Rg2(

Velocity of satellite s =

2e = 2/)gR2( ......... (1)

Further ] s =

r

GM =

hRgR2

2s =

hRgR2

h = R = 6400 km


7. T2 = 3

2

xGm4

Hence time period of revolution T is

T = Gmx

23

(Put x = 2R)

T = gR8

2

8. Now total energy at height h = total energy at earth's surface (from principle of conservation of energy)

0 � G M hR

m

= 21

m2 � GM Rm

or21

m2 = R

mGM �

R2GMm

( h = R)

v = gR

9 to 11Let the angular speed of revolution of both stars be about thecommon centre , that is, centre of mass of system. The centripetal force on star of mass m is

m2

3d2

= 2d

)m2(Gm. Solving we get T= 3

2

dGm3

4

The ratio of angular momentum is simply the ratio of moment of inertiaabout center of mass of system.

2

3d

m2

3d2

m

L

L2

2

M

m

M

m

I

I

Similarly, The ratio of kinetic energy is simply the ratio of moment of inertia about center of mass ofsystem.

+2

3d

m2

3d2

m

K

K2

2

2M

2m

M

m

I2

1

I2

1

12. Till the particle reaches the centre of planet, force on both bodies are in direction of their respective velocities,hence kinetic energies of both keep on increasing . After the particle crosses the centre of planet, forces onboth are retarding in nature. Hence as the particle passes through the centre of the planet, sum of kineticenergies of both the bodies is maximum. Therefore statement-1 is True, Statement-2 is True; Statement-2 isa correct explanation for Statement-1.

13._ It is minimum �ve (iii) Energy density = r0

2

2B

and B increases by a large factor..

15. for closed paths (circular or elliptical) the total mechanical energy is negative.14. (i) g� = g � R cos 2

At equator = 0 g� = g � R

0 = g � R


= Rg

= 3106400

8.9

= 1.24 × 10 �3 rad/s

(ii)dtdA

= mL

2= constant because angular momentum of planet (L) about the centre of sun is constant.

Thus, this law comes from law of conservation of angular momentum.

(iii) T r 3 / 2

1

2

TT

=

2/3

1

2

r

ror T

2 =

2/3

1

2

r

r T

1 =

2/3

R 7R 5.3

(24) h = 8.48 h

EXERCISE-4PART - I

1. Time period of a satellite very close to earth�s surface is 84.6 minutes. Time period increases as the distance

of the satellite from the surface of earth increase. So, time period of spy satellite orbiting a few hundred km,above the earth�s surface should be slightly greater than 84.6 minutes. Therefore, the most appropriate

option is (C) or 2 hrs.

2. (A) Gravitational field is a conservative force field. In a conservative force field work done is path independent. W

I = W

II = W

III

3. speed of particle at A VA = escape velocity on the surface of moon =

RGM2

At highest point B, VB = 0From energy conservation.

2AmV

21

= VB � VA = m

m

U�

m

U AB

or

mU

�mU

2V AB

2A

, also 3A

R2

GM�

mU

[3R2 � r2]

22

3 100R

�R5.0�R5.1R

GM��

hRGM�

RGM

orR1

10099

21

�R23

�hR

1�R1

2

or h = 99.5 R 99R Ans

4. AB

mA mBcom

rA rBC

2BA

BA

)rr(

mGm

= m

Ar

A 2

A

2

T

4 = m

B r

B 2

B

2

T

4

2B

BB2A

AA

T

rm

T

rm

As C is com mAr

A = m

Br

B

hence TA = T

B

5. (A) It is similar equation as v = 22 xa ù in SHM.

(B) Particle on positive x-axis move towards origin with speed decreasing as x decreasing.(C) It is spring mass system performing SHM.(D) Object moves away from Earth so its speed will decrease, since its speed is greater than escape velocityso it will never return back.


6. = 0

r < R = 0 r > RCase I r < R

FC =

rmV2

mgr

mVRr 2

(g = acceleration due to gravity at surface of sphere)

V = Rg

r for r < R

Case II r > R

2r

GMm =

rmV2

V = Rrg

rGM

So

7. If only gravitational force acts on astronaut (that is in state of free fall), he shall feel weightless. Thusstatement-2 is correct explanation of statement-1.

8. Wext = U � UP

Wext = 0 �

1.

xGdm

��

Wext = G 222rR16

rdr2

R7

M

= 2R7

GM2

22 rR16

rdr

= 2R7

GM2 z

zdz = 2R7

GM2 [Z] r

3R

4Rdr

P

x

Wext = 2R7

GM2 R4

R3

22 rR16

Wext = 2R7

GM2 R5�R24 Wext = 2R7

GM2 5�24 .

9.A Bc.m.

5d6

d6 11 Ms

2.2 Ms

c.m. about B of momentum Angularc.m. about momentum angular Total

=

6d

6d

)M 11(

6d

6d

)M 11(6d5

6d5

)M2.2(

s

ss

= 6.

10. g = 2R

GM = 2

3

R

R34

)G(

; g R

g'g =

R

'R

' =

3

2

R

'R =

116

Given,R

'R =

2263

Ve = R

GM =

R

R34

))()G( 3

Ve R ; Ve = 3 km/hr.


11. Ve = 0v2 KE = 2

emv21

= 20v2m21

= mv0

2

12. Ves

= RGM2

= R

R34

.G.2 3

= 3G4

R

Ves

RSarface area of P = A = 4R

P2

Surface area of Q = 4A = 4 RQ

2 RQ = 2R

p

mass R is MR = M

P + M

Q

3RR

34 =

3PR

34 +

3QR

34 R

R3 = R

P3 + R

Q3

= 9RP

3

RR = 91/3 R

P R

R > R

Q > R

P

Therefore VR > V

Q > V

P P

R

VV

= 91/3 and Q

P

VV

= 21

PART - II1. Electric charge on the moon = electric charge on the earth

2. V = RGM2

=

10R

10GM2

e

e = 10

e

e

R

GM2 = 110 k m/s

3. Acceleration due to gravity at leight h from earth surface.

g' = 2

Rh

1

g

2

Rh

1

g9g

h = 2R

4. 2x

Gm = 2)xr(

)m4(G

x1

= xr

2

r � x = 2x 3x = 3r

x = 3r

3/r2)m4(G

3/rGm

r

Gm6r

Gm3 =

rGm9

Ans.

5. 2

2

)R2(

Gm = m2R

3

2

R4

Gm = 2

= 3R4

Gmv = R v = 3R4

Gm × R =

R4Gm

6. W = 0 � R

GMm

R

GMm

= gR2 × Rm

= mgR = 1000 × 10 × 6400 × 103

= 64 × 109 J = 6.4 × 1010

RESONANCE SOLN_Work, Power & Energy - 59

TOPIC : WORK, POWER AND ENERGY

EXERCISE-1PART - I

SECTION (A)A 1. f = mg = F

Displacement = vt(a) W

mg = mg × vt cos90º = 0

(b) WN = N × vt cos90º = 0

(c) Wf = �mgvt

(d) WF = Fvt = mgvt.

A 6. m = 500 g = 21

kg

mg sin = fk

kfW = (mg sin ) (2) =

21

(10) 54

× 2 = 8 J

A 7. W1 = (mg sin )4

= (20 × 10 × 54

) (4) = 640 J

2FW + ravGW = K = 0

2FW � (mg sin ) (4) = 0

W2 =

2FW = 4 mg sin = 640 J

SECTION (B)B 2. W = Area under given graph from x = 0 to x = 35m

= 21

× (20 + 40) × 10 � 21

× 5 × 5

= 2

575J.

B 4. F at any moment

=

)x(mg

W =

0dx

)x(mg

=2

mg.

SECTION (C)

C 4. Work done by resistive force = WR = K

=21

× 20 × 10�3 (1002 � 8002)

= � 6300 J

So, average resistive force <R> = m1

J6300 = 6300 N.


C 6. Work done by the force = ds F = F × 21

at2

= F × 21

× mF

× t2

=m2tF 22

= 521020 22

= 4000 J

Now K = 21

m(v2 � u2)

= 21

m (2as)

= m × mF

×21

× mF

× t2

= m2tF 22

= 4000 J

WF = K.

C-10. U = �K

21

kx2 � 2mgx = 0

x = kmg4

.

C-11. (a) Since, gravitational force is conservative, So, work done by it in round trip is zero.

(b) sin = 105

= 21

= 30º

WF = mg(sin + cos) ×

= 0.3 × 9.8 10 23

15.021

= 18.519 J

(c) Wf = �f.s

= �mg cos (2)

= � 2mg cos = � 2 × 0.15 × 0.3 × 9.8 × 10 × 23

= �7.638 J.

(d) By W.E.T,K

f � K

i = W

F + W

f + W

g

Kf = (18.519 � 7.638)J = 10.880 J.

C 12. Displacement of 4kg block = 2 × 2m = 4m

4kg = 2 × 2m = 4m

Final speed of 4kg block = 2 × 0.3 = 0.6 m/s

4kg = 2 × 0.3 = 0.6 m/s

Wf + W

g = K

� × 4 × 10 × 4 + 2 × 10 × 2 = 21

× 4 × (0.6)2 + 21

× 2 × (0.3)2

160 = 40 � (0.72 + 0.09) = 160

81.0�40 = 0.2449


C-14. (i) w.r.t. person in the train

v1 = at =

mFt

(ii) w.r.t. person on ground,

v = vc + v

1 = v

c +

mFt

(iii) According to person in the train,

K1 =

21

mv1

2 = m2tF 22

(iv) According to person on ground,

K = 21

m 2c

2

c mv21

mFt

v

.

(v) S1 =

21

a1t2 =

m2Ft2

.

(vi) According to person on ground,

S = vct +

21

mF

t2 = m2

Ft2

+ vct.

(vii) According to person in the trainwork done by F = Fs

1

= m2tF 22

According to person on ground,Work done by F = F.s

=

tv

m2Ft

F c

2

.

(viii) Comparing Wg = K

g

and Wc = K

c .

(ix) Work�energy theorem holds in moving frame also.

SECTION (D)D 5. Let m

1 = 2m

2

a = g)mm()mm(

21

21

=

2

2

m3m

g = g/3

So, distance travelled by each block = 21

at2 = g/6

Also T = 21

21

mmgmm2

=

3gm4 2 = 16

m2 = g

12

Hence, loss in gravitation P.E. during first second= (m

1 � m

2)gh

= (2m2 � m

2) g ×

6g

= 6g

gg

12 = 2g.


D 6. a = 2323

g =

5g

Distance covered in fourth second =2

a)1n2(

= 52

g)142(

=

10g7

Hence, work done by gravity = (m2 � m

1)gh

= (3 � 2)g × 10

g7

= 2g

107

.

D 7. Ws + W

g + W

f = K

21

kx2 + mgx sin37º � mg cos37º × x = 0

21

× 100 × (0.1)2 + 1 × 10 × 0.1 × 53� × 1 × 10 ×

54

× 0.1

= 81

.

SECTION (E)

E 4. Power developed by motor = t

mgh =

60512010400

= 1600 W..

E 6. Power P =t

mgh

m = ghPt

= 1010

60102 3

kg = 1200 kg.

E 7. P = t

mgh

t = P

mgh =

1000104010200

sec. = 8 second.

E 8. 20 kg / minute = 20 kg / 60 sec = 31

kg/s

P =

3

1 g (20) =

32010

watt

746 W = 1 H.P

P = 1119100

HP


SECTION (F)

F 1. (a) w = sd.F

= )j�dyi�dx).(j�yxi�yx( 2222

= )dyyxdxyx( 2222

which is not a perfect integral and hence cannot be integrated without knowing y = f(x) or x = f(y). So, work

done by F

depends on path. So, it is non�conservative force.

(b) While moving along AB, y = 0 and along BC, x = a.

WABC

= a

0

a

0

2222 dyyxdxyx

= 0 + a2 × 3

a3

= 3

a5

While moving along AD, x = 0 and along DC, y = a

So WADC

= a

0

a

0

2222 dyyxdxyx

= 0 + a2 .3

a3

= 5

a5

Along ACx = y

So WAC

= a

0

a

0

2222 dyyxdxyx

= a

0

a

0

2222 dyyydxxx = 5a2 5

.

F 2. (a) F(y) = dydU

=

(b) F(y) = dydU

= � 3ay2 + 2by

(c) F(y) = dydU

= �U0 cos y..

F 5. At x = 0, total energy is in form of K.E. since U = 0and it turns back when its K.E. = 0So, total energy is in form of P.E. U = �K

21

kx2 = 1

x2 = 1 × 2 × 2

x = ± 2m Ans.


A 3. W = (force) (displacement ) = (force) (zero ) = 0

A 6. W = (2000 sin 15º) × 10 = 5176.8 J


A 9. S1 =

21

g 12 , s2 =

21

g 22 , S3 =

21

g 32

S2 � S

1 =

21

g 3, S3 �S

2 =

21

g 5

W1 = (mg) S

1, W

2 = (mg) (S

2 � S

1) , W

3 = (mg) (S

3 � S

2)

W1 : W

2 : W

3 = 1 : 3 : 5

A 10. T = mg + ma, S = 21

at2

WT = T × S

= 2

at)ag(m 2

A 11.* W = K, 0 = K, k remains constant, speed remains constant.

A 14.* W = K > 0 K ( = kinetic energy) increases

p = mk2 , p as k.

A 15.* Wf + W

G + W

N = K = 0

As WG = 0, W

N = 0 so W

f = 0.

SECTION (B)

B 1. W = 1x

o

cx dx = c 2

x21

B 2. F = K1x

1 , x

1 =

1KF

, W1 =

21

K1 x

12 =

1

2

K2

F

similarly W2 =

2

2

K2

Fsince K

1 > K

2 , W

1 < W

2

SECTION (C)

C 2. a = mF

, S = 21

2t

mF

, W

F = FS = F

m2Ft2

C 5. h = 21

gt2, W = mgh = mg 2

gt2, W = K

f � K

i

2tmg 22

= Kf �

21

mu2, Kf =

21

mu2 + 2

tmg 22

Hence Ans. is (A)

C-10. V dxdV

= � Kx,

x

0

2v

u

2

2Kx

�2

V

V2 � u2 = � Kx2

21

mu2 � 21

mV2 = 21

mK x2

Loss x2


C 12. (mg sin ) x �

x

0

cosmg dx = 0

sin x = o cos

x

0

dxx

x tan = 0

2x2

, x = 0

tan2

SECTION (D)

D 3. Ui + 0 = U

f +

21

mv2

Ui � U

f =

21

mv2

U = 21

mv2

m = 2v

U2

D 4.21

mu2 = mgh, u2 = 2gh ....(i)

mg

4h3

+ K.E. = mgh

K.E. = 4

mgh

.E.P

.E.K = 4/mgh3

4/mgh =

31

D 6. WF + W

S = 0, W

F � U = 0 , W

F = U = E

E = 21

KA x

A2 , Fx

A =

21

KA x

A2

AKF2

= xA ,

AKF2

= AKE2

, KA =

EF2 2

...(i)

similarly KB =

B

2

EF2

, KA = 2K

B

EF2 2

=

B

2

EF2

2

EB = 2E

Alter :F = K

A x

A = K

Bx

B

EA =

21

KA x

A2

EB =

21

KB x

B2

2

B

A

B

A

B

A

x

x

K

K

E

E

21

21

2EE

2

B

A


D 7. 100 = 21

K(2cm)2 , E = 21

K(4cm)2

so100E

= 4 , E = 400 J

E � 100 = 300 J

D 13.

22 mv

21

21

u)m2(21

.... (i)

21

(2m) (u + 1)2 =21

mv2 ....(ii)

From (i) and (ii) u = 12

1

D 14. W1 = work done by spring on first mass

W2 = work done by spring on second mass

W1 = W

2 = W (say)

W1 + W

2 = U

i � U

f

2W = 0 � 21

Kx2

W = � 4

Kx2

D 15. Wa + W

c = K = 0, W

a � mg

º60cos

2�

2

= 0

Wa =

4mg

= (0.5) (10)

41

= 4

5 J.

SECTION (E)

E 3. V = 0 + at, F � mg = ma , F = mg + ma,P = (mg + ma) at

E 5. P = TV = 4500 × 2 = 9000 W = 9KW

E 6. P1 = 80 gh/15 , P

2 = 80 gh/20

2

1

PP

= 1520

= 34

SECTION (F)

F-2. vedxdU

,vedxdU

BxAx

So, FA = positive, F

B = negative

F-5. WC

= WC + W

C = 5 + 2 = 7

P R P Q Q R

F-6.xU

= cos (x + y),

yU

= cos (x + y)

F = � cos (x + y) i� � cos (x + y) j�

= � cos (0 + 4

) i� � cos (0 + 4

) j�

| F | = 1


EXERCISE-2PART - I

1. a = )mm(F

21 f1 = m

1a = m

1 )mm(F

21

2F � f1 � f

2 = m

2a

� f2 = � 2F + f

1 + m

2a = m

1a + m

2a � 2F

� f2 = (m

2 + m

1) )mm(

F

21 � 2F = F � 2F = � F f

2 = � F

2F � K (m

2 + m

1)g = (m

2 + m

1)a

2F � (m2 + m

1) )mm(

F

21 = µ

K (m

2 + m

1)g g)mm(

F

21 =

K

W = work done by friction force on smaller block

= f1x =

)mm(Fm

12

1

x

2. mg = N + F sin .......(1)N = F cos .......(2) mg = F cos + F sin

F =

sincosmg

.......(3)

WF =

èsin ìècos)10(così mg

è =

tan540000

Ans.

F is min. if D = cos + sin is maximum and its maximum value is 21 ì

Fmin.

= 2

2

1

1 mg

= 21

mg

minFW = 21

mg

21

10

ì

= 0.2, mg = 4000 Nt

minFW = 22 ))2.0(1(

10)4000)(2.0(

=

2)04.01(

20400

=

04.18000

= 7692.307 J Ans.

