2012 t2 exam paper 1 and 2 answers

7/28/2019 2012 T2 Exam Paper 1 and 2 Answers

1/4

May 2012. Term Two Physics AS examination. MC answers.1 A 21 C2 C 22 C3 C 23 C4 B 24 C

5 D 25 C6 C 26 B7 B 27 A8 A 28 C9 B 29 B10 C 30 B11 C 31 A12 A 32 C13 C 33 D14 A

15 D16 B17 A18 C19 D

20 A

1. (a) (i)

F1 weight /mg(1)

F2 reaction or normal contact force(1)

F3 driving force(1)

F4 friction or air resistance(1)

(ii) zero acceleration (1)

zero resultant force(1) max 5

(b) (P =Fvgives) 18 103 =F 10(1) (andF =1.8 103 N) 1

(c) (i) 1800 250 =1.6 103 N (1) (1.55 103 N)

(ii) force =4 1.55 103 =6.2 103 N (1)(allow e.c.f. from(i))

(iii) total force =6200 +250(N) (1) (=6.45 103(N))

(P =Fvgives) P =6.45 103 20 =1.3 105 W (1) (1.29 105 W)(allow e.c.f. for value of total force) 4 [10]

2. (a) scalars havemagnitude(or size) (1)vectors havemagnitudeanddirection(1) 2


2/4

(b) (i) s=t(1)

s=100 =5 km(1)

(ii) 1.59 (1) km (or other correct unit) (1) 4[6]

3. (a) 2

(b) (i) taking moments (about A) (1)(400 4.0) +(300 2) =FB 1.6(1)

=FB soFB =1375(N) (1)

(ii) FA =1375 700 =675 (1) N (1) allow e.c.f. fromFB

(no marks if weight of board not used) max 4[6]

4. accept mirror image for (a) and (b)

(a) (b)

straight line sloping up (1) constant value shown(1)sudden change to negative velocity (1)smaller negative velocity (1)same gradient as positive line(1) 4

(c) (i) vertically down at P (1) (ii) vertically down at Q (1)

(iii) along tangent at P (1) (iv) along tangent at Q (1) 4

(d) horizontal component of velocity at Q = 15 cos 50(1) =9.64 m s1(1)

momentum at Q =0.15 9.64 =1.45 (1) Ns (or kg m s1) (horizontally) (1) 4[13]

60

3

F

F

A

B(1)

(1)

6.1

2200

velocity

time

acceleration

time


3/4

5. (a) (i)

B must make an appreciable angle with wall and bar

(ii) A weight of sign and bar (accept gravity) (1)B reaction of wall (1)C tension in wire(1) max 5

(b)

use ofmg(1) clockwise moments 118 0.375 (1)

=anticlockwise moments (Tcos40 (1)) 0.750(1) T=77 N (1) max 4[9]

6. (a) (i) equation showing momentum before =momentum after (1)

correct use of sign(1)

(ii) no external forces (on any system of colliding bodies) (1) 3

(b) (i) (by conservation of momentumm11 +m22 =0)

0.25 2.2 =()0.502(1)

2 =()1.1(0)msl(1)

(ii) =total k.e. = 0.25 2.22 + 0.5 1.12(1)

=0.91J (1) 4 [7]

7. (a) (i) gradient = =3.0 ms2(1)

(ii) distance is area under graph (to t=0.1 s)

or 0.7 2.1 0.3 (1) =1.4(2) m(1) 3

(b) (i) T mg=ma [orT=1500(9.8+3.0)] (1)

=1.9 104 N (1)

T=mg=l.5 104N (1)

(ii) EF (1) 4

(c) power =F or l.5 1042.5(1) =3.7[3.8] 104 W (1) 2[9]

B

C

A

(1)

(1)

(1)

0.375m

118N

40

2

1

2

1

7.0

1.2

2

1

+

2

5.21.2


4/4

8. (i) a= =11 ms2(1) F =ma=1.1 105 N (1) (ii) =236 m s1

a= =29.5 ms2(1)(iii) sone= t= 4.0 =88m(1)

stwo = t= 8.0(1) =1296(m) (1) total distance =1384 m(1)

[6]

9. (a) sum of clockwise moments equals sum of anticlockwise moments(1)for a body in equilibrium(1) 2

(b) point in the body through which the weight/mass (appears to) acts[or point where resultant torque/moment is zero]

[or point where body would balance] (1) 1

(c) (i) towards A (1)so that weight of ruler (1)provides balancing moment (1)

(ii) (moments about pivot give) 1.0 (0.30 d) =0.50 d(1)1.5d=0.30 andd=0.20 m(1) 5

[8]

10. (a) (i) (use ofv2 =u2 +2asgives) 0 =252 - 2 9.81 s(1)19.6s=625 ands=32 m(1)

(ii) t= =2.5 s(1)

(iii) (use ofv2 =u2 +2asgives) v2 =252 2 9.81 16(1)

(allow C.E. from (a)(i)) andv=18 m s1(1) max 4

(b) time to stop the ball is greater (1) rate of change of momentum is less(1)

[or work done on ball is the same but greater distance(1) less force(1) ] 2[6]

11. (a) kinetic energy not conserved (1)[or velocity of approach is not equal to velocity of separation] 1

(b) (i) (use ofp=mvgives) p=4.5 102

60 (1)=2.7 kg m s1 (1)

(ii) (use ofF = gives) F = (1)

=180 N (1)

[or a= = =4000 (ms1)

F =(ma) =4.5 102 4000 =180 N] 4

[5]

0.4

44

0.8

236

av

+

2

044

av

+

2

44280

81.9

25

t

mv

)(31015

7.2

t

uv31015

60

2012 t2 exam paper 1 and 2 answers

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