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Your Name:

PHYSICS 101 MIDTERMOctober 23, 2007 2 hours

Please circle your section1 9 am Galbiati 2 10 am McDonald3 11 am McDonald 4 11 am Staggs5 12:30 pm Sondhi

Problem Score1 /102 /103 /154 /155 /156 /207 /15

Total /100

Instructions: When you are told to begin, check that this examination bookletcontains all the numbered pages from 2 through 18. The exam contains 7 problems.Read each problem carefully. You must show your work. The grade you get dependson your solution even when you write down the correct answer. BOX your finalanswer. Do not panic or be discouraged if you cannot do every problem; there areboth easy and hard parts in this exam. If a part of a problem depends ona previous answer you have not obtained, assume it and proceed. Keepmoving and finish as much as you can!

Possibly useful constants and equations are on the last page, which you maywant to tear off and keep handy

Rewrite and sign the pledge: I pledge my honor that I have not violated the Honor Codeduring this examination.

Signature

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Problem 1: Grab Bag

(a) This problems deals with collisions in a 1-dimensional system. Given the masses of thetwo colliding bodies (m1, m2), their initial velocities (v1,i, v2,i), and the kind of collisionspecified in each line, you are requested to calculate the final velocities: v1,f and v2,f .

m1

m2

v1,i

v2,i

a.1 [2 pts] Perfectly elastic collision; m1=m2=10 kg; v1,i=12 m/s; v2,i=−5 m/s.

The velocity of the center of mass vCM is:

vCM =10 kg · 12 m/s + 10 kg · (−5 m/s)

2 kg= 3.5 m/s

The initial velocities in the frame of the center of mass frame of reference are obtained bysubtracting the velocity of the center of mass:

u1,i = 8.5 m/s, u2,i = −8.5 m/s.

Since the collision is elastic, the final velocites in the center of mass frame of reference areobtained from the initial velocities (in the same frame) by flipping the sign:

u1,f = −8.5 m/s, u2,f = 8.5 m/s.

We finally find the final velocities in the original frame of reference by adding the velocity ofthe center of mass:

v1,f = -5 m/s

v2,f = 12 m/s

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a.2 [1 pts] Perfectly elastic collision; m1=20 kg; m2=10 kg; v1,i=12 m/s; v2,i=−5 m/s.

The velocity of the center of mass vCM is:

vCM =20 kg · 12 m/s + 10 kg · (−5 m/s)

2 kg= 6.3 m/s

The initial velocities in the frame of the center of mass frame of reference are obtained bysubtracting the velocity of the center of mass:

u1,i = 5.7 m/s, u2,i = −11.3 m/s.

Since the collision is elastic, the final velocites in the center of mass frame of reference areobtained from the initial velocities (in the same frame) by flipping the sign:

u1,f = −5.7 m/s, u2,f = 11.3 m/s.

We finally find the final velocities in the original frame of reference by adding the velocity ofthe center of mass:

v1,f = 0.6 m/s

v2,f = 17.6 m/s

a.3 [2 pts] Perfectly inelastic collision; m1=20 kg; m2=10 kg; v1,i=12 m/s; v2,i=−5 m/s.

The velocity of the center of mass vCM is:

vCM =20 kg · 12 m/s + 10 kg · (−5 m/s)

2 kg= 6.3 m/s

Both balls move with the velocity of the center of mass after the collision.

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(b) [5 pts] A PHY101 student on winter break snowboards down the backside of SphericalHalf Dome, starting form rest on the top of the Dome. At what angle does she becomeairborne?

R

R

R

The take off condition requires the normal force to be null. At takeoff, the only force providingthe centripetal acceleration is the component of the weight force in the radial direction:

mg cos θ = mv2

R

The kinetic energy can be obtained by applying the energy conservation principle between thestart of the slide and the point at of takeoff at the angle θ. Note the height at the start of theslide is R, and at the takeoff point is R cos θ.

mgR = mgR cos θ +12mv2,

Combining the two equations:

mgR = mgR cos θ +12mgR cos θ,

cos θ = 2/3

θ = 48◦.