4. fK =

m ( � x)g

W =

4x

g)x�(m� ì dx

W =

mgì×

2

])x[(4

2

= � 2

mgì ×

169 2

= � 32mg9


7. (a) Taking F = 40 N, m = 4 kg , = 53º

ax = (F cos � mg sin )/m

= (40 cos � 40 × 54

)/4 = 10 cos � 8

ay =

msinF á

= 4sin40 á

= 10 sin

x = 0 × 2 + 21

(10 cos � 8) (2)2 = 20 cos � 16

y = 21

(10 sin ) (2)2 = 20 sin

WF = (F cos ) x + (F sin )y

WF = (40 cos ) (20 cos � 16) + (40 sin ) 20 sin

= 800 cos2 � 640 cos + 800 sin2 W

F = 800 � 640 cos

WF 800 � 640

WF 160 J

(b) If WF = 160 J then 160 = 800 � 640 cos cos = 1

y = 0 and x = 20 � 16 = 4

WG = (�mg sin ) (4) = (�4 × 10 ×

54

) 4

= �128 J

(c) F acts along the x-axis.W

G + W

F = K

�128 + 160 = K Kf = 32 J.

9.

Work energy theorem (Between A & C)W

f + W

G + W

sp = K

mg cos (5 + 3) + mg 2 sin = 0

= 82

tan 37o = 163

work energy theorem (bet. A & B)W

sp + W

G + W

f= K

mg 5 sin 37o � mg 5 cos � 21

K (0.4)2 = 0

(4 × 10)

54

)5(163

53

5 = 21

× 10016

K

K = 9000/ 8 N/m so x = 9

10. Work energy Theorem on �m�W

G + N + W

T + W

f =K

� mg R + O + WT �

2

0

)sinmg( R d = 0

WT = mgR ( + 1)


11. WF + W

Sp + W

fric = K

Fx � 21

Kx2 � m1g x = 0 & Kx = m

2g

F � 21

m2g � m

1g = 0

F = m1g +

2gm2

13. mg = kx

K = x

mg =

2.0100

= 500 N/m

21

K (0.2)2 + 21

mv2 = m × 10 × 0.2

21

× 500 × 4 × 10�2 + 21

× 10 v2 = 10 × 10 × 0.2

10 + 5v2 = 20v2 = 2

v = 2 m/s

Since u is 4 m/s ( ) so block will compress the spring.Let x be the compression of spring.

2mu21

+ 2)2.0(K21

+ 0 = 21

m(0)2 + 21

Kx2 + mg (x + 0.2)

21

× 10 (4)2 + 21

× 500 ×

100

4 =

21

× 500 (x)2 + 10 × 10 (x + 0.2)

80 + 10 = 250x2 + 100 x + 2025 x2 + 10 x � 7 = 0 solving this

x = 0.36 mSo from initial position distance is ( 0.2 + 0.36) m = 56 cm

15. (i) mg = T cos

mg = amg2

aaxa 22 ×

22 xa

x

x = 2a

(ii)21

K 2a2 + mga = 21

K (2a � a)2 +21

mv2

21

amg2

(2a2 � a2) + mga = 21

mv2

ga4 = v


(iii)21

K 2a2 + mg a = 21

K2

22 ya

� mg y

21

K2a2 + mg a = 21

K(a2 + y2) � mgy

21

amg2

2a2 + mg a = 21

amg2

a2 + 21

amg2

y2 � mg y

3 mg a � mg a = ay mg 2

� mg y

2 mg a = ay mg 2

� mg y

2a2 = y2 � ay

y2 � ay � 2a2 = 0y2 + ay � 2ay � 2a2

y (a + y) = 2a (y + a) y = 2a

16. (a) P = extF

. V

Where V

is the vel. of point of application

Fext

+ m, g = T & m2g =T

Fext

= m2g � m,g = (m

2�m

1) g

P = (m2�m

1) g v Ans.

(b) Fext

+ m,g � T = m,a

T�m2g = m

2a

_____________________________F

ext = (m

1+ m

2) a (m

2�m

1)g

= m2(g+a) � m

1(g � a)

P = (Fext

) (0 + at)= {m

2(g+a) � m

1 (g � a)} at Ans.

19.

� g [ma + m´a + M2a

] = � g [ m2a + m´0 + Ma] + 21

(M + m + m´)v2

'mmM

]a´m2

MaMamama2[g2

= v

v = ´mmM´)m�m(2M

ag

Ans.


21. U (x) = 20 + (x � 2)2

dxdu

= 2(x � 2)

� F = 2(x � 2)

F = � 2(x � 2)

m (x � 2) = � 2 (x � 2)

Let x = x � 2

mx = � 2 x

1 x = � 2 x

x = � 2 x Simple Harmonic MotionMean position is x = x � 2 = 0 x = 2 W2 = 2 ,

Kinetic energy =21

mv2

= 21

(1) (2) (A2 � x2) = x � 2, x = 5 � 2 = 3

20 = 21

(1) (2) {A2 � 32}

20 = A2 � 9 A2 = 29 A = 29Aliter :for mean position

F = � dxdU

= � 2(x � 2) = 0 x = 2

At x = 5K.E. = 20 JU

(x = 5) = 20 + (5 � 2)2 = 29 J

Total energy, T.E. = 20 + 29 = 49 JAt amplitudeU(x)

max = 49 J = 20 + (x � 2)2

29 J = (x� 2)2

x = 2 ± 29

x = 2 + 29 , 2 � 29

xmin

= 2 � 29 = �3.38

xmax

= 2 + 29 = 7.38

K.E.max

when U(x)

is minimum at x = 2U(x)

min = 20 J

KEmax

= 29 J

22. Using work energy thoerem,

Wf + mg

2

2R

Rmg4

21

2R3

=

21

m 2gR3

Wf =

21

mgR

f = N (as kinetic friction)

Wf = f

dx = F

s cos d

x ( x = 2 R tan ; dx = 2 R sec2 d)

Fs = k 2 R (sec 1)

Wf = k 2 R (sec 1) cos 2 R sec2 d


Wf = 4 R2 k

0

0

(sec2 sec ) d

0 = tan 1 (3/4)

= 4 R2 k tan (sec tan )

n0

0= � 4 R2 k [ tan

0 ln (sec

0 + tan 0)] =

2mgR

4 R2 k 3

4

5

4

3

4

n = R2 k [3 4 ln

2] =

mg R

2

= mg

Rk n2 3 4 2( )

= 1

8 3 4 2( ) n Ans.

23. Velocity will be maximum when a = 0For a = 0, F = 0This situation occurs for v

e following arrangement of springs.

150 150

Natural length is c = 150 mmNow , U

i + K

i = U

f + K

f

Ui =

21

K{ 5 c � c}2 + 21

K{ 2 c � c}2

Ki = 0

Uf = 2.

21

K{ 2 c � c}2

21

K{ 5 c � c}2 + 21

K{ 2 c � c}2

=21

mv2 + 2.21

K{ 2 c � c}2

Solving the equation & putting the valueswe have

v = 2/1

22 )12()15(2

15

m/s = 3.189 ms�1 .

PART - II3. W

agent + W

G = K = 0

Wagent

= � WG, But W

G is independent of the path joining initial and final position. W

G is independent of time

taken.

5. Wf + W

G = K

�mgd � mgh = 0 � 21

m v0

2

gd + gh = 21

(v0

2)

(0.6) (10) d + 10(1.1) = 18 d = 67

= 1.1666 1.17

7. WS + W

f = K

� U + Wf = � K

i

� Uf � mgx = � K

i

21

K x2 + mgx = 21

mu2

100 x2 + 2(0.1) (50) (10) x = 50 × 4

x2 + x � 2 = 0

x = 1 m


8. v = s sdtds

,

t

0

s

0

dts

ds s2 = t

s = t/2 ....(1)

W = workdone by all the forces = K

= 21

mv2 = 21

m 2s = 21

m2

4t22

10. K.E. + P.E. = constant fu;r = C (say)

K � mg (tu sin � 21

gt2) = C

K = mg [tu sin � 21

gt2] + C [= parabolic]

C 0 so answer is (B)

12.dxdU

= positive constant

For x < a, F = negative constant and for x > a, F = 0so, ans. (C)

14. E = m2

p2

, )E(

P1

= m2

1 = constant

Rectangular hyperbola (C)

17. System is block & string. Applying work energy theorem on system

(200)10 � 10g(R � R cos60º) = 21

(10)v2

2(200 � 10 × 5) = v2

v = 300 = 310 .

19. dW = F

. sd

where sd

= i�dx + j�dy

and F

= � K ( i�y + j�x )

dW = � K ( ydx + xdy = � K d (xy)

W = )a,a(

)0,0(dW = � K )xy(d

)a,a(

)0,0( = � K [xy] )0,0()a,a(

W = � Ka2

20. From given graphs :

ax =

43

t and ay =

1t

4

3 v

x =

83

t2 + C

At t = 0 : vx = � 3 C = � 3

vx =

83

t2 � 3 dx = dt3t8

3 2

.... (1)

Similarly; dy = dt4tt83 2

.... (2)

As dw = ds.F = )j�dyi�dx.(F

4

0

22W

0

dtj�4tt83

i�3t83

.j�1t43

i�t43

dw


W = 10 JAlternate Solution :Area of the graph ;

dtax = 6 = )3(V f)x( V(x)f

= 3.

and dtay = �10 = )4(V f)y( V(y)f

= � 6. Thus, u = 5 m/s and v = 45 m/s.

Now work done = KE = 10 J

22.* WG = K, mgh =

21

mv2 � 21

mu2, 21

mu2 + mgh = 21

mv2

so v > u and v depends upon u.

23.* dWF = sd.F

, if F

perpendicular to sd

then

dWF = 0, sd

is displacement of point of application of force, v =

dtsd

.

(A), (C), (D) are true.

EXERCISE-3

1. The displacement of A shall be less than displacement L of block B.Hence work done by friction on block A is positive and its magnitude is less than mgL.And the work done by friction on block B is negative and its magnitude is equal to mgL.Therefore workdone by friction on block A plus on block B is negative its magnitude is less than mgL.Work done by F is positive. Since F>mg, magnitude of work done by F shall be more than mgL.

2. The FBD of block isAngle between velocity of block and normalreaction on block is obtuse work by normal reaction on block is negative.As the block fall by vertical distance h,from work energy TheoremWork done by mg + work done by N = KE of block

|work done by N| = mgh � 21

mv2

21

mv2 < mgh

|work done by N| < mgh(B) Work done by normal reaction on wedge is positiveSince loss in PE of block = K.E. of wedge + K.E. of blockWork done by normal reaction on wedge = KE of wedge. Work done by N < mgh.(C) Net work done by normal reaction on block and wedge is zero.(D) Net work done by all forces on block is positive, because its kinetic energy has increased.Also KE of block < mgh

Net work done on block = final KE of block < mgh.

3. If the particle is released at the origin, it will try to go in the direction of force. Here dxdu

is positive

and hence force is negative, as a result it will move towards � ve x-axis.

4. When the particle is released at x = 2 + it will reach the point of least possible potential energy (�15 J)

where it will have maximum kinetic energy.

2maxvm

21

= 25 vmax

= 5 m/s


6. (A) WCL

+ Wf = KE W

CL = KE � W

f

(a) During accelerated motion negative work is done against friction and there is also change is kineticenergy. Hence net work needed is +ve.(b) During uniform motion work is done against friction only and that is +ve.(c) During retarded motion, the load has to be stopped in exactly 50 metres. If only friction is consid-ered then the load stops in 12.5 metres which is less than where it has to stop.Hence the camel has to apply some force so that the load stops in 50m (>12.5 m). Therefore the workdone in this case is also +ve.

7. WCL

|accelerated motion

= KE � Wfriction

where WCL

is work done by camel on load.

= 50.mg0mv21

k2

= 501000101.05100021 2

=

2125

1000

similarly, WCL

|retardation

= KE � Wfriction

2mv2

10 � [

k mg.50] =

2

751000

motion retardedCL

motion daccelerateCL

|W

|W =

75125

= 35

5 : 3

8. Maximum power = Fmax

× V

Maximum force applied by camel is during the accelerated motion.We have V2 � U2 = 2as

25 = 02 + 2 . a . 50a = 0.25 m/s2 ; for accelerated motion

FC � f = ma

FC = mg + ma = 0.1 × 1000 × 10 + 1000 × 2.5

= 1000 + 250 = 1250 NThis is the critical point just before the point where it attains maximum velocity of almost 5 m/s.Hence maximum power at this point is = 1250 × 5 = 6250 J/s.

14. Potential energy depends upon positions of particles

15. (i) The net force on the body may have acute angle with its velocity, but one of the constituent forcemay have obtuse angle with the velocity. Such a force shall perform negative work on the body eventhough the kinetic energy of the body is increasing.

(ii) A net force that is always perpendicular to velocity of the particle does no work but changes thedirection of its velocity.

(iii) A force which is always constant is also conservative.(iv) From Work - Energy theorem

Wall forces

= KEfinal

� KEinitial

EXERCISE-4PART - I

1. Power P = F

. V

= FV

F = V

dtdm

= V

dtvolume(d = density

= V

dtvolume(d

= V (AV)

= AV2

Power P = AV3

or P V3


Alternate SolutionPower output is proportional to number of molecular striking the blades per unit time [which dependson the velocity V of wind] and also proportional to energy to striking molecules or proportional to squareof velocity V2 Therefore, power output P V3

2. F = � dxdU

dU = � F . dx or ;k U(x) = �

x

0

3 dx)axkx(

U(x) = 2

kx2

� 4

ax4

U(x) = 0 and x = 0 and x = ak2

U(x) = negative for x > ak2

From the given function we can see thatF = 0 at x = 0 i.e. slope of U-x graph is zero at x = 0. Therefore, the most appropriate option is (d).

3. Let x be the maximum extension of the spring. From conservation of mechanical energy :decrease in gravitational potential energy = increase in elastic potential energy

Mgx =21

kx2

or x = kMg2

4. From F = dxdU

x

0

x

0

)x(U

0

dx )kx(FdxdU

U(x) = 2

kx2

as U(0) = 0Therefore, the correct option is (A).

5. In horizontal plane Kinetic Energy of the block is completely converted into heat due to Friction but in thecase of inclined plane some part of this Kinetic Energy is also convert into gravitational Potential Energy. Sodecrease in the mechanical energy in second situation is smaller than that in the first situation. So state-ment-1 is correct.Cofficient of Friction does not depends on normal reaction, In case normal reaction changes with inclinationbut not cofficient of friction so this statement is wrong.

6.


As springs and supports (m1 and m

2) are having negligible mass. Whenever springs pull the massless

supports, springs will be in natural length. At maximum compression, velocity of B will be zero.

And by energy conservation

21

(4K) y2 = 21

Kx2

21

xy Ans. (C)

7. T = gmmmm2

21

21

=

36.072.036.072.02

× 10

T = 4.8 N

a = gmmmm

21

21

=

3g

s = 2at21

= 21

3g

(1)2 = 6

10

Work done by T = (T) (S)

= (4.8) × 6

10 = 8 J Ans.

8. pFdt

21

× 4 × 3 � 21

× 1.5 × 2 = pf � 0 pf = 6 � 1.5 = 29

K.E. = m2

p2

= 224

81

;K.E. = 5.06 J Ans.

PART - II1. Let initial velocity is u and retardation is a

So, (vr%) 4

u2

= u2 � 2a × (0.03) ...(i)

0 = 4

u2

� 2a × S ..(ii)

here S is required distancefrom equation (i) & (ii)S = 0.01 m = 1 cm

2. WC = � U

= � (Ufinal

� Uinitial

)

= �

22 5k21

�15k21

]

WC = 8 Joule

3. K = 5 × 103 N/mx = 5 cm

W1 = 21

k × 21x =

21

5 × 103 × (5 × 10�2)2 = 6.25 J

W2 = 2

k(x1 + x2)

2

= 2

× 5 × 103 (5 + 10�2 + 5 × 10�2)2 = 25J

Net work done = W2 � W1 = 25 �6.25 = 18.75 J

= 18.75 N-m


4. Mass per unit length = LM

= 24

= 2 kg/m

The mass of 0.6 m of chain = 0.6 × 2 = 1.2 kg

The centre of mass of hanging part = 2

06.0 = 0.3 m

Hence, work done in pulling the chain on the tableW = mgh= 1.2 × 10 × 0.3

= 1.2 × 10 × 0.3

= 3. 6 J

7. F = ma = T

m

T

0a

Instantaneous power = F= ma

= T

m. at =

Tm

. T

. t

= 2

2

T

m.t

8. Maximum height attained by the particle

m45

1025

g2u

H22

Wg = -MgH = -0.1 × 10 × (5/4) = -1.25 J

9. Velocity of ball just after throwing

v = gh2 = 2102 = 40 m/s

Let a be the acceleration of ball during throwing, then

v2 = u2 + 2as = 02 + 2as a = s2

v2

= 2.02

40

= 100 m/s2

F - mg = ma F = m(g + a) = 0.2(10 + 100) = 22 N

(2) is correct

10. kmv21 2

4K

mv21

41

4v

m21

)60cosv(m21 2

22

11. Assuming mass of athlete is between 40 kg to 100 kghere we will consider mass of athlete m = 50 kg

V = S/t = 10

100 = 10 m/sec

So, K = 1/2 mv2 1/2 × (50 ×102) = 2500 JSo Answer is (C)

12. K.E. = ct

21

mv2 = ct

m2P2

= ct

P = ctm2

RESONANCE SOLN_Circular Motion - 79

TOPIC : CIRCULAR MOTION

EXERCISE-1PART - I

SECTION (A)A 1. Given v

= 2i � 2j

(a) when moves in clockwise `Ans. : First quadrant

(b) When moves in counter clockwise

Ans. : Third quadrant

A 3. Given 0 = 0 , = const

= 0t +

21t2

for first two seconds

1 = 0 +

21×(2)2 = 2

for next two seconds

2 =

4 �

2 =

21 (4)2 �

21 (2)2 = 6

2 /

1 = 3 : 1 Ans.

A 5. Given = R = 1 cm , t = 15 Second

12 V�VV

V = 2 V

V = R

V = 1602

= 30

cm/sec. V = 302

cm/sec.

a = tV

=

15302

cm/sec2. Ans.

SECTION (B)

B 1. R = 0.25 m , = 2 rev./sec. = 4 rad/sec. (at = 0)

ac = 2R

= (4)2 × 0.25

= 42 m/s2. Ans.


B 3. R = 1.0 cm , V = 2.0 tat t = 1 sec V = 2.0 cm/sec.

ac =

Rv2

= 4 cm/sec2.

at =

dtdv

= 2.0 cm/sec2.

a = 2t

2c aa = 22 24 = 52 cm/sec2. Ans.

SECTION (C)C 1. m = 200 g = 0.2 kg , g = 2 m/s2

Time period = 2g

cos

= 22

2.1

= 2

56

Ans.

Tension = cos

mg =

13/122.0 2

= N6

13 Ans.

C 3. N = r

mv2

given r = 5 m , v = 55 m/s

for no slipping mgf

µmin

N = mg

µmin

= N

mg = 2v

rg

µmin

= 2)55(

105 =

52

Ans.

C 5. = 2n = 6015002

rad/sec

r = 2d

= 60 cm = 0.6 m

m = 1 g = 10�3 kg

F = m2r = 10�3 ×

2

6015002

× 0.6

= 10

15 2 = 14.8 Ans.

This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude butopposite in direction.

SECTION (D)

D 1.

R =

av 2

= gsinu 22

Ans.


SECTION (E)E 1. Tension is maximum in circular motion in vertical plane at lowest position.

At lowest positionT

max � mg = m2R 30 � 0.5 ×10 = 0.5 2 × 2

2 = 25.0

25

= 5 rad/sec. Ans.

E 3. When string become slack apply equation for centripetal force.

amv2

= mg cos 60º v = 2

ga....(i)

apply energy conservation

21

mu2 = 21

mv2 + mga(1 + cos) ....(ii)

from equation (i) & (ii)

u = 2ga7

apply equation for centripetal force at lowest position.

T � mg = a

mu2

put the value of u and we getT = 9mg/2

E 5. Using energy conservation :

2Bmv

21

= mgh vB =

mmgh2

vB = hg2 .....(1)

Also to complite vertical circle

vB = gR5 .....(2)

R = h52

= 2 cm

Section (F)

F 1. For safe driving vmax

= rg

10 = rg

for wet road v´ = rg2

= 2

10 = 5 2 m/s Ans.

F 4. v = 48 km/hr = 40/3 m/s.For safe turn without friction

tan = rgv2

= xh

given x = 1m h = rgv2

= 10400)3/40( 2

=

452

m Ans.

F 7. T = 2.effg

cos

= 2 .effg

h

geff.

= g + a ; T = 2 put geff

= 20 g + a = 20 a = 10 m/s2.

Ans. Retardation = 10 m/s2

Ans. 10 m/s2



A 1. Speed v1 =

tr2ð

v2 =

tr2ð

1

= 1

1

t2

rv

...(i)

2 =

2

2

t2

r2v

...(ii)

From eq. (i) and (ii)1

2

2

1

tt

1 =

1

2

tt

A 3. r =

20 m, a

t = constant

n = 2nd revolutionv = 80 m/s

0 = 0,

f =

rv

= /20

80 = 4 rad/sec

= 2 × 2 = 4

from 3rd equation2 =

02 + 2 (4)2 = 02 + 2 × × (4) = 2 rad/s2

at = r = 2 ×

20 = 40 m/s2 Ans.

A 5.* In curved path, may be circular or parabolic.In circular path speed and magnitude of acceleration are constant.In parabolic path acceleration is constant.

A 7. second

= T2

= 602

rad/sec.

v = .r = 602

× 0.06 m/s = 2 mm/s Ans.

if vvv

= 2 v = 2 2 mm/s Ans.

SECTION (B)B 1. Angular velocity of every particle of disc is same

aP = 2r

p , a

Q = 2r

Q

rP > r

Q a

P > a

QAns.

B 3. ac =

rv2

, radius is constant in case (a) and increase in case (b). So that magnitude of acceleration is

constant in case (a) and decrease in case (b).

SECTION (C)C 1. r = 144 m, m = 16 kg, T

max = 16 N

T = r

mv2

v = MTr

= 16

14416 = 12 m/s Ans.


C 3. Uniformly rotating turn table means angular velocity is constant. New radius is half of the original value.r´ = r/2 and = constantv´ = r= r/2 = v/2 = 5 cm/s Ans.a´ = 2 r = r/2 = a/2 = 5 cm/s2 Ans.

C 5.

T1 � T

2 =

2M

2 2L

T1 > T

2Ans.

SECTION (D)

D 1. At t = 0 a = g cos , R =

av2

= cosg

u2

SECTION (E)E 1. Let the car looses the contact at angle with vertical

mg cos � N = R

mv2

N = mg cos � R

mv2

During descending on overbridge is incerese. So cos is decreasetherefore normal reaction is decrease.

E 3. T � mg cos = r

mv2

....(1) (from centripetal force)

from energy conservation.

21

mu2 = 21

mv2 + mgr (1 � cos ) (here u is speed at lowest point)

from (1) and (2)

T = r

mu2

+ 3mg cos � 2mg for = 30º & 60º T1 > T

2

E 5.* For normal reaction at points A and B.

mg � N = r

mv2

N = mg � r

mv2

NA > N

Band normal reaction at C is N

C = mg, so N

C > N

A > N

BAns.

E-7._ a.T

= | T | | a | cos = 0

either | T | = 0 or | a | = 0 or = 90º

a = 0r

V2 for whole motion there is velocity.

So T = 0, T = 0 for

T + mg =

2mVT =

2mV � mg

mg 2 + 21

mV2 = 21

mu2

V2 = u2 � 4 gl T =

2mu � 5 mg T = 0 or T < 0 g5u

E 9_ V.T

= | T | | V | cos

= 90º every time.

So V.T

= 0 for every value of u.


Section (F)F 1. Here required centripetal force provide by friction force. Due to lack of sufficient centripetal force car thrown

out of the road in taking a turn.

F 3. When train A moves form east to west

mg � N1 =

R)Rv(m 2

N

1 = mg �

R)Rv(m 2

N1 = F

1

When train B moves from west to east

mg � N2 =

R)Rv(m 2

N

2 = mg �

R)Rv(m 2

N2 = F

2F

1 > F

2Ans.

F 5_ mg = m2 R , = Rg

EXERCISE-2PART - I

1. Change in velocity when particle complete the half revolution : v = vf � v

i = 2v

Time taken to complete the half revolution t = vR

average acceleration = tv

= v/R

v2

= Rv2 2

=

552 2

=

10 m/s2 Ans.

3. ac = a cos 30º = 25

23

m/s2 Ans.

ac =

Rv2

v2 = aCR = 25

23

× 2.5

v =

2/1

43

125

m/s Ans.

at = a sin 30º =

225

m/s2 Ans.

5. (i) The normal reaction by wall on the block is N = R

mv2

(ii) The friction force on the block by the wall is f = µN = R

µmv2

(iii) The tangential acceleration of the block = mf

= R

µv2

(iv)dtdv

= � R

µv2

or vdsdv

= � R

µv2

v

v0v

dv = � ds

Rµ

R2

0

integrating we get n 0vv

= � µ 2 or v = v0 e�2µ


7. Centripetal accelerationm2 r = T

1 cos + T

2 cos .... (1)

apply Newton law in vertical directionT

1 sin = mg + T

2 sin .....(2)

given m = 4 kg, T1 = 20 kgf = 200 N, r = 3m

cos = 53

, sin = 54

Put in equation (2) T2 = 150 N Ans.

Put in equation (1) we get

2 = 34

210

= 2

35 =

235

rad/s

n =

2 =

21

2

35 rev/sec. n =

30

235

rev/min. Ans.

9. Time take by ring to fall on ground.

T = gh2

from centripetal force

m2x = ma = mv dxdv

2x = vdxdv

v

00

2 vdvxdx

2v

2L 22

2 vx = L

x = . T = gh2

vy = L y = T = g

h2

distance of one ring from center is = 22 )x(y

distance between the point on the ground where the rings will fall after leaving the rods.

= 2 22 )x(y where x = y = gh2

12. (i) CP = CO = Radius of circle (R)

COP = CPO = 60º

OCP is also 60º

Therefore, OCP is an equilateral triangle.Hence, OP = RNatural length of spring is 3R/4. Extension in the spring

x = R � 4R3

= 4R

Spring force, F = kx =

Rmg

4R

= 4

mg

The free body diagram of the ring will be a shown.C

O FP

mg

Here, F = kx = 4

mg

and N = Normal reaction


(ii) Tangential acceleration ar = The ring will move towards

the x-axis just after the release. So, net force along x-axis :

Fx = F sin 60º + mg sin 60º =

4mg

23

+ mg

23

Fx =

835

mg

Therefore, tangential acceleration of the ring.

aT = a

x =

mFx =

835

g

aT =

835

g

Normal Reaction N : Net force along y-axis on the ring just after the release will be zero.F

y = 0

N + F cos 60º = mg cos 60º

N = mg cos 60º � F cos 60º = 2

mg �

4mg

21

= 2

mg �

8mg

N = 8mg3

14. (a) at equatorT + m2 R = mg.

% TT

= gR2

= 100

8.9)606024(

1000640042

2

= 0.65 % Ans.

(b) T = 2

mg....(1)

T + m2R = mg ....(2)from (1) and (2)

2R = g/2 = R2g

T =

2 = g

R22 = 2hr Ans.

16. Block B rotate in vertical plane. Tension is maximum in string at lowest position. When block B at lowestposition and block A does not slide that means block A not slide at any position of B.At lowest position

T � mg =

2mv T = mg +

2mv ....(1)

From energy conservation

mg(1 � cos ) = 21

mv2 ...(2)

from equation (1) and (2)T = mg + 2mg (1 � cos )

= 3mg � 2mg cos

for no slipping.T = mg = 3mg � 2mg cos

min

= 3 � 2 cos Ans.


18. Constant speed = 18 km/hr = 5m/sec.m = 100 kg, r = 100 m

(a) at B mg � NB =

rmv2

= 100

5100 2 = 25 N

B = 975 N Ans.

at D ND � mg =

rmv2

ND = 1025 N Ans.

(b) at B & D friction force act is zero.

at C f = mg sin 45 = 100 × 102

1( v = constant) = 707 N Ans.

(c) for BC part

mg cos 45 � NBC

= R

mv2

NBC

= 682 N

for CD part

NCD

� mg cos 45 = R

mv2

NCD

= 732 N

(d) f N Nf

position where its maximum and N is minimum which is in part BC at C position.

rmv

º45cosmg

º45sinmg2

682707

= 1.037 Ans.

PART - II1.

QP = 2 � 5 = � 3 rad/s

RP

= 3 � 5 = � 2 rad/s

Time when Q particle reaches at P = t1 =

32/

= 61

sec.

t2 =

32/5

= 65

sec. t3 =

32/9

= 23

sec.

Time where R particle reaches at P. t1 =

2 =

21

sec. t2 =

23

= 23

sec.

Common time to reaches at P is23

sec. Ans.

3. at loose contact N = 0

mg cos = R

mv2

....(1)

from energy conservation

mgR(1 � cos ) = 21

mv2 ....(2)

from (1) & (2)

cos = 32

sin = 35

tangential acceleration = g sin = 3

g5Ans.


6. For M to be stationaryT = Mg .... (1)

Also for mass m,T cos = mg .... (2)

T sin = sin

mv2

.... (3)

dividing (3) by (2)

tan = sing

v2

v =

sin.cos

gM

m

Mg

T Tsin

Tcos

mg

Time period = vR2

=

sin.cos

g

sin2

From (1) and (2) cos = Mm

then time period = 2 Mg

m

9. F = kx, T1 = ka = m2 2a =

m2k

Time period =

2 =

km2

2 = T

T2 = 2ka = m23a =

m3k2

Time period = k2m3

2 = T´ T´ =

23

T Ans.

12. (i) at angle at = g sin

from centripetal acceleration T � mg cos =

2mv...(1)

From energy conservation :

0 + mg cos = 21

mv2 v = cosg2 ....(2)

from (1) & (2) T = 3mg cos aC = 2g cos

a = 2c

2t aa = g

2cos31

(ii) Vertical component of sphere velocity is maximum when acceleration in vertical is zero that meansnet force in vertical direction is zero.Net force in vertical at angle

T cos = mg T = cos

mg...(3)

and tension also from equationT = 3mg cos ....(4)from (3) & (4)

3 mg cos = cos

mg cos =

3

1

T = mg 3 Ans.

(iii) Total acceleration is directed along horizontal that means avertical

= 0

cos = 3

1Ans.


14. For vertical circular motion, in lower half circle tension never be zero anywhere. Tension is maximum atlowest point of oscillation. Tension decrease both side in same amount. Therefore correct option is (D).

16. Maximum retardation a = gFor apply brakes sharply minimum distance require to stop.

0 = v2 � 2gs s = g2v2

For taking turn minimum radius is

g = r

v2

, r = gv2

, here r is twice of s

so apply brakes sharply is safe for driver.

19. = 2 dtd

= dtd2

= 2 × 0.4 = 0.8 rad/s

vAC

= r = 0.8 × 21

= 0.4 m/s

aC = 2r = (0.8)2

21

= 0.32 m/s2

a = aC = 0.32 m/s2 (a

t = 0)

21. Energy conservation from initial and final position

mgr + mg r

2

11 =

21

mv2 + 21

mv2 v = gr2

1gr2 Ans.

Normal reaction at bottom position A

N � mg = r

mv2

N = r

mv2

+ mg = r

2

grgr2m

2

+ mg = 3 mg � 2

mg = 2.29 mg

23. The acceleration vector shall change the component of velocity u|| along the acceleration vector.

r = n

2

av

Radius of curvature rmin

means v is minimum and an is maximum.

This is at point P when component of velocity parallel to accelera-tion vector becomes zero, that is u

|| = 0.

u|| = 0

R = a

u2 =

242

= 8 meter..

25.2

3T =

)2/3(

mv2

........(1)60o

/ 2T

T cos 60o

60o

VT sin 60o

mg

3 / 2

2T

= mg .......(2)

Hence T = 2 mg , So (B) holdsFrom (1) & (2) V2 = 3 g/2

V = 2

6.18.93

V = 2.8 3 m/s . So (C) hold


ac = V2/r = )2/3(

)2/g3(

= 3 × g = 9.8 3 m/s2

(D) holds

t = v

r2 =

)2/g3(

2/32

t = 4/7 (A) holds.

27. Speed of cage = gr = const.

Normal reaction at (weight reading)

NA � mg =

rmv2

NA = 2mg = 2w Ans.

Weight reading at G & C = mg = w Ans.weight reading at E

mg � NE =

rmv2

NE = 0 Ans.

29. Tangential acceleration = at = gsinNormal acceleration = an = g cosat = ang sin = g cos = 45°

vy = vxuy � gt = ux20 � (10)t = 10

t = 1 sec.During downward motionat = anvy = � vx20 � 10 t = � 10 t = 3 sec.

EXERCISE-31. From graph (a) = k where k is positive constant

angular acceleration =

dd

= k × k = k2

angular acceleration is non uniform and directly proportional to . (A) q, s

From graph (b) 2 = k . Differentiating both sides with respect to .

2

dd

= k or

dd

= 2k

Hence angular acceleration is uniform. (B) p

From graph (c) = kt

angular acceleration = dtd

= k Hence angular acceleration is uniform (C) p

From graph (d) = kt2

angular acceleration = dt

d = 2kt Hence angular acceleration is non uniform and directly proportional to t.

(D) q,r


2. v = 2t2

Tangential acceleration at = 4t

Centripetal acceleration ac =

Rv2

Rt4 4

Angular speed = Rv

= Rt4

, tan =

c

t

a

a= 34 t

R

t4

tR4

Sol. 3 to 5.The angular velocity and linear velocity are mutually perpendicular

v = 3x + 24 = 0 or x = � 8

The radius of circle r =

v =

105

= 21

meter

The acceleration of particle undergoing uniform circular motion is

va

= )j�4i�3()j�6i�8( = k�50

6. mg = r

mu20

u0 = gr

Now, along vertical

r = 2gt21

t = gr2

Along horizontal ; OP = 2u0t = 22 r

7. As at B it leaves the hemisphere, N = 0

mg cos = r

mV2

r

u /30

O

hv

B

N

mg cos

A

mg rh

= r

mV2

mv2 = mgh .............(1)By energy conservation between A and B

mgr + 21

m2

0

3

u

= mgh +

21

mv2

Put u0 and mv2 h =

27r19

8. As ac =

rv2

= g cos

at = g sin

anet

= gAlternate Solution :when block leave only the force left is mg. a

net = g.

9. aggeff

�a

ggeff

Tension would be minimum when it (tension) is along effg

tan = mg

43mg

= 34

= 53º .


10.

Vmin = effg = g45

= 2

g5.

11. Tmax = 6 mgeff (geff = g45

) = mg2

15

12. For conical pendulum of length , mass m movingalong horizontal circle as shownT cos = mg .... (1)T sin = m2

sin .... (2)

From equation 1 and equation 2, cos = 2

g

cos is the vertical distance of sphere below O point of suspension. Hence if of both pendulums aresame, they shall move in same horizontal plane.Hence statement-2 is correct explanation of statement-1.

13. The normal reaction is not least at topmost point, hence statement 1 is false.

14. Let the minimum and maximum tensions be Tmin

and Tmax

and the minimum and maximum speed be u and v.

Tmax

= R

mu2

+ mg

Tmin

= R

mv2

� mg

T =

Rv

Ru

m22

+ 2 mg.

From conservation of energy

Rv

Ru 22

= 4g is indepenent of u.

and T = 6 mg. Statement-2 is correct explanation of statement-1.

15. Statement-2 is wrong. R = a

v 2

, where a is acceleration component perpendicular to velocity..

and as particle goes up, v2 decreases and a increases so radius of curvature R decreases hence statement-1 is true

16. (i) False. It has tangential as well as radial acceleration. The angle is less than 180°.

(ii) True. The angle between velocity and radial acceleration is 90°.

(iii) True. It has no acceleration in verticall direction

(iv) False. = 2

finalinitial is valid only for constant angular acceleration.

(v) False. aT =

22c dt

dva

> a

c


17. (i) Given that tangential acceleration = at = 3 m/s2

Centripetal acceleration = ac = r

v2 =

10020 2

= 4 m/s2

` Now a = 2t

2c aa = 22 34 = 5 m/s2

(ii) <a> = average acceleration = tv

=

R/RR2

=

2R2

Instantaneous acceleration = 2R

aa

=

2Ans.

(iii) Tension before cuttingT sin = mg

T1 =

sinmg

Tension after cutting.T

2 = mg sin

1

2

TT

= sin2 Ans.