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Problem 2: Ladder [10 pts]

A ladder of length L = 2.3m and mass M = 8.7 kg leans against a wall at angle θ = 27◦

to the vertical. What is the minimum coefficient µs of static sufficient friction between thefoot of the ladder and the floor to keep the ladder from slipping, supposing that there is nofriction between the ladder and the vertical wall?

Static equilibrium ⇔∑

F = 0 =∑

τ.

The force diagram is:

Since the wall exerts no vertical force on the ladder, the (upward) normal force NF of the flooron the ladder obeys:

NF = Mg (= 85.3 N).

The normal force NF is perpendicular to the floor, not to the ladder. Similarly, the normalforce NW of the wall on ladder is perpnedicular to the wall, not to the ladder. Because thereis no friction between the ladder and the wall there is no vertical force between the ladder andthe wall.

The force of friction at the floor is Ffriction ≤ µNF = µMg, which must balance the (horizontal)normal force NW of the wall on the ladder. If the coefficient of friction is minimal, we havethat:

NW = Ffriction = µminMg.

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For static equilibrium, the sum of the torques on the ladder must be zero, so if we calculatethe torques about the foot of the ladder, we have:

0 = Mg(L/2) sin θ −NW L cos θ,

noting that the force of gravity acts on the center of mass (CM) of the ladder, at distance L/2from its ends. Thus,

NW = (Mg/2) tan θ = Ffriction = µminMg (= 21.7 N),

and so,µmin = 0.5 tan θ = 0.255 (= NW /Mg = 21.7/85.3).

Taking torques about the CM of the ladder:

0 = NF (L/2) sin θ − (NW + Ffriction)(L/2) cos θ,

or:Mg sin θ = 2µminMg cos θ,

which also leads to µmin = 0.5 tan θ.

We could also calculate the torques about the upper end of the ladder:

0 = NF L sin θ − FfrictionL cos θ −Mg(L/2) sin θ,

or:Mg

2sin θ = µminMg cos θ,

which also leads to µmin = 0.5 tan θ.

We could even calculate the torques about the axis where the floor meets the wall:

0 = NF L sin θ −NW L cos θ −Mg(L/2) sin θ,

orMg

2sin θ = µminMg cos θ,

which also leads to µmin = 0.5 tan θ.

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Problem 3. Pulley.

A mass M = 7.5 kg is suspended from a pulley of mass m = 1.3 kg, radius r = 0.053 m,and moment of inertia I = 2mr2/3. A (massless) rope is attached to a roof beam and runaround the pulley, as shown in the figure. The tension T on the free end of the rope is 10%more than that needed to keep the system at rest.

a. [5 pts] What is the tension T?

Two segments of the rope hold the masses at rest, so Trest = (M + m)/2. Then,

T = 1.1(M + m)g/2 = 1.1(7.5 + 1.3)9.8/2 = 47.4 N.

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b. [5 pts] What is the relation between the angular acceleration α of the pulley and thevertical acceleration a of mass M?

α = a/r.

This is straightforward. However, if tension T is applied by your hand, then the verticalacceleration of your hand must be 2a not a.

c. [5 pts] What is the acceleration a of mass M?

The (upward) acceleration a of mass M is the same as the acceleration a of thecenter of mass of the pulley.Let T ′ be the tension in the portion of the rope that is connected to the roofbeam. Then, the vertical component of f = ma on the system of pulley plus massM is:

T + T ′ − (M + m)g = (M + m)a.

If we only analyze the acceleration of the pulley, we have:

T + T ′ − T ′′ −mg = ma,

while f = ma for mass M alone gives:

T ′′ −Mg = Ma,

so if we add these two equations we get the first equation, which is for masses mand M combined.The torque equation for the motion of the pulley is:

τ = (T − T ′)r = Iα =2mr2

3a

r,

so that:T − T ′ =

23ma.