(iv) tan = (v2/rg) = 22 hb

h

]

Ans :

2/1

22 hb

ghr

(v) Acceleration at lowest position

aL =

Rv2

From energy conservation

mgR (1 � cos ) = 2

mv2

Rv2

= 2g(1 � cos)

aL = 2g (1 � cos)

acceleration at highest position.a

H = g sin

according to problema

L = a

H

2g(1 � cos ) = g sin 2 (1 � cos ) = sin 2(1 � 1 + 2 sin2 /2) = 2 sin /2 cos /2

tan 2

= 21

tan =

41

21

222

1

2

tan1

tan2

= 34

= 53º Ans.


EXERCISE-4PART - I

1. Net acceleration a

of the bob in position B has two components. //////////////////////////

A

na

a

at

B

(i) na

= radial acceleration (towards BA)

(ii) ra

= tangential acceleration (perpendicular to BA)

Therefore, direction of a

is correctly shown in option (C).

2. (a) h =

2d

R (1 � cos)

velocity of ball at angle is

v2 = 2gh = 2

2d

R (1 � cos)g .......(1)

Let N be the total normal reaction (away from centre) at angle . Then

mg cos � N =

2d

R

mv2

Substituting value of v2 from equation (1) we get

mg

v

h

mg cos � N = 2mg (1 � cos) N = mg (3 cos � 2) Ans.(b) The ball will lose contact with the inner sphere when

N = 0 or 3cos � 2 = 0 or = cos�1

32

After this it makes contact with outer sphere and normal reaction starts acting towards the centre. Thus for

cos�1

32

:

NB = 0

and NA = mg (3 cos � 2) and for cos�1

32

NA = 0

and NB = mg (2 � 3cos)

The corresponding graphs are as follows

2/3 +1-1

mg

NA

cos2/3 +1-1

5mg

NB

cos

2mg

3. By energy conservation,

21

mu2 = 21

mv2 + mg(1 � cos)

V2 = U2 � 2g (L � L cos)

4gL5

= 5gL � 2gL (1 � cos)

5 = 20 � 8 + 8 cos

cos = � 87

43

< < Ans. (D)


4. T sin = m Lsin2

324 = 0.5 × 0.5 × 2

2 = 5.05.0

324

= 5.05.0

324

= 5.0

18 = 36 rad/sec.

5. Since distance of particle P from point O is initially decreasing then in-creasing so, its angular velocity will initially increase then decrease. So,angle swept by P is more than angle swept by disc. So it will fall in un-shaded portion.Since distance of particle Q from O is continuously increasing so its iscontinuously decreasing. So angle swept by Q is less than angle swept bydisc. So it will fall in unshaded portion.

6.

vr = |2 v sin )| = |2v sin t)|

PART - II1. For a particle moving in a circle with constant angular speed, velocity vector is always tangent to the circle

and the acceleration vector always points towards the centre of circle or is always point towards the centreof circle or is always along radius of the circle. Since, tangential vector is perpendicular to radial vector,therefore, velocity vector will be perpendicular to the acceleration vector. But in no case acceleration vectoris tangent to the circle

2. When a force of constant magnitude acts on velocity of particle perpendicularly, then there is no change inthe kinetic energy of particle. Hence, kinetic energy remains constant.

3. S = t3 + 5Linear speed of the particle

= dtdS

= 3 t2 at t = 2 s v = (3 × 22) m/s = 12 m/s

Linear acceleration a1 = dtd

= 6 t at t = 2 s, a1 = 12 m/s2

The centripetal acceleration

a2 = R

2

= 20

122

m/s2 = 7.2 m/s2

anet = 22

21 aa = 22 2.712 = 14 m/s2

4. aC = �RV2

cos i� �RV2

sin j�

5. They have same .centripetal acceleration = 2r

2

1

aa

= 2

1

22

12

r

r

r

r

RESONANCE SOLN_Centre of Mass - 96

TOPIC : CENTRE OF MASS

EXERCISE-1 SECTION (A)

A 1. xcm

= 127

32160cos131201

ycm

= 43

1233

660sin130201

r =

22

43

127

=

619

12192

14476

163

14449

m

A 3.43

A1 = M, A =

34

M

xcm

= 2mmmxmx

1

2211

=

M

3/M�2a3

M34

a

= 3a4

+2a�

xcm

= a

2x33�8

= a65

Similarly ; ycm

= a65

Ans.

A 5. M1 = (2R)2 × M

2 = (R)2 × x

1 = 0, x

2 = R

So Xcm

= 21

2211mm

xmxm

Xcm

=

22

22

RR4

RR0R4 =

5R

towards smaller disc

A 8. length of the shaded region = 2y = 2kx2

dm = 2y dx ×

dm = 2kx2 × dx

M = a

o

2a

odxkx2dm =

3ak2 3

Xcm

=

3a

k2

4a

k2

dm

dxkx2

dm

dmx

3

4

a

0

a

0

3

a

0

a

0 Xcm

= 4a3

By symmetry the y-coordinate of the shown plate is zero.

SECTION (B)

B 2. 1 = 40

x30710 get x = � 1 cm

B 3. Xcm

= g

cossinu2 2

= 10

2

1

2

120202

= 40 m.

x1 =

240

= 20 m xcm

= 21

2211

mmxmxm

40 =

m2xm20m 2

get x2 = 60 m


B 5. So, ms × R = (40 + 60) × x

g

cossinu21 2

= 100 x get x = 0.1 m

B - 7 initially Ycom

= mM

Mh

since no external force is acting COM should be at rest.

yCM

= 21

2211

mmymym

Let baloon descend by a distance x.

0 = Mm

)h�x(M)x(m

Mh = (m + M) x

x = Mm

Mh

(Distance decend by ballon)

h � x = Mm

mh

(Distance raised by man)

SECTION (C)C 1. 238 × 0 = 4 × 1.17 × 107 + 234 × v2

V2 = �2 × 105 m/ses

C 3. (a) P1 = 2.4 × 10�26 kg�m/sec.

P2 = 7.0 × 10�27 kg�m/sec

P1 + P2 + P3 = 0 P3 = � (24 ×10�27 + 7.0 × 10�27)P3 = 31 × 10�27

V3 = 27�

27�

1067.1

1031

= 18.6 m/sec.

(b) Pe = 2.4 × 10�26 i�

P

an = 7.0 × 10�27 j�

P

p = �(P

e + P

an) = � (24 × 10�27 i� + 7.0 × 10�27 j� )

PP

= 22 )0.7()24( ×10�27 Vp = p

p

m

|P|

= 15.0 m/sec.

C 4. P1 =20 × 20 i� P3 = 40 × 20 k� P2 = 30 × 20 j�

Pi = P1 + P2 + P3 = Pf 400i +600 j� + 800 k� = 30 (10i + 20 k� ) + 40v

get v = 40

k�200j�600i100 = 2.5 i� + 15 j� + 5 k� Ans.

C 6. mc = 20 kg mT = 180 kg

Pi = 200 × 36 × 185

= 2000 kg m/sec

just before jump Vbg = bTV + TgV = (10 + VT)

So MTVT + mc Vc = Pf = Pi180VT + 20 × (10 + VT) = 2000

VT = 200

1800 = 9 m/sec. Time taken to cover 10 m t =

1010

= 1 sec.

distance covered by trolly = 9 × 1 = 9 m.


C 8. Net ext force = 0 F = dtdp

= 0 , p = constt

COM remain at rest

mg R = 21

mv2 + 21

MV2

mv = MV

v = m

MV or V =

Mmv

mgR = 21

× mv2 + 21

M 2

22

M

vm2mgR = 2

2

vM

mmM 2gR

mMM

= v2

Mm

1

gR2

= v

SECTION (D)

D 1. Energy Conservation

Total change in length of spring = 2x { 21 2

extkx = 2compkx

21

}

Time is sameno external force centre of mass is at rest

hence m1x

1 = m

2x

2

2

1

xx

= 1

2

mm

& x1 + x

2 = 2d

or, m1x

1 = m

2x

2 & x

1 + x

2 = 2d

x1=

1

22

mxm

1

22

mxm

+ x2 = 2d x

2

1

mm

1

2 = 2d

x2

1

12

mmm

= 2d x2 =

21

1

mmdm2

& x1 =

21

2

mmdm2

D-3. By momentum conservation

mAVA = mBVA (i) VB = B

AAm

Vm

K.EA = 2Vm

m2Vm

m2P 2

AA

A

2A

2A

A

2 ...(i)

Similiarly K.EB = 2

Vm 2BB ...(iii)

dividing (ii) by (iii) we get.

B

AE.KE.K

= 2BB

2AA

Vm

Vmput VB =

B

AAm

Vmwe get

B

AE.KE.K

= a

Bmm

. hence proved.

SECTION (E)

E 2. Pi = 200 × 10�3 ( j��i3 )

Pf = 200 × 10�3 ( j�i3 )

|Pi| = |Pf|P = |Pf| � |Pi| = 0

|P| = |Pf � Pi| = |(200 × 10�3 3 i � 200 × 10�3 j� ) � (200 × 10�3 3 i + 200 × 10�3 j� )|

|P| = |2 × 200 × 10�3 j� | = 0.4 kg m/sec.


E 4. v = gh2 = 4102 = 80

(a) J = P = 2mv = 2 × 21

× 80

J = 54 N-s

(b) N dt = dP N × 2 × 10�3 = 54

N = 52 ×103N.

SECTION (F)F 1. from momentum conservation

mu + 0 = (m + m) v v = 2u

from energy conservation P.E. = 21

mu2 �

21

2m 2

2

u

21

mu2 = K P.E. =

2K

F 3.

After first collisionAfter collision of B from wallvB = � v + 2 × 0 = �v

so

F 5. Particle B is a restmv + 0 = mv1 + 2mv2 v = v1 + 2v2 .....(i)

0vv�v 12

= 1

v2 � v1 = v .....(ii)Adding (i) + (ii)3v2 = 2v

v2 = 32

v v1 = v2 � v = 3v�

Now, (iii) + (iv)

t = 12 vv

r2

=

3v

3v2

r2

t =

vr2

Ans.

SECTION (G)G 1. m

0 = 20 kg ; m = 180 kg.

� Fth = (m +M)g = 2×103 N

� Fth = v

r

dtdm

So dtdm

= 3106.1

2000

= 1.25 kg/s. Ans.

v = vr n

mm0

� gt.


(i) t1 = dt/dm

m =

2180

= 90 s.

v1 = 1600

n

20200

� 10 × 90 v1 = 2.784 km/s. Ans.

(ii) t2 = dt/dm

M =

20180

= 9 s.

v2 = 1600

n

20200

� 10 × 9 v2 = 3.59 km/s.


A-2. A1 = R2 A2 = 16R2

x1 = 0 x2 = 4R3

xcen = 20R

�

16R

�R

4R3

16R

�0

22

2

A-4. A1 = 2r × r = 2r2 A2 = 2r2

x1 = 2r

x2 = 3r4

xcm =

2r

�r2

3r

2r

�2r

r2

22

22

= ]�4[3r2

2�4

r

32

�1r

2

3

A-9. ycm = 081

× 0.14 + 87

×h = 0

8h7

= �814.0

h = �0.02 below x-axis.

SECTION (B)

B-2. vcm = 21

2211

mmvmvm

vcm =

m2)j�2(m)i�2(m

acm = m2

)0(m)ji(m .

vcm has same direction as of acm straight line.

B-3. a = mnm

)m�nm(

g = )1n()1�n(

g

a1 = a2 = a

acm = )mnm(ma�nma 21

= a

)1n()1�n(

acm = g)1n(

)1�n(2

2

.


B 9.

a cm =

21

2211

m ma m a m

= m) (m

am 0 m

= 2a

given m1 = m

2 = m

1a = 0

2a =

a

SECTION (C)

C-3. mv i� + mv j� + 2m v

3 = 0

3v

= 2

)j�vi�v(�

= �

2v

( i� + j� ) = �2

v. kf =

21

mv2 + 21

mv2 + 21

2m2

v2.

kf = 2

mv3 2.

C-4. 500 × 10 = 550 × v v = 55

500 = s/m

11100

.

C 6. Vcom = V cos

V cos = m2

mv0m� 2

v2 = 2V cos

C 8. vM

= m. 0 + (M � m) ´v

´v

= )mM(

vM

SECTION (D)

D 1. Pi = mv1 + mv2 Pf = (m + M) v Pi = Pf v = )Mm(Mvmv 21

By energy consarvation

21

mv12 +

21

Mv22 =

21

(M + m) v2 + 21

kx2 mv12 + Mv2

2 = (M + m) 2

2

221 kx

)mM(

)Mvmv(

solving x = (v1 �v2) k)mM(mM

.

SECTION (E)

E 1. v1 = gh2 = 10102 = 210

k2 = 41

k1 v22 =

41

v12

v2 = 2v1 = 25

|P| = |�mv2 � (mv1)| = m |�v2 � v1|

|P| = 50 × 10�3 × 23

× 210 = 2

1015 1�

J = P = 1.05N-s.


E 3. From momentum conservation

mu = 2mv v = 2u

from energy conservation

21

× 2m ×

2

2

u

= 2 mgh h = g8

u2

SECTION (F)

F 5. 0.05 × vp + m × 0 = 5.05 v

i

f

vv

= 505.0

= 10�2 2

i

2f

)v(m21

)v(m21

= (10�2)2 = 10�4.

F-6. gh2m1 + 0 = (m1 + m2) v

v = )mm(

gh2m

21

1

v2 � u2 + 2g × 9h

= 6 + 2g × 4h

= 2

gh

v = 2gh

Also, 2gh

= 21 mm

gh2m

2m1 + m1 + m2

1mm

2

1 .

F-8. MA = × 34r3 e =

21

MB = × 34 (2r)3 = 8MA

mA v + 0 = mAv1 + mBv2 .........(i)ev = v2 � v1 .........(ii)

Adding (i) + (ii) = 9v2 = v + 2v

= 2v3

v1 = v2 � 2v

= 6v�

2v

= � 3v

. 2

1

vv

= 6/v3/v

= 2.

F-9. V2 = Z

0

Vel. of Sep = Vel of approach ( elastic) 20 + 5 = V � 5

V = 30 m/s Ans.vb = �(v0 + 2v) m1 > > m2vb = �(20 + 10) = �30 m/sec.

F-11. t = 0vd2

(time for succeesive collision)

N × t = dP = mv0 � (�mv0)

N × 0vd2

= 2mv0

N = d

mv20


F-14. If mass = mfirst ball will stop v = 0so K.E. = 0 (min) (K.E. can't be negative )

SECTION (G)

G 1. F = dtdm

210 = 300 × dtdm

dtdm

= 0.7 kg/s.

EXERCISE-2PART - I

1. T = gH2

= 10

802 = 16 = 4 s. t =

gh2

= )6080(102

= 1040

= 2 s.

t = T � t = 4 � 2 = 2 s V = t

d

= 22

= 1 m/s

Mv = mv v = 2

160 = 30 m/s Ans.

R = vt = 30 × 2 = 60 m Ans.

4.

By momentum conservationMu = mV (i)

V = mM

u

By energy conservation

mgh = 21

Mu2 + 21

mV2 = 21

Mu2 + 21

m 2

mM

u2

mgh = 21

Mu2 + 21

m

M2 u2 =

21

u2

mMMm 2

2m2gh = u2 (Mm + M2)

2

2

MMm

ghm2

= u2.

u = m 2MMm

gh2

...(ii)

By momentum conservationmV = (M + m) V

1

V1 =

mMmV

...(iii)

By energy conservation

21

mV2 = 21

(m + M) V1

2 + mgh1

21

mV2 = 21

(m + M) 2

mMmV

+ mgh

1

21

mV2 � 21

)mM(

Vm 22

= mgh

1


21

V2

mMm

�m2

= mgh1

21

)mM(

MmV2

= mgh

1...(iii)

Put V = �

mM

× u and u = m 2MMm

gh2

V = M 2MMm

gh2

....(iv)

put value of V from eqn (iv) to (iii) h' = 2

2

)mM(

hM

6.

By mechancnical energy conservation

2r

2b mV

21

V)m2(21

= 2 mgl

2 Vb

2 + Vr2 = 4gl ....(1)

using momentum conservationmV

r = 2 mV

bV

r = 2 V

b....(2)

2Vb

2 + 4Vb

2 = 4 gl 6Vb

2 = 4gl Vb = gl

32

(a) Vr = 2 V

b = gl

32

2

when string be comes vertical velocity of block wrt to string.

Vbr

= Vb � (� V

r) = 3V

b = gl

32

3

(b) T � 2 mg =

2br )V(m2

T = 2 mg + 3

m)2(gl29 = 14 mg

8. m = 20 × 10�3 kg ; M = 5 kgu = 400 d = 0.2 mV = 200 = ?P

Bullet = P

Block

m (u � v) = 20×10�3 (400 � 200) = 4 kg. m/s.

KEBlock

= M2

P2

= 52

42

= 1.6 J = Mgd

= Mgd6.1

= 2.0105

6.1

= 0.16 Ans.

10. string will taut when A waves a distance of (.7 � .25) m

at that Pt VA = gh2 and Now B starts on using with same velocity as A.

let us suppose it is u.

dtT = mB u + 0 ....(1)

� dtT = mA (u �V

A) ....(2)

from (1) and (2) u = BA

AAmm

Vm

= 56

56

3Tdt = 3.6 m/s


12. R1 = V cos T

1

R2 = V cos T

2

R3 = V cos T

3

R = R1 + R

2 + R

3 = u cos [T

1 +T

2 + T

3]

= V cos

gsinue2

gsineu2

gsinu2 2

R = g2sinV)ee1( 22

14. V2 = U2 + 2as 0 = (V)2 � 2as s = a2)V( 2

(a) e = 1 so after collision VA = 0 and V

B = 5 m/sec

So mg = ma s = 102.02

)5( 2

s = 6.25 m

(b) when e = 0 applying momentum conservationm × 5 + 0 = (m + m) × V V = 2.5m/sec

so V2 = u2 + 2as s = 102.022)5.2( 2

a =

2g

s = 3.12 m.

16. Px = 5×2 = 10 ; P = P

xi� + P

yj�

Py = 10 3 = 10 i� + 10 3 j�

= (5 + 10) V

= 1510

j�3i�

V = 32

j�3i�

V = 34

m/s. Ans.

H = E = Ei � E

f =

2

22211 um

21

vm21

� 21

(m1+m

2) V2

= 21

× 5 × 22 + 21

×10 × 23 � 21

× (10+5) (4/3)2 = 25 � 340

= 3

35Ans.

18. (a) V = gl2

In x dir

2m × 2

v = 3 mV v =

2

3 V = gl2

2

3 v = gl3 Ans.

(b) V = )Cos�1(gl2�V2

= glCos2gl2�gl2

For = 60° V = gl

Vx = gl Cos 60° (at heighest point) V

x =

2

gl

20. Applying momentum conservation in horizontal direction mV

0 = Mu M = 2m

u = 2

VM

mV 00

Eqn of e along normal

e =

sinVsinucosV

0 =

sinV

sin2

VcosV

0

0

e = 0V

V cot +

21

...(i)


Along incline surface of wedge friction is negligible so change in momentummV

0 cos = mV sin

0VV

= cot ...(ii)

Put value of (ii) in (i)

e = cot2 + 21

given tan = 2 = 43

21

41

Ans.

(b) h = (ut) tan

By (2)nd eq. of motion � h = Vt � 21

gt2 � (ut) tan = Vt � 21

gt2

or � u tan = V � 21

gt 21

gt = V + u tan

t = g2

(V0 cot +

2V0 tan ) t = g

2 V

0 tan (cot2 +

21

)

t = gtanV2 0

(e) = gtanVe2 0

substituting values : 10

21043

2

= 3sec

22. (a) At highest pointV = 50 cos ...(i)After striking bullet get embedded with bobso by momentum conservation.

MV = 4Mu u = 4V

....(ii)

from (i) u = 4cos50

By energy conservation after collission

21

(4m) u2 = 4mg (1 + cos 60º)

2

4cos50

21

=

21

13

1010

16cos5050 2

= 100 cos2 =

2516

cos = 54

= 37º

(b) Max height y = g2

)º37sin50( 2 =

53

53

5050201

= 5 × 9 = 45

(c) x = 102º37cosº37sin250

g2sinu

21

2R 22

= 120 m Ans (a) = 37º (b) x = 120 m and y = 45 m

PART - II1. COM can lie anywhere, within or at the radius r.

3. Since no external force is acting on the system hence VCM

remain constant.

5. when cylinder reaches pt B.then block get shifted by x but since than there is no extforce therefore com remain at its position[(R�r) � x]m = Mx

x = mM

)r�R(m

8. Pi = 0 ...(i) Pf = MV � mV1 ....(ii)

MV � mV1 = 0 u = mM

V..

using V12 = u2 + 2ax.

a = g.2

mMV

= 0 + 2g x. x =

gm2

VM2

22


10. Taking the origin at the centre of the plank.m1x1 + m2 x2 + m3x3 = 0( xCM = 0)(Assuming the centres of the two menare exactly at the axis shown.)60(0) + 40(60) + 40 (�x) = 0 ,

x is the displacement of the block. x = 60 cm

i.e. A & B meet at the right end of the plank.

12. yCM = 0

yCM = 1y4m

+ 2y4m3

y1 = + 15

y2 = �5 cm

14. I. Since velocity of both R and S is positive they will move in same direction.II. At mid point velocities of R and S are same.III. Change in velocity of R is small as compare to change in velocity of S. But change in momentum issame for both in magnitude. Hence mass of R should be greater than S.Hence all three are correct.

16. If we treat the train as a half ring of mass 'M' then its COM will be at a distance

R2 from the centre of

the circle. Velocity of centre of mass is :VCM = RCM .

=

R2. =

R

V.

R2( =

RV

)

VCM =

V2 MVCM =

MV2

As the linear momentum of any system = MVCM

The linear momentum of the train =

MV2Ans.

18. I = f × t and F = t

)gh2hg2(m 12

F = 01.0

)5.28.92625.08.92(10100 3�

F = 105 N

20. Using momentum conservation

0pppp 4321

4321 p�p�p�p

24

23

221 pppp

K. E1 = m2

ppp

m2

p 24

23

22

21 = E0 + E0 + E0

Total energy = 3E0 + E0 + E0 + E0 = 6E0

22. e =

cosgh2

sinv

apply conservation of momentum

m gh2 sin = m vcos ......(i)

e gh2 cos × m = mv cos ......(ii)

etan

= cot.

e = tan2 on solving


25. m2vcos = 3vy

cosv

vy =

32

Also e = cosv

vy =

32

.

27. sin = R

2/R; = 30º

Both have equal mass it means alongLOI particle transfer it velocity to disc which is vcos.

so VD = Vcos = Vcos 30º = 2V3

29. mcr vv

vr = mv � cv = v � u = 0.

since vr = 0 so Ft = dt

vrdm = 0.

Fnet = mdtdv

F + 0 = (m0 � t) dtdv

F = (m0�t) dtdv

.

30. Neglecting gravity,

v = un

t

0

m

m;

u = ejection velocity w.r.t. balloon. m0 = initial mass mt = mass at any time t.

= 2n

2/m

m

0

0 = 2n2.

34. mv = nvm v = nv

time for first collisen is t1 = VL

(2nd block)

2nd collisions t2 = V2

= 2t1 (3rd block)

so t = t1 + 2t1 + 3t1 + at1 ...........(n�1) t1.t = t1 [1 + 2 + 3] .......................(n�1)]

= 2

)11�n()1�n( =

2)1�n(n

so t = V2L

n (n � 1).

36. a = mf

for elastic collission e = 1

v12 = 0 + 2ad

vb12 = d.

mF2

vb1 = mFd2

after collisin vb2 = 0.


39.

Pi = mv (i) Pf = (m + m) vat maximum conservationPi = Pf v' = v/2By energy compression

21

mv2 + 0 = 21

(2m) (v)2 + 21

kx2 kx2 = 2

mv2 x = v

k2m

.

at maximum compression k = 21

× 2m × 4v2

k = mv'2 = mv2/4.

EXERCISE-31. (A) If velocity of block A is zero, from conservation of momentum, speed of block B is 2u. Then K.E. of

block B = 2

1m(2u)2 = 2mu2 is greater than net mechanical energy of system. Since this is not possible,

velocity of A can never be zero.(B) Since initial velocity of B is zero, it shall be zero for many other instants of time.(C) Since momentum of system is non-zero, K.E. of system cannot be zero. Also KE of system isminimum at maximum extension of spring.(D) The potential energy of spring shall be zero whenever it comes to natural length. Also P.E. of springis maximum at maximum extension of spring.

2. (A) Initial velocity of centre of mass of given system is zero and net external force is in verticaldirection. Since there is shift of mass downward, the centre of mass has only downward shift.

(B) Obviously there is shift of centre of mass of given system downwards. Also the pulley exerts aforce on string which has a horizontal component towards right. Hence centre of mass ofsystem has a rightward shift.

(C) Both block and monkey moves up, hence centre of mass of given system shifts verticallyupwards.

(D) Net external force on given system is zero. Hence centre of mass of given system remains atrest.

3. (a) The acceleration of the centre of mass is

aCOM

= m2F

The displacement of the centre of mass at time t will be

x = 21

aCOM

t2 = m4

Ft2

Ans.

4 & 5Suppose the displacement of the first block is x

1 and that of the second is x

2. Then,

x = m2mxmx 21 or, 2

xxm4

Ft 212 or, x

1 + x

2 = m2

Ft2

...(i)

Further, the extension of the spring is x1 � x

2. Therefore,

x1 � x

2 = x

0...(ii)

From Eqs. (i) and (ii), x1 =

21

0

2

xm2

Ftand x

2 =

21

0

2

xm2

Ft


6. During collision, forces act along line of impact. As collision is elastic and both the balls have samemass, velocities are exchanged along the line of impact. Therefore ball B moves with velocity VB||, thatis equal to u cos 30°. Ball A moves perpendicular to the line of impact with velocity VA = u cos60°.

Along the line of impact, ball A does not have any velocity after the collision.Therefore velocity of ball A in vector form after the collision

30°

30°

60°

R

u

VA

VA||

VB||

x

y

= VA cos60°i + VA cos 30°j

= (u cos 60°) cos60°i + (u cos 60°) cos 30°j

= i.21

.21

.4 + j.23

.21

.4 = )j3i( m/s

7. Using impulse-momentum equation for ball B x

VB||

B

dtNif ppdtN

and as 0pi

fpdtN

= (mu cos 30°) cos 30 i � (mu cos30°) cos 60° j

= i.23

.23

.4.m � j.21

.23

.4.m = (3 m i � 3 m j) kg sm

8. Suppose V2 is velocity of ball B along the line of impact and V1 is velocity of ball A along the line ofimpact, after the collision, as shown.

Then 21

(Velocity of approach) = Velocity of separation

u.

23

21

= V2 � V1 .... (1)

B

V1A

V2

Conserving momentum along the line of impact

m. 23

u = m. V2 + mV1 .... (2)

Solving and using u = 4 m/s

V2 = 233

m/s j60cos233

i30cos233

V2

=

j

433

i49

m/s

EXERCISE-4PART - I

1. vCOM

= 21

2211

mmvmvm

= 410

041410

= 10 m/s.


2. Angular speed of particle about centre of the circle,

= Rv2

, = t = Rv2

t

pv

= (� v2 sin i� + v

2 cos j� ) or pv

=

j�t

Rv

cosvi�tRv

sinv 22

22

and mv

= v1 j�

linear momentum of particle w.r.t. man as a function of time is

pmL

= )V�V( mp

= m

j�vt

Rv

cosvi�tRv

sinv 12

22

2

3. (i) X1 = V

0 t � A (1� cost)

Xcm

= 21

2211

mmxmxm

= V

0 t X

2 =

0 t +

2

1

mm

A (1�cos t) Ans.

(ii) a1 = 2

12

dt

xd = � 2 A cos t

The separation X2 � X

1 between the two blocks will be equal to

0 when a

1 = 0 or cos t = 0

x2 � x

1 =

2

1

m

mA (1�cos t) + A (1� cos t)

0 =

1

m

m

2

1A (cos t = 0)

Thus the relation between 0 and A is,

0 =

1

m

m

2

1 AA

8. According to Newton�s Law

e = 21

12

uuvv

For elastic collision cofficient of restitution e = 1 so

12 vv

= 21 uu

Statement - 1 is correct

Linear momentum is conserved in both elastic & non elastic collision but it�s not the explanation of

statement -1 so it is not the correct explanation of the statement A.

9. i�pP1

i�pP2

as there is no external force so momentum will remain conserved

2121 PP'PP

0PP 21

Now from option

(A) 21 PP

= k�cj�)bb(i�)aa( 12121

(B) 21 PP

= k�)cc( 21

(C) 21 PP

= j�)bb(i�)aa( 2121

(D) 21 PP

= j�b2i�)aa( 121

and it is given that a1 b

1 c

1 , a

2, b

2, c

2, 0

in case of A and D it is not possible to get 21 PP

= 0

Hence Ans. (A) and (D)


10. At point B there is perfectly inelastic collision so componentof velocity to incline plane becomes zero and componentparallel to second surface is retainedvelocity immediately after it strikes second incline

V = 30cosgh2 = 3102 × 23

= 4

9102

V = 45 m/s

11. At point �C�

gh2VV 2B

2C

VC

2 = 45 + 2 × 10 × 3

VC = 105 m/s

12. The block coming down from incline AB makes an angle 30° with incline BC. If the block collides with

incline BC elastically, the angle of block after collision with the incline shall be 30°.

Hence just after collision with incline BC the velocity of block shall be horizontal. So immediately afterthe block strikes second inclined, its vertical component of velocity will be zero.

13. ycm

= 54321

5544332211mmmmm

ymymymymym

ycm

= m6mmmm

)a(�m)a(m)0(m)a(m)0(m6

= 10a

.

14. Since masses of particles are equal and collisons are elastic, soparticles will exchange velocities after each collision. The firstcollision will be at a point P and second at point Q again andbefore third collision the particles will reach at A.

15. from momentum conservation :9m = (2m) V

1 � (m)V

2

9 = 2V1 � V

2..... (1)

e = 19

VV 21

......(2)

from eqn(1) and eqn(2) V1 = 6 m/sec.

for second collision between second block and third block :(2m) 6 + m(0) = (2m + m) V

C V

C = 4 m/sec.

16*. Since collision is elastic, so e = 1Velocity of approach = velocity of separationSo, u = v + 2 .............(i)By momentum conservation :

1 × u = 5v � 1 × 2

u = 5v � 2

v + 2 = 5v � 2

So, v = 1 m/sand u = 3 m/s


Momentum of system = 1 × 3 = 3 kgm/s

Momentum of 5kg after collision = 5 × 1 = 5 kgm/s

So, kinetic energy of centre of mass = 21

(m1 + m2)

2

21

1

mm

um

=

21

(1 + 5)

2

631

= 0.75 J

Total kinetic energy = 21

× 1 × 32 = 4.5 J.

17. R = gh2

u 20 = 10

52V1

and 100 =

1052

V2

V1 = 20 m/s , V

2 = 100 m/sec.

Applying momentum conservation just before and just after the collision(0.01) (V) = (0.2)(20) + (0.01)(100) V = 500 m/s

18. = 0.1 2mu

21

= mg × 0.06 + 21

kx2

21

× 0.18 u2 = 0.1 × 0.18 × 10 × 0.06

0.4 = 10N

N = 4 Ans.

PART - II

14. If initial momentum of particles is zero, then they loss all their energy in inelastic collision but here initialmomentum is not zero.Principle of conservation of momentum holds good for all collision.

RESONANCE SOLN_RIGID BODY DYNAMICS - 114

TOPIC : RIGID BODY DYNAMICS

EXERCISE-1PART - I

SECTION (A)A 1.

i = 0 t = 5 sec = 50 (2) rad.

= it + 21

t2

(50) (2) = 0 + 21

(5)2

(50) (2) = 0 + 2

25

= 25

)2)(2)(50( = 4 (2) = 4 rev/ se2

f = i + tf = 0 + 4(5) = 20 rev/ sec

SECTION (B)

B 1.

For first solid spheresIAB = Iam + Md2

IAB = 52

MR2 + MR2 =

2MR5

7

Similar way for second sphere

IAB = 57

MR2

I = 2 IAB =

2MR5

14

B 4. I0 =

2mR2

0 = cm + md2

2mR2

= cm + m

2

3R4

cm = 2

mR2

� m

2

3R4

ICM

=

22

3R4

M2

MR


SECTION (C)

C 2.

1F = 2 i� � 5 j� � 6k� at point

2F = � i�+ 2 j� � k� at point

r0(�1,0,5)

1r = (1 i� + j� + 0 k� ) � (� i� + 0 j� + k� )

`

1r = k��j�i�2

1 =

11 Fr = k��j�i�2 × k�6�j�5�i�2

1 = ( � 10 k� + 12 j� � 2k� � 6 i� � 2 j� � 5 i� )

1 = (�111 i� + 10 j� � 12 k� )

2 =

22 Fr = k�j�i� × k��j�2i��

Total

T =

1 +

2

T = k��j�i�2 × k�6�j�5�i�2 + k�j�i� × k��j�2i��

T = k�9�j�10i�14�

C 4. (a) 0 = mg R/2 = mg

g22sinv2

0 = mg g2

2sinv2

=

22sinmv2

0 = (mv2 sincos)

(b)

0 = mgR

= (2mv2 sincos)


SECTION (D)D 2. The F.B.D. of rod is as shown

For rod to be in translational equilibriumN

1 = P ....(1)

N2 = W = mg ....(2)

For rod to be in rotational equilibrium, net torque on rodabout any axis is zero.

Net torque on rod about B is zeroi.e.,

mg 2

cos � N2 cos + P sin = 0 .......(3)

from equation (2) and (3) solving we get

P = 2

mg cot

D 3. For translational equilibriumFx = 0 Fy = 0N1 = f N2 = 75g + 24g = 99g = 990 NRotational equilibrium = 0 (about any point)

B = 0

N1 × 6 = 24g (5cos 37º) + 75g (8cos 37º)

N1 × 6 = 24g (5 × 54

) + 75g (8 × 54

)

N1 × 6 = (96g + 480 g)

N1 = 96g = 960 Nf = N1N2 = N1

= 2

1

NN

= g99g96

= 3332

Ans. 990 N, 960 N , 3332

SECTION (E)

E 1. (a) Torque about hinge

(m1g � m2g)

2

= .

= 2

2

2

1

21

2m

2m

)2/(g)mm(

= )mm(g)mm(2

21

21

= )36(210)36(2

=

310

rad/sec2.


(b) If mass of rod is 3 Kg Torque about hinge

(m1g � m2g)2

= ''

' =

12m

2m

2m

2 g)mm(

23

2

2

2

1

21

' =

3m

mm

g)mm(2

321

21

=

33

362

10)36(2= 3 rad/s2

For m1 blockm1g � T1 = m1a

T1 =

2

mgm 1

1

T1 = 60 � 2

326 = 42 N

For m2 blockT2 � m2g = m2a

T2 = m2g + m2 2

= 30 + 2

323 T2 = 39 N

E 3. Let be the angular acceleration of the pulley system.For 6 kg block6 g � T

1 = 6 (2) .........(i)

for 3 kg blockT

2 � 3g = 3 .........(ii)

for pulley system 2T

1 � T

2 = = 3 .........(iii)

From equation (i) and (ii) putting the values of T1 and T

2.

2[6g � 12] � [3g + 3] = 3 12 g � 24 � 3g � 3 = 3 30 � 9g

= 3090

= 3 rad/s2 Ans.

SECTION (F)

F 2. initial positial final position

Using Energy conservationKi + Ui = Kf + Uf

0 + 3mg 2

= 21

2

= (1 + 2)

[ = 3

m 2

+ m2]


3mg 2

= 21

22

m3

m

2 + 0 3g=

22

3

2

3g = 34

2 = 4g9

F-4. Initial and final positions are shown below

Decrease in potential energy of mass �m � = mg

4R5

2 = 2

mgR5

Decrease in potential energy of disc = mg

4R

2 = 2

mgR

Therefore, total decrease in potential energy of system

= 2

mgR5+

2mgR

= 3 mgR

Gain in kinetic energy of system = 212

Where = moment of inertia of system ( disc + mass ) about axis PQ.= moment of inertia of disc + moment of inertia of mass

=

22

4R

m4

mR+ m

2

4R5

=

8mR15 2

From conservation of mechanical energy -Decrease in potential energy = Gain in kinetic energy

3 mgR = 21

8mR15 2

2 = R5g16

Therefore, linear speed of particle at its lowest point

v =

4

R5 =

4R5

R5g16

or v = gR5

SECTION (G)

G 2. 3x + 4y = 5

45

4x3�

y

P = mv= 2 × 8 = 16 (kg � m/s)

L = (5/4) × mv cos 37º

L = 5/4 × 2 × 8 × 54

= 16 kg m2/s


G 3.

initial position Final position

No external torque so

L = cont.