Adding this equation to the 1st equation to eliminate T ′, we find:

2T − (M + m)g =(

M +53m

)a.

Inserting the relation T = 1.1(M + m)g/2, the acceleration of mass M is:

a =0.1(M + m)M + 5m/3

g =0.1(7.5 + 1.3)7.5 + 1.67 · 1.3

9.8 = 0.89 m/s2.

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Problem 4. Sledding.

Two girls, each of mass M = 20 kg, are sledding on identical plastic sleds each of massm = 2.0 kg. At the bottom of a hill, they find themselves sliding next to each other withoutfriction on a horizontal area at the same velocity v0 = 5.0 x m/s. Suddenly, child A leapsoff her sled and lands on the sled of child B. While she is in the air, the x-component of hervelocity is v1x = v0 = 5.0 m/s and the y-component is v1y = v0/2 = 2.5 m/s.

Note: in this problem, the x and y axis cover the horizontal area where the two sleds slideafter coming down the hill. The y axis is not vertical!

a. [8 pts] What is the final velocity v2 of A’s sled after she leaps off it?

The initial momentum of A and her sled is P0 = (M + m)v0x = 11mv0. The total horizontalmomentum of A and her sled just after she leaps is the sum of her momentum, P1 = Mv1 =10m(v0x + (v0/2) y) and the sled’s final momentum which we may write as P2 = mv2. Sincethere are no horizontal forces acting on the isolated system of A and her sled, momentum isconserved so that P0 = P1 + P2. In other words,

P2x = P0x − P1x = 11mv0 − 10mv0 = mv0,

so that v2x = v0 since the mass of the sled alone is just m. For the y-component we find

P2y = −P1y = −5mv0,

so that v2y = −5v0.

Thus, v2 = (5.0 m/s)x− (25.0 m/s)y.

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b. [7 pts] What is the final velocity vF of A, B and B’s sled after A lands on it? If you didnot get part a, leave your answer in terms of v2x and v2y as needed.

Here we can either consider the initial state to be that of A flying through the air with v1 whileB and her sled slide in the x direction, or we can consider the original initial state, in whichwe found above for one sled-plus-child P0 = 11mv0. We’ll do the latter, noting that there areno external horizontal forces acting on the isolated system comprising the two children and thetwo sleds. We then consider the total initial momentum (for two children and two sleds) to beP0,tot = 2P0 = 22mv0x.

Using the notation from above for A’s sled’s final momentum, we can write: P0,tot = P2 +Pf ,where the final momentum of the two children plus sled is Pf = (2M + m)vf = 21mvf .

Therefore:Pfx = 22mv0 − P2x = 22mv0 −mv0 = 21mv0 = (21m)vfx

so that vfx = v0. And for the y-component,

Pfy = −P2y = +5mv0 = (21m)vfy,

so that vfy = (5/21)v0.

Thus, v2 = (5.0 m/s)x + (1.2 m/s)y.

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Problem 5. Kayaking.

A man stands on a cliff, ready to toss a water bottle down to his wife. His wife paddles pastin a kayak. As she draws even with him, he tosses the bottle up and forward (by “forward”we mean here the direction of the motion of the kayak), from a height h = 2.6 m above her.At the same instant, the kayak speed is v0k = 5.0 m/s and beginning to slow, so that itsconstant acceleration is a = −2.0 m/s2. After the kayak has gone forward d = 4.0 m, thebottle lands in the kayak, as shown in the figure.

h

d

a. [5 pts] How long is the bottle in the air?

The time is entirely determined by how long it takes the kayak to get to d to receive the bottle!Solve the equation d = v0kt + 1

2at2. Then,

t =v0k ±

√v20k − 2|a|d|a|

=5.0 m/s±

√25 m2/s2 − 2(2.0 m/s2)(4.0 m)

2.0 m/s2= (2.5± 1.5) s.