Li = Lf(i0 = f0)

4mr

4mr 22

0 = ( + mr2 + mr2)

02

2

mr22

mr

G 5. Angular momentum conservation about Ov

mR

M o

= mvR

2MR2

= mvR

MR = 2mv

v =

m2MR

With respect to bord man's rotation v + R velocity so in one rotationwhen velocity v + R angle taken by man (2).

RV

R2t

Angular velocity bord is so at the same time angle covered by disc = .

R

VR2

t .

m2Mm4

Rm2

MRR2

SECTION (H)H 1 VA = (Vcm � /2)

= 50 � 5 × 5 = 25 m/s

VB =

2Vcm

= 50 + 25 = 75 m/s

H 4 (a) vA sin = v

0 cos

vA =

tanv0 =

3v4 0

(b) =

cosvsinv A0 = 5

3v4

4v3 00

= 3

v515

v16v9 000

(c) vx =

2

BxAx vv =

2v0

vy = ByAy vv

21

= 3

v2 0


SECTION (I)I-3. For linear motion :

mg � T = ma ............(i)For angular motion :

T.R. =

2mR2

T = 2

mR............(ii)

For no sliping :a = R ............(iii)

From equation (i), (ii) & (iii)

a = 32

g.

I-4. Let R & r be the radii of hemispherical bowl & disc respectivelyFrom energy conservation,

mg(R � r) = 21

mv2 + 212

For pure rolling,v = r

mg(R � r) = 21

mv2 +

22

rv

mr21

21

mg(R � r) = 43

mv2 ...........(i)

From FBD of bottom :

N � mg = )rR(

mv 2

...........(ii)

From equ. (i) & (ii),

N = 37

mg.

I-5. Let v1 & v

2 be minimum speed of ring of bottom & top of cylindrical part

At top of path

N + mg = )r�R(

mv22

for minimum speed N = 0v

22 = g (R � r) .......... (i)

From energy conservation between bottom & top point of cylindrical part

21

mv1

2 + 21

1

2 = 2 mg (R � r) + 21

mv2

2 + 21

22

For pure rolling 1 =

rv1 ,

2 =

rv2

21

mv1

2 + 21

(mr2) 2

21

r

v = 2 mg (R � r) +

21

mv2

2 + 21

(mr2) 2

22

r

v

mv1

2 = 2 mg (R � r) + mv2

2 .......... (ii)from equation (i) & (ii) mv

12 = 2 mg (R � r) + mg (R � r)

v1 = )r�R(g3


I-6. For linear motion,F = ma ..........(i)

For angular motion,

F.R. =

mR

5

2 2

= mR2F5

..........(ii)

= 0t +

21

at2

2 = 0 +21

mR2

F5 t2 t2 =

F5

mR8

Distance covered by sphere during one full rotation

S = ut +21

at2 =0 + 21

mF

F5mR8

S = 5R4

SECTION (J)J 2. (a) Pi = m2v

Pf = (m1 + m2) Vcmm2v = (m1+m2) Vcm

Vcm =

21

2

mm

vm

(b) v1 = (u � Vcm)

V1 = v � 21

2

mmum

=

21

1

mm

um

(c) V1 = � Vcm =

21

1

mm

um�

(d) Xcm = )mm(2L

m)0(m

21

21

= )mm(2

Lm

21

2

L1 = 2L

� )mm(2Lm

21

2

L1 =

21

21

1

mm

Lm

momentum of particle

Pi =

21

2

21

12

1cm2 mm

umu

)mm(2Lm

21

mL)Vu(m =

)mm(2umm

21

212

Momentum for rod = 221

221

cmcm1)mm(

um

2

LmLVm

(e) For particle :

1 = m2L2 = 2

21

2212

)mm(4Lmm

2 =

2

21

21

21

)mm(2Lm

m12

Lm

= 1 + 2 = )mm(12

L)m4m(m

21

2211


(f) Velocity of centre of mass

=

21

2

mm

vm

Using angular momentum conservationm2v × Lcm = cm

= )mm(2Lm

um21

12

= cm .

= )mm(2Lm

um21

12

= )mm(12

L)m4m(m

21

2211

× = L)m4m(

vm6

21

2

.

SECTION (K)

K 1. Force balanceN= mg cos f = mg sin Torque balance (about centre of mass)

Nx = f × 2a

= 2sinamg

and x =

cosmg2sinmga

= 2

tana

Torque of normal force Nx = mg sin 2a

PART - IISECTION (A)A 1. 0 = 3000 rad/min

0 = 60

3000 rad/sec = (50 rad/sec)

t = 10 secf = 0f = 0 + t = 50 � (10) = 5 rad/sec2

= o t + 21

t2

= (50) (10) + 21

(�10) (10)2

= 500 � 250 = 250 rad

A 3.* Sphere is rotating about a diameterso , a = Rbut, R is zero for particles on the diameter.

SECTION (B)

B 3. B > AB > Aso, If the axes are parallel

B 6. Moment of inertia of the elliptical disc should be less than that of a circular disc having radius equal to themajor axis of the elliptical disc.Hence (D)


B 7. 0 = 1 + 2

0 =

32

2/m2

+

32

2/m2

= 12

m 2

SECTION (C)

C 1.

1F = 2i + 3j + 4k

2F = �2i � 3j � 4k

Net force 0FFF 21net

the body is in translational equilibrium.

1r = 3i + 3J + 4k

2r = i

1 =

1r ×

1F = )k�4j�3i�3( × )k�4j�3i�2(

1 = i�12�j�8i�12j�6�j�12�k�9

1 = k�3j�4�

2 =

2r ×

2F = ( i� ) × ( k�4�j�3�i�2� )

= �3k� + 4 j�

0j�4k�3�k�3i�4�21

body in rotational equilibrium

C 3.

F = 2 i� + 3 j� �k� at point (2,�3,1)

torque about point (0, 0, 2)

r = k�j�3�i�2 � k�2

=

r ×

F = )k��j�3i�2()k��j�3�i�2(

= )k�12i�6(

= )56(

SECTION (D)D 2. N

1 = N

2 ,

N1 + N

2 = mg ,

A = o

3 N2 � 4 N

1 �

23

mg = o

Hence = 31

Ans.

AliterUsing force balancef1 = �N1 N1 + f2 = mg (1)f2 = N2 N2 = f1

N2 = N1 (2)Using aq (1)N1 + N2 = mgN1 + N1 = mg

N1 +

21

mg

torque about point B B = 0 For rotational equilibriumf1 × 4 + mg (5/2 cos 53º) = 3N1


4N1 + 2mg3

= 3N1 2mg3

= (3 � 4) N1

2mg3

= (3 � 4)

21

mg

23

=

21

4�3

3 + 32 = 6 � 8

32 + 8 � 3 = 0

32 + 9 � � 3 = 0

3( + 3) �1 ( + 3) ( = 1/3)

D 4 �x x

w1w

weight of object = w

w ( � x) = w1x ...........(i)If weight is kept in another pan then :w2( � x) = wx ...........(ii)By (i) & (ii)

2ww

= ww1

w2 = w1 w2

w = 21ww .

SECTION (E)

E-3.

N =

2m 2

E-4. Initial velocity of each point onthe rod is zero so angular velocity of rod is zero.Torque about O =

20g (0.8) = 3

m 2

20g (0.8) = 3

)6.1(20 2

2.3g3

= = angular acceleration

= 16

g15

SECTION (F)

F 2. By energy conservation

mg4

= 21

.22 m

487

[ (about O) =

22

4m

12m

]

0 = 487

ml 2 = 7g24

Ans.


SECTION (G)

G 3. x = v0 cos 45º × t = 2

tv0

= mgx = 2

tmgv0 =

dtdL

L = 2

mgv0

g/v

0

0

dtt = g22

mv30

G 5. external torque ext

= 0

11 = 22when he stretches his arms so 1 < 2then (1 > 2)so, (L = constant)

G 7.* External force will act at hinge so linear momentum of system will not remain const. but torque of external

force is zero about hinge so

L = const., collision is elastic so K.E = const.

SECTION (H)

H 3*. for pure rollingV = RVA = 2V

VB = 2 V

(VC = 0)

SECTION (I)-3. mg sin � f = ma

a =

m

f�sinmg.......(i)

a is same for each body.

f.R = 2mk

R.f

For solid sphere k2 = 52

R2 is minimum there fore is maximum hence, k.E. for solid sphere will be max

at bottom.

-5. mg sin � f = ma

a = m

f�sinmg

a is equal for each body so all the object will reach at same time.

-7. There is no relative motion between sphere and plank so friction force is zero then no any change in motionof sphere and plank.

SECTION (J)

J-2.* at the moment when ring is placed friction will act between them due to relative motion. Friction is internalforce between them so angular momentum of system is conserved.I11 = I22

2mR2

0 =

2

2mR

2mR

= 30


J-3. Conservation of angular momentum about C.O.M. of m and loop of mass m gives

2mVR

=

222

2R

m2R

mRm

V = 3 R = R3V

J-4.

velocity of COM after collision is V friction will act such that = o at some intant after some time (V = R)

SECTION (K)

K-2. For no slippingµmg cos mg sin .........(1)

For toppling

mg sin 2h mg cos.

2a

.........(2)

for minimum µ (by dividing)

µ.a2

= h2

µmin

= ha

.

[ Ans.: a/h ]Sol.(2) If f > mg sin

mg cos > mg sin ( > tan ) block will topple before slidingtorque about point A A =0

mg sin 2h = mg cos

2a

tan = ha

> ha

If > tan (block will slide)

K-3.

b/2

a

N

mg

a/2

The block will not topple if mg acts from within the base area of the block. So,

2b

cos2a

ab

cos


EXERCISE-2PART - I

3. linear density =

m

dm =

dx

m

AB

= 2x·dm =

o

2)45cosx(·dxm

m dx

2

x

0

2

= 2

m

3x3

0

2

6

m

4. dm = (2xdx)

= 2x·dm =

R

o

2x.)xdx2(

=

R

0

3 dxx2

=

R

0

3 dx)x·(x2 =

R

o

R

o

43 dxxdxx2

= 2

5R

4R 54

5.ry

Rh

r = hR

y

dm = (r2dy)

dAB

= 21

(dm) r2

AB

=

h

0y

22 rdyr21

= 2

4

4

h

R

5h5

= 10

R4 h .

hR31

m

2 ...........

hR31

m

2 =

103

mR2


7. =

2R

m

For small ring friction forcedfr = K(2rdr)gTorque of the friction = (�rdfr) = � 2krgr2dr

= �2krg R

0

2drr = � 32

krgR3

For rotation about z-axis( = )

� 32

krgR3 = 2

))(R( 2

R2 =

R3kg4�

From equation of motion =

0 + t

0 = 0 +

R3kg4�

t t =

kg4R3 0

8. =

m m

1 = x =

mx

(a = R) m1g � T = m

1a ............ (i)

T R =

2MR2

+ (m � m1) (R2) ............(ii)

T = 2

MR + (m � m

1) R T =

2Ma

+ (m � m1) a

m1g �

2Ma

� (m � m1) a = m

1a m

1g �

2Ma

� ma + m1a = m

1a

m1g =

ma

2Ma

= )m2M(am2 1

= R)m2M(

gmx

2

R)m2M(mgx2

10. Using energy conservation

mgh = 21

kx2 + 212 +

21

mv2

String does not slipSo(V = r)

mg x = 21

kx2 + 21

2

2

r

v +

21

mv2

x = 0.1m = 0.1 kg � m2 r = 0.1 m K = 100 N/mm = 11 kg

11 × 10 × 0.1 = 21

× 100 × (0.1)2 + 21

× 0.1 2

2

)1.0(

V +

21

× 11 × V2

22 = 1 + 10 V2 + 11 V2

21 V2 = 21V = 1 m/s


11. (a) Energy conservationloss in P.E. = gain in rotational K.E.

mg 2

(1 � cos ) = 21

3

m 2

2

2 =

g3 (1 � cos )

= )cos1(g3

(b) = mg 2

sin = 0

mg 2

sin = 3

m 2

= 2g3

sin =

4g3

sin

fy = ma

y

mg � N2 = ma

t sin N

2 = mg � ma

t sin = mg � m

4g3

sin2

N2 = mg

4sin3

12

fx = ma

x

N1 = ma

1 cos =

4g3m

sin . cos

Ans. Normal reaction = 22

21 NN

where N1 =

4mg3

sin cos N2 =

4sin3

1mg2

12. (a) About the axis of rotation of rod, the angular momentum of the system is conserved velocity of the flyingbullet is V

mv =

3M

m2

2

=

3M

m

mv =

M

mv3 (m <<< M) ................. (i)


conservation of mechanical energy of the system (rod + bullet)

21

3M

m2

2

2 = (M+m)g 2

(1� cos ) ��(ii)

From (i) and (ii)

V = mM

3g2

sin 2

(b) P = mv�2

M)(m

From v and w

P = 21

mv =

2sin6g

M

mvx =

2

2mx

3M

´ ´ = 2M

mvx3

final momentum

pf = mx ´ +

0

´y

M dy =

2M

´ =

xmv

23

p = pf � p

i = mv

1�

2x3

= 0

32

x

5. a = R (Pure rolling)v = u + at (v = at)For pure rolling = (v = R)(a) After 2 secV

A = V + R = 2V = 2at

VB = V

i + R (�j) = ( 2 V) = 2 at

V0 = V � R = 0

(b) a = R

aA = 2a i� + 2R (� j� )

aA = 2a i� +

RR22

(� j� )

aA =

2222

Rta4

)a2(

2

442

R

ta16a4

aA = 2

42

R

ta41a2

aB = (a � 2R) i� + (R) (� j� )

aB =

Rta4

�a22

i� + a (� j� )

aC = 2R

aC =

Rv2

=

Rta 22


16. V = R = (For pure rolling)(linear acceleration = 0)

rolls with out slipping soacc. only centripetal acc.

aA = R

v2

V = R

VA = (V � V cos ) i� + V sin j�

VA = j)èsinv()ècosv�v( 2

VA = (2V sin /2) = t

dtds

= 2 V sin \2 = 2 V sin

2t

s

0

ds =

/2

O

dt2

tsinv2 =

v8 = (8R)

18. Kinetic energy can become zero only for the case shown in figure ;Torque equation :

(mg).R = .2

MR2

= R

g2

Therefore , t =

0 = g2

R0

............(1)

For translational motion

t = g

v0

............(2)

From (1) & (2) g2

R0

= g

v0

0 =

R

v2 0 = 2.0

)10(2 = 100 rad/sec. Ans.

21. (i)

f

f

MM

M M

(a) mg � 4f = ma ......... (i)

fR = = Ra

fR2 = a

fR2 = 2

MR2

a

f =

2Ma

mg =

ma

2Ma4

= (2M + m)a

M = 2kg, m = 5 kg

a = 9g5

()


(b) If M = 0 (c) m = 0f = 0 mg = (2M + m) amg = ma 0 = aa = g() a = 0

(ii)(a) (m + 4M)g � 4f = (m + 4M)a

Torque about centre of disk ( = a / R)

f . R = 2

MR2

. Ra

f =

2Ma

(m + 4M) g � 2Ma = (m + 4M) a

(m + 4M) g = (m + 6M) a(5 + 8) g = (5 + 12) a

a =

17g13

()

(b) If M = 0 If M = 0mg = ma 4Mg = 6Ma

a = g() a = 3g2

()

22. (a) mg sin � f = ma �(i)

Torque about comfR = I

fR = 52

mR2 . For pure rolling a = R

f = 52

m (R)

f = 52

m (R) =

ma

5

2

mg sin � 52

ma = ma

mg sin = 52

ma + ma = 5ma7

a =

7

sing5

f = 52

m

7

sing5 =

7

sinmg2

f = N

= Nf

=

cosmg7sinmg2

=

tan

72

(b) torque about com

f.R = 52

mR2 .

NR = 52

mR2.


tan

7

1 (mg cos ) R =

52

mR2

= R14

sing5

mg sin � f = ma

mg sin � 71

tan . mg cos = ma

mg sin � 51

mg sin = ma

a =

sing

76

22

21

mv21

E.K

v2 = u2 + 2as

v2 = 0 + 2 76

g sin

v2 =

sing

7

12

s = ut + 21

at2

= 0 + 21

× 76

g sin t2

t =

½

sing37

K.E. =

sing37

R14sing5

mR52

21

sing7

12 m

21

22

K.E. = 76

mgsin + 845

mgsin

K.E. = 1211

mgsin.

23. Given mass of disc m = 2Kg and radius R = 0.1 m(i) FBD of any one disc is

Truck a = 9m/s 2

×

z

yx


Frictional force on the should be in forward direction.

a0

Pf

Let a0 be the acceleration of COM of disc the angular acceleration about its COM. Then �

a = 9m/s 2

f

Q

a0

= mf

=2f

......(i)

= I

= 2mR

21

R.f =

mRf2

= 1.02

f2

= 10 f .......(2)

Since there is no slipping between disc and truk therfore.Acceleration of point P = Acceleration of point Q a

0 + R = a

or

2f

+ (0.1)(10 f) or23

f = a f = 3a2

= 3

0.92 N

f = 6NSince this force is acting in positive x-direction.Therefore, in vector form

f

=( 6 i� ) N Ans. 3 (i)

(ii)

= r

× f

Here f

= ( 6 i� ) N ( for both the discs

Pr

= 1r

= 0.1 j� � 0. 1k� and

Qr

= 2r

= 0.1 j� � 0. 1k� and

20cm = 0.2 m

1z

Oy

x

2

QPf f

Therefore, frictional torque on disk 1 about O (centre of mass ) �

= r

× f

= ( �0.1 j� � 0.1k� ) × ( 6 i� ) N-m

= ( 0.6k� � 0.6 j� )

or1r

= 0.6 ( k� � j� ) N-m 0.6 ( k� � j� )

and |1r

|= 22 )6.0()6.0( = 0.85 N-m

Similarly,1r

= 2r

× f

= ( 0.1 j� �0.1k� ) × ( 6 i� ) N-m

1r

= 0.6 ( � j� �k� ) 0.6 ( k� � j� )

and |2r

| = |1r

| = 0.85 N-m Ans. 3 (ii)


24. (a) The cylinder rotates about the point of contact. Hence, the machanical energy of the cylinder will beconserved i.e.,

R

R

Rco

s

(PE + KE ) 1 = ( PE + KE )

2

mgr + 0 = mgr cos + 21

I 2 + 21

mv 2

V�

but = v / R ( No slipping at point of contact. )

and I = 21

mv 2

Therefore,

mgR = mgR cos + 21

2

21

mR

2

2v

R + 21

mv 2

or43

v 2 = gR ( 1 � cos )

or v 2 = 34

gR ( 1 � cos )

or R

2v =

34

gR ( 1 � cos ) ...........(1)

mg

V

N = 0

mg cos

At the time of leaving contact, normal reaction N = 0 and = c hence,

mg cos = R

2mv

orR

2v = g cos ...........(2)

From Eqs. (1) and (2)

34

g ( 1 � cosc )

= g cos

c

or47

cos c = 1 or cos

c = 4 / 7 or

c = cos � 1 ( 4 / 7 )

(b) v = )cos1(34

gR [From Eq. (1)]

At the time of losing contactcos = cos

c = 4 / 7

v =

74

134

gR

v = gR74


Therefore, speed of COM of cylinder just before losing contact is gR74

Therefore, rotational kinetic energy K R

= 21

I 2

or K R

= 21

mR2

12

2v

R =

41

mv 2 = 41

m

gR

7

4

or K R

= 7

mgR

Now, once the cylinder losses its contact, N = 0, i.e., the frictional force , which is responsible for its rotation,also vanishes. Hence, its rotational kinetic energy now becomes constant, while its translational kineticenergy increases.Applying conservationdecrease in gravitational PE = Gain in rotational KE + translational KE Translational KE (K

T) = Decrease in gravitational PE � K

R

or KT = (mgR) �

7mgR

= 76

mgR

From Eqs. (3) and (4)

R

T

KK

=

7mgR

76

mgR

orR

T

KK

= 6

28. (i) (a) CM

=

124C

12C

M22

= 412

CM5 2

A =

CM + Mx2

A =

412CM5 2

+

16MC5 2

= 48MC20 2

A mg ×

2C

= A = C5

g6

(b) acm

= x = 4

C5C5g6 = x

C5g6

ax = � a

cm cos

=� x

C5g6

. x.4C

= � 20

g6

= � 0.3 g

ag = � a

cm sin = x

2/C.x

C5g6

= � 0.6 g

j�g6.0i�g3.0a

(ii) (a) Mg � T = M acm

.... (1)

CM

T × 2C

= 48MC5 2

T = 24MC5

.... (2)

As aA = 0

(we know : acc. along the string is zero)a

cm �

x cos(90 � ) = 0


acm

= x sin = ax.

x2C

acm

= 2C

.... (3)

T = C

a2

24MC5 cm =

12

aM5 cm .... (4)

Mg = M acm

+ 12

aM5 cm

= 12

aM17 cm , acm

= 17

g12

(a) = C

a2 cm = C17g24

(iii)

(a) cm

= 2C

2mg

= cmI

4Cmg

= 48MC5 2

= C5g12

(b) FA =

2Mg

Mg � FA = m a

cm a

cm =

2g

= 0.5g

31. Coefficient of restitution m

m

e = = 1

2

V

sin2

V

(V1 = V

2 +

2

sin ) ..... (i)

angular momentum about point A

Li = mV

1

2

sin

L� = L

CM + L

A =

sin

2mVI 2CM

Li = L

�

mV1 2

sin = 12

m 2

� mV2 2

sin ..... (ii)

Put equation (i) in (ii) equation

mV1

2

sin = 12

m 2

� m

sin

2V1

2

sin

mV1

2

sin = 12

m 2

� mV1

2

sin + 4

m 2

sin2

222

1 sin4

m12

msinmV

V1 sin =

12

+ 4

sin2

1sin3

)sin12(V2

1


PART - II1. The given structure can be broken into 4 parts

For AB = CM

+ m × d2 = 22

4m5

12m

;

AB

= 34

ml 2

For BO = 3

m 2

For composite frame : (by symmetry)

= 2[AB

+ OB

] =

3m

3m4

222

= 3

10ml 2.]

3. M of the system w.r.t an axis to plane & passing through one corner

= 3

ML2 +

3ML2

+

22

2L3

M12

ML

= 3

ML2 2

+

4ML3

12ML 22

= 3

ML2 2

+ 12ML10 2

= 3

ML3 2

= 12ML18 2

= 23

ML2

Now 23

ML2 = 3k2M k = 2

[ Ans.: 2

]

4. = 1 +

2 +

3

1 =

2 =

23

mr2

3 =

2mr 2

= 1 +

2 +

3 =

27

mr2

Moment of inertia = 3mk2 where k is radius of gyration.

3mk2 = 27

mr2 k = 6

7r

5. Taking mass of plate m = 6M

Then M of two plates through which the axis is passing = 6am 2

× 2 = 3am 2

M. of 4 plates having symmetrical position from the axis

= 4 ×

22

2a

m12am

= 4 ×

3am 2

Total M = 3am4 2

+ 3am 2

= 3am5 2

using 6M

= m = M = 18Ma5 2


6.

Taking cylindrical element of radius r and thickness dr

dm = )RR(

M21

22

× (2r dr)

AB

= ed =

2rdm =

2

1

R

R

321

22

drr.)RR(

M2= )RR(M

21 2

122

Using parallel axis theorem

IXY

= )RR(M21 2

122 + MR

22 = )RR3(

2M 2

122

10.

For rotational equilibrium

N1 ×

4

= N2 ×

6

N1 : N

2 = 4 : 3

11. Balancing torque about the centre of the rod :

N1 .4

� N2 . 4

= 0

N1 = N2.

12. j�)200200(i�)100400(Fnet

= j�400i�300 |F|

= 500 N

Angle made by netF

with the vertical is = tan�1

400

300 = 37°

also = 500 R therefore point of application of the resultant force is at a distance R from the centre.Hence (C).

14. Immediately after string connected to end B is cut, the rod has tendency to rotate about point A.Torque on rod AB about axis passing through A and normal to plane of paper is

3m 2

= mg 2

= 2g3

Aliter : Applying Newton�s law on center of mass

mg � T =ma .....(i)Writing = I about center of mass

T 2

= 12

m 2

....(ii)

Also a = 2

....(iii)

From (i) , (ii) and (iii)

= 2g3


16. For rigid body separation between two point remains same.v

1 cos60° = v

2 cos30°

2v1 =

2v3 2 v

1 = 3 v

2

disc

= d

60sinv30sinv 12 = d

2v3

2v 12

= d2v33v 22

= d2

v2 2 = d

v2

disc =

dv2

18. When ball at maximum height block and ball has equal velocitySo Using momentum conservation

Pi = mv

Pf = 2mv

0(v

0 final velocity)

Pi = P

t

mv = 2 mv0

V0 =

2V

Using energy conservation

212 +

21

mv2 = 21

2 + 21

2mv0

2 + mgh

( = mR2) v = R

21

mv2 = 21

2mv0

2 + 2mgh

v2 � 2 4v2

= 2gh

g4v

`h2

20. As torque = change in angular momentum F t = mv (Linear) ..... (1)

and

2

F

t = 12

m 2

(Angular) ..... (2)

Dividing : (1) and (2)

2 =

v12 =

v6

Using S = ut :

Displacement of COM is 2

= t =

v6t and x = vt

Dividing

x2 =

6

x = 12

Coordinate of A will be

0,

212

Hence (D).

24. = dtdL

= 4

from figure 22r m

Hence = Fr

4 = 22 .F F = 2 N Ans.


28. By conservation of angular momentum about hinge O.

L =

mv2d

=

22

2d

m12

Md

4md

2md

2mvd 22

2md

43

2mvd

dv

32

30. FBD for sphere & block

a1

fr

mm

fr

a2

a1 =

mfr =

mmg

a2 =

mfr =

mmg

i�ga1

i�ga2

i�g2aaa 21rel

arel

= 2g.

31. Using Energy conservation,(at maximum distance V = 0 V

0 = 0)

21

Kx2 = (mg x sin )

x =

Ksinmg2

33. Since the two bodies have same mass and collide head-on elastically, the linear momentum gets interchanged.Hence just after the collision 'B' will move with velocity 'v

0' and 'A' becomes stationary but continues to rotate

at the same initial angular velocity

R

v0.

Hence, after collision.

(K.E.)B = 2

0mv21

and (K.E.)A = 2

21

= 2

02

R

v.mR

32

21

2

3.)E.K(.)E.K(

A

B Hence (D).

Note : Sphere 'B' will not rotate, because there is no torque on 'B' during the collision as the collision is head-on.

35. Decrease in PE =

Increase in rotational K.E

2mg. 2

� mg. 2

= 21 . 2 =

21

4.m

4m2

2

2

2mg

= 21

. 4

m3 2

. = 8

m3 2

2

= 3g4

and v = r = 2

3g4

=

3g

[ Ans.: (a) V = g / 3 , = 4 3g / ]


36. Just before collision Between two Ballspotential energy lost by Ball A = kinetic energy gained by Ball A.

2h

mg = 2cm

2cm mv

21

21

= 2cm

2cm2 mv

2

1

R

vmR

5

2

2

1

= 51

2cmmv +

21

2cmmv

mgh75

= 2cmmv

7mgh

= 51

2cmmv

After collision only translational kinetic energy is transfered to ball B

So just after collision rotational kinetic energy of Ball A = 51

2cmmv =

7mgh

39. Torque about COMf.R = · (a = R)

f.R = 2

mR2

a =

R·

2mR2

2ma

f

41. Here, u = V0, 0 = R2

V0

At pure rolling ;

V = V0 �

m

Fft

&RV

=

R.m

F

R2

V f0t (In pure rolling V = R) ( =

= 2

f

mR

R.F)

V0 � V = V + 2

V0

2V = 2

V0 V =

4

V0 Ans.

42. As the disc is in combined rotation and translation, each point has a tangential velocity and a linearvelocity in the forward direction.From figurevnet (for lowest point) = v � R= v � v = 0.

and Acceleration = Rv2

+ 0 = Rv2

(Since linear speed is constant)

43. Since there is no slipping at any interface, the velocities of bottom and upper most point of lower and uppercylinder are shown in figure.

Angular velocity of upper cylinder = R2

VV2 =

R2V3

Angular velocity of lower cylinder = R2

0V =

R2V

The ratio is 13


44. For maximum a, normal reaction will shift to left most position. for rotational equilibrium

P = = 0 [in frame of truck]

ma 2

= mg 2b

a =

gb

45. Torque about point A

TA = Fr 2

d + F2 (d)

1A = Fr

4d3

+ F1 (d)

(F1 + F

2)

2d

+ F2d = (F

1 + F

2)

4d3

+ F1d

2FF 21 + F

2 =

121 FF

43

F43

2F1 �

43

F1 � F

1 =

2F

�FF43 2

21

1

1 F�4F�

=

2F

�4F� 22

4F5 1 =

4F3 2

5F1 = 3F

2

2

1

FF

=

53

.

47. By angular momentum conservation ;

L = mv 2R

+ mvR = 2mR2

23

mvR = 2mR2

= R4v3

Also at the time of contact ;

mgcos � N = R

mv2

N = mg cos � R

mv2


when it ascends decreases so cos increases and v decreases.

mgcos is increasing and R

mv2

is decreasing

we can say N increases as wheel ascends.

48. Torque about point A

w0

v0

A

mg

( mg) R = .mR52 2

=

R2g5

v = u + at0 = v

0 � gt

t =

g

v0 =

0 + t 0 =

0 �

P2g5

. g

v0

0 =

R2

v5 0

EXERCISE-31. Since all forces on disc pass through point of contact with horizontal surface, the angular momentum

of disc about point on ground in contact with disc is conserved. Also the angular momentum of disc inall cases is conserved about any point on the line passing through point of contact and parallel tovelocity of centre of mass.The K.E. of disc is decreased in all cases due to work done by friction.From calculation of velocity of lowest point on disc, the direction of friction in case A, B and D istowards left and in case C is towards right.The direction of frictional force cannot change in any given case.

2. (A) Speed of point P changes with time(B) Acceleration of point P is equal to 2x ( = angular speed of disc and x = OP). The acceleration is

directed from P towards O.(C) The angle between acceleration of P (constant in magnitude) and velocity of P changes with time.

Therefore, tangential acceleration of P changes with time.(D) The acceleration of lowest point is directed towards centre of disc and remains constant with time

3. Let the angular speed of disc when the balls reach the end be . From conservation of angular momentum

21

mR2 0=

21

mR2 + 2m

R2 + 2m

R2 or = 30

4. The angular speed of the disc just after the balls leave the disc is

= 30

Let the speed of each ball just after they leave the disc be v.From conservation of energy

21

2mR21

0

2 =

21

2mR21

2 + 21

2m

v2 +

21

2m

v2

solving we get

v = 3

R2 0

NOTE : v = 2r

2 v)R( ; vr = radial velocity of the ball

5. Workdone by all forces equal Kf � K

i =

2m

21

v2 = 9

mR 20

2


6 to 8The free body diagram of plank and disc isApplying Newton's second law

F � f = Ma1

.... (1)f = Ma

2.... (2)

FR = 21

MR2 .... (3)

from equation 2 and 3

a2 =

2R

From constraint a1 = a

2 + R

a1 = 3a

2.... (4)

Solving we get a1 =

M4F3

and = MR2F

If sphere moves by x the plank moves by L + x. The from equation (4)

L + x = 3x or x = 2L

9. The moment of inertia about both given axis shall be same if they are parallel. Hence statement-1 isfalse.

10. As x increases, the required component of reaction first decreases to zero and then increases (withdirection reversed). Hence statement-1 is false.

11. For a disc rolling without slipping on a horizontal rough surface with uniform angular velocity, theacceleration of lowest point of disc is directed vertically upwards and is not zero( Due to translationpart of rolling, acceleration of lowest point is zero. Due to rotational part of rolling, the tangentialacceleration of lowest point is zero and centripetal acceleration is non-zero and upwards). Hencestatement 1 is false.

12. The acceleration of any point on the disc rolling with uniform angular velocity in the frame of ground is equalto centripetal aceleration of that point in the frame of centre of mass of disc. Hence Statement-1 is True,Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

13. Let v be the speed of centre of mass, be its angular speedand R be the radius of disc respectively. v = R, because the disc rolls without slippping.The velocity of any point P on disc distant x from centre

O due to translation is Tv

= (vT = v) in horizontal direction and its

velocity due to rotation is Rv

(vR = x) as shown.

Since the magnitude of Rv

is less than or equal to magnitude of Tv

, the horizontal component of Rv

cannot

cancel Tv

to make velocity of any point P vertically upwards. Hence statement-1 is true.

In statement-2 R may be more than v. Hence condition of statement-1 may not be satisfied. Otherwisestatement-2 is true.

14. The applied horizontal force F has tendency to rotate the cube in anticlockwise sense about centre ofcube. Hence statement-2 is false.

15.

(i) = I

= F × r :

Torque is same in both the cases but moment of inertia depends on distribution of mass from theaxis.Distribution of mass in both the cases is different.Therefore,moment of inertia will be different or theangular acceleration will be different.

= I

= F × r :


(ii) S1 : The line of action of each action reaction pair is same. Therefore magnitude of couple of each such pair

is zero. Hence net torque of all internal forces is zero.S

2 : The direction of angular acceleration and angular velocity shall be opposite if the body is slowing down.

S3 : If net torque on a rigid body is zero, its angular velocity will be constant. The constant may necessarily

not zero.S

4 : Since the centre of mass is fixed, that is, at rest; hence net force on rigid body is zero. Therefore torque

on this rigid body about any point is same.

(iii) I 1

1 = I

2

2

2 =

2

1

II

. 1 =

2

.42

222

2

RMMR

MR

= 54

(iv) In case of ring : T

R

KK

= 1

In case of pure rolling

or K R

= K T =

2K

21

(0.3) v1 2 =

2K

.........(1)

In case of disc :T

R

KK

= 21

or K T =

32

K 21

(0.4) v2

2 = 32

K .........(2)

From Eqs. (1) and (2) ,

2

1

vv

= 1

i . e ., v1 = v

2

(v) TrueAngular momentum will be conserved if the net torque is zero .Now for the sphere to move down:mg sin > mg cosLet x be the perpendicular distance of the point (as shown in figure)about which torque remains zero.for = 0 ; x > R as shown

Note: As mgsin > mgcos, the point should be inside the sphere.

16.(i) Let the mass of disc without hole = m

The mass of cut out hole of radius 2R

is 4m

m � 4m

= M or m = 34

M.

Moment of inertia of given body about axis passing through O= MI of complete disc � MI of cut out hole.

= 2R

3

M4

2

1

�

22

2R

3M

2R

3M

21

= 2413

MR2


(ii) mvr = K ( a constant ) v = mrk

T = r

mv 2

=

rm 2

mrK

= mK 2

. 3

1

r

= Ar � 3

mwhere

2k A ,

Hence , n = � 3

(iii) under the given conditions only posibility is that friction is upwards and it accelerates downwards as shownbelow :

The equations of motion are :

a = m

f�sinmg =

mfº�30sinmg

= 2g

� mf

.......(1)

= I

=

I

fR =

2mR

fR2 =

mRf2

......(2)

For rolling (no slipping)a = R or g/2 � f/m = 2f/m

mf3

= g/2 or f = mg/6

(iv) Talking moments about of point O :

Moment of N (normal reaction) and f (force of friction) are zero. In critical case normal reaction will passthrough O. To tip about the edge, moment of F should be greater than moment of mg. or,

F a

4

> (mg) 2a

F > 2mg


EXERCISE-4PART - I

1. mg sin component is always down the plane whether it is rolling up or rolling down. Therefore, for noslipping, sense of angular acceleration should also be same in both the cases. Therefore, force of friction falways act upwards.