We identify the appropriate time as t = 1.0 s. (The longer time solution would only result ifthe kayak decelerated until it was going backwards, and in that way returned to the spot adistance d from the starting place.)

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b. [5 pt] At what angle θ above the horizontal was the bottle launched?

We can find the angle from tan θ = v0y/v0x, if we can find the inital x- and y− compo-nents of the velocity of the bottle. We solve for v0x from d = v0xt, so that v0x = d/t =(4.0 m)/(1.0 s) = 4.0 m/s = v0x. Next we solve for v0y from the equation

−h = v0yt− 1/2gt2,

which can be rearranged to give

v0y =gt

2− h

t=

(9.81 m/s2)(1.0 s)2

− 2.6 m1.0 s

= 2.3 m/s .

Then we write

θ = tan−1

(v0y

v0x

)= tan−1

(2.34.0

)= 30◦.

Thus, θ = 30◦ .

c. [5 pts] What was the maximum height the bottle attained above the river?

Note that the maximum height does not occur at t/2, where t is the time for the bottle tomake its complete trip. Instead, we use the fact that at the ball’s maximum height, vy = 0.The kinematic variables in y that we know are then a = −g, v0y = 2.3 m/s, and vy = 0, andthe unknown variables are t′ and y′. We find t′ from

vy = v0y + at′ = v0y − gt′ = 0,

so that t′ = v0y/g = (2.3 m/s)/(9.81 m/s2) = 0.23 s. The height above the initial position thenfollows from

y′ = v0yt− 12gt2 = (2.3 m/s)(0.23 s)− (0.5)(9.81 m/s2)(0.23 s)2 = 0.27 m.

Finally then, the maximum height above the water is y′ + h = 2.6 m + 0.27 m = 2.9 m .

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Problem 6. Sliding Mass.

A mass m = 1.00 kg slides in a circle of radius r1 = 1.00 m on a frictionless table as shown.The mass is attached to a massless string which passes through a hole in the table and isheld by by an experimenter. The mass has an angular velocity ω1 = 10.0 rad/s.

mr1

a. [5 pt] What is the moment of inertia of the mass-string system?

I = mr21 = 1.00 kg ·m2.

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b. [5 pts] What is the tension in the string?

The tension supplies all of the centripetal acceleration and thus equals:

T1 = mω21r1 = 100 N

c. [5 pt] The experimenter now slowly pulls the string until the mass is at radius r2 =0.500 m. How much work is done on the mass-string system in this process?

As the force is always radial, it produces no torque about the center of the (instantaneous)orbit which is always at the hole in the table. Thus angular momentum is conserved. Hence,

L1 = I1ω1 = mr21ω1 = L2 = I2ω2 = mr2

2ω2

whence:

ω2 = ω1r21

r22

= 40 rad/s.

The increase in kinetic energy is:

∆KE =12I2ω

22 −

12I1ω

21 =

12m

(r22ω

22 − r2

1ω21

)=

12m

(r22ω

21

r41

r42

− r21ω

21

)=

12mr2

1ω21

(r21

r22

− 1)

.

The increase in kinetic energy comes from the work done on the mass-string system by theexperimenter pulling on the string. This can be rewritten as:

W = ∆KE =12mr2

1ω21

[r21

r22

− 1]

=121 kg (1 m)2 (10 rad/s)2

[(1 m

0.5 m

)2

− 1

]= 150 J .

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d. [5 pts] The string will break if its tension exceeds 9.00 × 102 N. What is the smallestradius rmin to which the mass can be safely pulled by the experimenter?

By our earlier reasoning the angular velocity at radius r is:

ω(r) =r21

r2ω1.