2. Since, there is no external torque, angular momentum will remain conserved. The moment of inertia willfirst decrease till the tortoise moves from A to C and then increase as it moves from C and D. Therefore will initially increase and then decrease.Let R be the radius of platform m the mass of disc and M is the mass of platform.Moment of inertia when the tortoise is at A

1 = mR2 +

2MR2

and moment of inertia when the tortoise is at B

2 = mr2 +

2MR2

O

ar

B C D

vthere r2 = a2 + 222 ]vtaR[

From conservation of angular momentum

0

1 = (t)

2

substituting the values we can see that variation of (t) is nonlinear.

3. (a) The distance of centre of mass (COM) of the system about point A will be :

r = 3

Therefore the magnitude of horizontal force exerted by the hinge on the body isF = centripetal force

or F = (3m) r2

or F = (3�m)

3

2

or F = 3 m2 Ans.(b) Angular acceleration of system about point A is

= A

A

I

= 2m2

23

)F(

COM

B

F

,A

C

y

x

3/2

= m4F 3

Now acceleration of COM along x-axis is

X = r=

3

m 43

or ax =

m4F


Now let Fx be the force applied by the hinge along x-axis. Then :

Fx + F = (3m) a

x

or Fx + F = (3m)

m4

F

or Fx + F =

43

F or Fx = � 4F

Ans.

Further if Fy be the force applied by the hinge along y-axis. Then :

Fy = centripetal force

or Fy = 3 m2 Ans.

4. In uniform circular motion the only force acting on the particle is centripetal (towards center). Torque of thisforce about the center is zero. Hence angular momentum about center remain conserved.

5. Let �� be the angular velocity of the rod.

Angular impulse = Change in angular momentum about centre of mass of the system

J · 2L

= C M M

J=MV (MV)

2L

= (2)

4ML2

· = LV

6. From conservation of angular momentum ( = constant), angular velocity will remain half. As,

K = 212

The rotational kinetic energy will become half. Hence, the correct option is (B).

7. In case of pure rolling bottommost point is the instantaneous centre of zero velocity.

QC

P

O

Velocity of any point on the disc, v = r, where r is the distance of point from O.r

Q > r

C > r

P

vQ > v

C > v

P

Therefore, the correct option is (A).

8. 0 =

1 �

2

where 1 = (M.. of full disc about O)

2 (M.. of small removed disc about O)

since mass area

totalofmassdisccutofmass

= 2

2

R9

R

= 91

mass of cut disc = m

0 =

2R)m9( 2

� m

2

2

3R2

23R

(by theorem of parallel axis.)

= 2

mR9 2

� mR2

9

4

18

1=

2mR9 2

� 2

mR2

= 2

mR8 2

= 4mR2.


9. Only direction of L

(angular momentum) is constant because the direction of rotation is unchanged.

10. From equilibrium,friction = mg N = Fabout centre of mass = 0 mg a = torque due to normal Normal will produce torquesince F passes through centre its torque is zero.

11. Initial angular momentum about fixed point = mvL

final angular momentum = =

22

mL3

ML

where is moment of inertia of the system about the fixed point and is angular velocity about the fixedpoint.angular momentum before collision = angular momentum after collision

mLv = L2

m

3M

=

m

3M

L

mv = L)m3M(

mv3

12. 2/5 MR2 = 1/2 Mr2 + Mr2

2/5 MR2 = 3/2 Mr2

r2 = 154

R2

r = 15

R2

13.* necessary torque for rolling = fr, (frictional force provides this torque)as mg sin � f = ma

but a = r mg sin � f = mr

as = fr = = fr/

mg sin � f = mrfr/ = 5f/2

5mr2 2

q

r f

mg sin = 2f7

thus friction increases the torque in hence the angular velocity and decreases the linear velocity.If decreases friction will decrease.

14*. As total mechanical energy at points A,B and C will be constant

A =

B =

C

mghA + K

A = K

B = mgh + K

C

KB > K

A(mgh

A + K

A = K

B)

and KB > K

C(mgh

C + K

C = K

B)

Also hA � h

C = mg

KK AC when mgh

A + K

A = mgh

C + K

C

if hA > h

C K

C > K

A

(if LHS is positive then RHS have to be positive)

15. (As collision is elastic)

F = mV21

mV2dtdP

torque about hinge = 2mV ×

4

b

2

b × 100 = 2mV

4b3

× 100 = Mg 2b

V = 10 m/s


16. 21

2 kx21

)2(21

22

2 kx21

221

2

1

xx

= 2

17. Apply conservation of angular momentum( × 2) + (2 × ) = ( + 2) �

� = 3

4

For Disc A

t = × (2 � �) = t3

2

18. Initial Kinetic Energy k1 =

21

× × (2)2 + 21

×2 × 2

Final Kinetic Energy k2 =

21

× × �2 + 21

× 2 �2

Loss of Kinetic Energy= k1 � k

2

= 3

2

19. From the conservation of energyloss in KE of body = Gain in potential energy

21

mv2 + 21

2

rv

= mg

43

2

gv

on solving

= 2

mr 2

The body is a disc

20. If torque external = 0, then angular momentum = constant =

21. The acceleration of centre of mass of either cylinder

a =

2

2

R

K1

sing

where K is radius of gyration.

So acceleration of centre of hollow cylinderis less than that of solid cylinder.Hence time taken by hollow cylinder will be more.So statement-1 is wrong.Ans. (D)

22. (A) Since there in no resultant external force, linear momentum of the system remains constant.(B) Kinetic energy of the system may change.(C) Angular momentum of the system may change as in case of couple, net force is zero but torque is not zero. Hence angular momentum of the system is not constant.(D) Potential energy may also change.

23*. )i�(R)i�(VVA

; i�VVB

; i�Ri�VVC

i�R2VV AC

2 )]i�(R)i�(V)i�(V[2VV CB

= �2R( i� )

Hence AC VV

= )VV(2 CB


so |VV| AC

= |)VV(2| CB

BC VV

= R( i� )

AB VV

= R( i� ) ABBC VVVV

Hence )i�(R2VV AC

ABBC VVVV

; BV4

= 4V( i� ) = 4R ( i� )

Hence )V(2VV BAC

24. Angle of repose 0 = tan�1 = tan�1 3 = 60º

tan = 2/15

5 =

32

. < 45º.

Block will topple before it starts slide down.

25. 2Ff 22' ...(i)

FR � f 'R = 2mR2

Ra

F � f ' = 2ma = 1.2 ...(ii)From (i) & (ii)(1.2 + f ')2 + f '2 = 22

2f '2 + 2.4f ' + 1.44 = 4f '2 + 1.2f ' + 0.72 � 2 = 0

f '2 + 1.2f ' � 1.28 = 0

f ' = 2

)28.1(444.12.1

= 0.6 ± 28.136.0

= �0.6 ± 64.1 = 0.68

From eq. (2)F = 1.88

= 88.168.0

= 10P

P = 3.61 4 Ans.

Note : In Hindi friction force is aksed, so the answer is P = 6.8. (for Hindi)

Note : But if only normal reaction applied by the rod is considered to be 2 N Law 2 � f = 2 [0.3]

f = 2 � 0.6

f - 1.4 Nx ...(i) a = R 0.3 = [0.5]

= 53

rad/s ....(ii)

c =

c

fR � 2R = mR2 f � 2 = mR

1.4 � 2 = 22

5

3 1.4 � 0.6 = 2µ

0.8 = 2µ = 0.4 = 10P

P = 4 Ans.


Note : In Hindi friction force is aksed, so the answer is P = 8. (for Hindi)

26. =

2MR5

2 2 + 2MxMR

5

2 22

= 2MR52 2

+ 2MR

52 2

+ (Mx2) 2

= 4

2MR52

+ 2mx2

= 2MR

58

+ 2mx2

= 4

2

10)24()5.0(225

5.058

=

8

55

× 10�4

= 9 × 10�4 = N × 10�4

So, N = 9 Ans.

27. Friction force on the ring.28. L = [m(vt)2]

L = mv2t2

So = dtdL

= 2mv2t

t straight line passing through (0, 0)

30.

I0 = 2

)R2()m4( 2 �

23

mR2

= mR2 [8 � 23

] = 2

13mR2


IP =

23

(4m) (2R)2 �

]R)R2[(m

2mR 22

2

= 24 mR2 � 2mR

211

= 2mR2

37

2132

37

I

I

O

P = 3

1337

Ans. 3

31. Angular Velocity of rigid body about any axes which are parallel to each other is same . So angular velocityis .

32. Since z- coordinate of any particle is not changing with time so axis must be parellel to z axis.

33. IP > I

Q

aP = 2

P mRI

sing

aQ = 2

Q mRI

sing

aP < a

Q V = u + at t

a1

t P > t

Q

V2 = u2 + 2as v a VP < V

Q

Translational K.E. = 21

mV2 TR KEP < TR KE

Q

V = R V P <

Q

34. V0 = 3R i�

VP (3R �

2R

cos 60º) i� + 2R

sin 60 j�

= i�4

R3i�

4R11

PART - II

1. AC

=

6M

21 2

= 12

M 2

, EF

= 12

M 2

, AC

= EF

.


2. mg sin � � = maCM

..........(i)�.R = ..........(ii)a

CM = R ..........(iii)

On solving (i),(ii) & (iii)

aCM

=

2MR1

sing

.

3. Central force is directed towards a point, therefore torque of the central force is zero.

4. IA = I

cm + m

2

2

a

= 6

ma2

+ 2

ma2

= 32

ma2

5.21

2 = mgh

21

3m 2

2 = mgh

h = g6

22

6.

Angular Momentum = m

2

0000 gt21

tsinV)cosV()tcosV)(gtsinv(

= � 21

mg V0 t2 cos 0 k�

7.

From angular momentum conservation about vertical axis passing through centre. When insect is comingfrom circumference to center. Moment of inertia first decrease then increase. So angular velocity inecreasethen decrease.


8. mg � T = ma

TR = 2

mR2

T = 2

mR =

2ma

mg � 2

ma = ma

2ma3

= mg a = 3g2

Ans.

9. To reverse the direction 0d (work done is zero)

= (20 t � 5t2) 2 = 40t � 10t2

= 22

tt410

t10t40

= t

0dt = 2t2 �

3t3

is zero at

2t2 � 3t3

= 0

t3 = 6t2

t = 6 sec.

= dt = dt)3t

t2(6

0

32

6

0

43

12t

3t2

= 216

2

1

3

2 = 36 rad.

No of revolution 2

36 Less than 6

10. L0 = Pr

L0 = mv cos H

= 23

mg . g2º30sinv 22

= g16mv3 3

.

RESONANCE SOLN_UNIT & DIMENSIONS - 157

TOPIC : UNIT AND DIMENSIONS

EXERCISE-1PART - I

3**. (i) U = AT4 [] = ]T][A[

]U[4 = 42

32

KL

TML

= MT�3 K�4

(ii) T = b [b] = [] [T] = LK

(iii) F = 6rv [] = ]v][r[]F[

= 1

2

LT .L

MLT

= ML�1 T�1

(iv) = AP

= 2

32

L

TML

= MLº T�3

(v) Energy = 21

Mi2 [M] = ]i[

]E[2 = ML2T�2 AA�2

(vi) ]V[]U[

= ]2[]B[

0

2

= [

0] =

]U[]V[ ]B[ 2

Also , F = qVB B = qvF

[0] =

]U][vq[

]V[)F(22

2

= MLTT�2A�2 Ans.

8. We have the equation

221

r

mGm = F 2

2

]L[

]M][G[= MLTT�2

[G] = M�1L3T�2 .......... (i)

hc= Energy ][

]c[]h[

= ML2T�2 [c] = LTT�1

[] = L[h] = ML2T�1 .............. (ii)

[c] = LT�1 ................ (iii)

taking the product of (i) & (ii)

[G] [h] = L5T�3

[c]3 = L3T�3

3]c[

]h][G[ = L2

2/32/12/1 chG = L

again from (iii)

[T] = ]c[]L[= 12/32/12/1 chG = 2/52/12/1 chG

From (ii)[h] = ML2T�1

[h] = 2/52/12/1

3

chG

MGhc

[h] = MG1/2h1/2c�3+5/2 or G-1/2 h1/2 c1/2 = M


10. Let, w = KMarbGc where K is a dimensionless constant.Writing the dimension of both the sides and equating then we have,T�1 = MaLb(M�1L3T�2)c

= Ma�c Lb+3c T�2c

Equating the exponents

� 2c = � 1 or c = 21

,

b + 3c = 0 or � 3 c = b = � 23

a � c = 0 . c = a = + 21

Thus the required equation is = K 3r

Gm

PART - II

5. All the terms in the equation must have the dimension of force [A sin C t] = MLT�2

[A] [M0L0T0] = MLT�2 [A] = MLT�2

Similarly, [B] = MLT�2

]B[]A[

= MºLºTº

Again [Ct] = MºLºTº [C] = T�1

[Dx] = MLTº [D] = L�1

]D[]C[

= MºL1T�1 .

6. [P] =

2V

a [a] = [P] [V2]

10.** V = R

V has the dimensions of

[V] = ]eargch[]work[

= TTML 22

= ML2 T�3 �1

[R] = ][]v[

= ML2 T�3 �2

11. [v] = [k] [a b gc] LT�1 = La Mb L�3b Lc T�2c

LT�1 = Mb La � 3b + c T�2c a = ½, b = 0, c = ½

so, v2 = kg

13. G = 6.67 × 10�11 N m2 (kg)�2

= 6.67 × 10�11 × 105 dyne × 1002 cm2 / (103)2 g2 = 6.67 × 10�8 dyne-cm2-g�2

14.

1ax

sinaxax2

dx 1n

2 .

L.H.S. is the dimensionless as

denominator 2ax � x2 must have the dimension of [x]2


(we can add or substract only if quantities have same dimension)

2xax2 = [x]

Also, dx has the dimension of [x]

2xax2

dxx

is having dimension L

Equating the dimension of L.H.S. & R.H.S. we have

[an] = M0L1T0 { sin�1

1

ax

must be dimensionless}

n = 1

16. [a] =

ma ....(i)

ma = M0 L0 T0

ma = [] = ][ = L

17. [g] = LT�2 and numerical value unit

1

18.** [] = ][

]h[4

= 443�

12

K.KMT

TML

= L2 T2

So, unit of will be m2s2.

Tesla)Farad()()weber( 22

= T

F.Tm 222 = m2s2

EXERCISE-2PART - I

1.** Reynold�s number and coefficient of friction are dimensionless quantities. Curie is the number of atoms

decaying per unit time and frequency is the number of oscillations per unit time. Latent heat and gravitationalpotential both have the same dimension corresponding to energy per unit mass.

2.** X = 3YZ2

[X] = [Y] [Z]2

[Y] = 2]Z[

]X[ = 222

2221

TQM

TQLM

= M�3L�2Q4T4

3.* Torque and work have same dimensions = ML2 T�2

Light year and wavelenth have dimension of length = L

4. Micron, light year & angstrom are units of length and radian is unit of angle.

5.* (A) L = i

or Henry = AmpereWeber

(B) e = � L

dtdi

L = dt/di

eor Henry = Ampere

ondsecVolt

(C) U = 21

Li2 L = 2i

U2or Henry = 2)Ampere(

Joule

(D) U = 21

Li2 = i2 Rt [L] = [Rt]. or Henry = ohm-second


6.* we have F = 04

1 2

21

r

qq

[0] = ]r[]F[

]q[]q[221

= 22

22

LMLT

T

= M�1L�3T42

Also C (speed of light) = 00µ

1

[µ0]1/2 = ][]c[1

0

[0] = MLT�2 �2

7.** (None of the four choices) 21

0E2 is the expression of energy density (Energy per unit volume)

3

222

0L

TML E

21

[ML�1 T�2] From this question, students can take a lesson that even in IIT-JEE, questions may be wrong or there

may be no correct answer in the given choices. So don�t waste time if you are confident.

8.** X = tv

L0

[X] =

tv

]L[][ 0

[v] = [Electric field] [Length]

= ]eargch[]Force[

[Length] = Q

LMLT 2

= MQ�1L2T�2

[0] = M�1 L�3T42 (as in question no. 6)= M�1L�3Q2T2

[X] = M�1L�3Q2T2 L T

TLMQ 221

= QT�1 = [x] is that of current

9.

kZ

= [M0 L0 T0] [] =

Zk

Further, [P] =

[] =

P =

ZP

k

Dimensions of k are that of energy. Hence,

[] =

21

22

TLML

TML = [M0L2T0]

Therefore, the correct option is (A).

10.** [Dipole moment] = LIT, [E] = ML3 /T3 [E] = ML/T3 .

11.** (A) 2e

se

R

MGM = Force

[GMeM

s] = [Force] [R

e2]

= MLT�2 L2 = ML3T�2


Hence S unit of GMeM

s, will be (kilogram) (meter3)(sec�2)

ie same as (volt) (coulomb) (metre)

(B)MRT3

= VR.M.S.

0MRT3

= [VR.M.S.

]2 = L2T�2

Hence S unit will be (metre)2 (second)�2 ie same as (farad) (volt)2 (kg)�1

(C)]Bq[

]F[22

2

= ]Bq[

]Bvq[22

222

= [V2] = L2T�2

Hence S unit (metre)2 (second)�2 i.e. same as (farad) (volt)2 (kg)�1

(D)

e

e

R

GM = ]Mass[

]R[]Force[ e =

MLMLT 2

= L2T�2

Hence S unit will be (meter)�2 (second)�2 i.e. same as (farad) (volt)2 (kg)�1

PART - II1. The dimensions of torque and work are [ML2 T2]2. h = Planck�s constant = J�s = [ML2T�1]

P =momentum = kg m/s = [MLT�1]

3. As we know that formula of velocity is

v = o0

1

v2= o0

1

= [LTT�1]2

o0

1

= [L2 T�2]

4. From Newton�s formula

= z/vAF

x

Dimensions of =

gradient�velocityofensionsdimareaofensionsdim

forceofensionsdim

= ]L[]L[

]MLT[1�2

2�

= [ ML�1 T�1]

5. I = mr2

[I] [ML2]

and = moment of force = Fr

= [L] [MLTT�2 ]

6. Energy stored in inductor

2LI21

U 2I

U2L 2

2

22

2�2

Q

ML

T/Q

TL]L[

Since Henry is unit of inductance L (4) is correct.

7. From F = qvB [MLT�2] = [C] [LT�1] [B] [B] = [MC�1T�1]

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