Hence the tension required to supply the centripetal acceleration is:

T = mω2(r)r = mω21

r41

r4r = mω2

1

r41

r3= mω2

1r1r31

r3= T1

r31

r3= 100 N

r31

r3,

We can now solve for r by requiring that T = 900 N. We obtain:

r = r1

(T1

T

)1/3

= r1

(100 N900 N

)1/3

= 0.481 m ,

which is then the farthest in that the mass can be pulled without endangering those nearby.

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Problem 7. Sliding Block.

A mass m=1.00 kg slides down the edge of the block of mass M=8.00 kg shown in the figure.The block of mass m slides down from its initial position, located at an height h=0.100 mabove the ground. There is no friction between the two masses m and M . The two massesare initially at rest on a frictionless table.

m

Mh

a. [5 pt] What are the velocities of m and M right after m reaches the table?

Let v be the speed of m to the right and V be the speed of M to the left after m slides off M .Conservation of energy and momentum lead to the equations:

mv + M(−V ) = 0 and mgh =12mv2 +

12MV 2,

which imply:V = −m

Mv.

Substituting into the energy derived from the energy conservation principle, we obtain:

v =

√2gR

1 + (m/M)=

√2 · 9.8 ms−2 · 0.1 m

1 + 1 kg/8 kg= 1.32 m/s .

Then:V = −m

Mv = −1 kg

8 kg1.32 m/s = - 0.165 m/s,

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b. [5 pts] Subsequently, m slides further on the table and makes an elastic collision with awall, which is of course at rest and unmovable. What is the final velocity of m?

The mass m collides elastically with an “infinite” mass. The velocity of the center of massframe of reference is zero, before and after the collision: i.e., the system of the observer isalready the center of mass frame of reference!

Therefore, the velocity of mass m is simply reversed:

v2 = −v = - 1.32 m/s .

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c. [5 pt] Show that m catches up with its old friend M. What is the highest point on M ’sback that m climbs up to now?

As m starts back to the left with a speed v2 = −v = −1.32 m/s it will overtake M which isahead but is trudging along at a mere speed −v m

M = −0.165 m/s. After reaching M it willstart to climb up its back again. At the highest point, whose height above the table we willdenote by h′, m will be at rest relative to M , i.e. they will both be moving at the same velocityu towards the left. Our favorite conservation laws applied before m reaches M and at thispoint of maximum height imply that:

M(−v

m

M

)−mv = (M + m)u,

and therefore:u = −v

2m

m + M.

Applying the principle of energy conservation we obtain:

12M

(−v

m

M

)2

+12m (−v)2 =

12

(M + m) u2 + mgh′,

which becomes:12mv2

(1 +

m

M

)=

12

(M + m)(− 2mv

m + M

)2

+ mgh′,

and:mgh′ =

12mv2

(1 +

m

M

)− 1

2mv2 4m

m + M,

Therefore:

h′ =v2

2g

(1 +

m

M− 4m

m + M

)=

v2

2g

(1 +

18− 4

9

)=

4972

v2

2g=

4972

(1.32 m/s)2

2 · 9.8 ms−2= 0.061 m .

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POSSIBLY USEFUL CONSTANTS AND EQUATIONSYou may want to tear this out to keep at your side

L = Iω I = Σmir2i x = x0 + v0t + at2/2

PE = mgh KE = 12Iω2 KE = 1

2mv2

ω = ω0 + αt ω2 = ω20 + 2α∆θ ∆θ = ω0t + 1

2αt2

v = v0 + at ~F∆t = ∆~p F = −GMm/r2

F = µN s = Rθ τ = F` sin θΣτ = Iα v = Rω ~p = m~vac = v2/r W = Fs cos θ v2 = v2

0 + 2a∆xWnc = ∆KE + ∆PE a = Rα I = 1

2mr2 [disk]

I = 25mr2 [sphere] I = 1

3Mw2 [thin sheet] L = mvr

REarth = 6400 km MEarth = 6.0× 1024 kg G = 6.67× 10−11 Nm2/kg2