physics-40a-final-exam-review

1. Physics 40A - Final Exam Solutions - Winter 2013
Problem 1 [25 points]
Two boxes of masses m1 = 1.65 kg and m2 = 3.30 kg slide down an inclined plane (with  = 30°)
while attached by a massless rod as shown. The coefficient of kinetic friction between box 1 and
the incline is 1 = 0.226; that between box 2 and the incline is 2 = 0.113. (a) Draw free body
diagrams for box 1 and box 2. Compute (b) the tension in the rod, and (c) the common
acceleration of the two boxes.
Solution:
(a) See diagrams above. T is the magnitude of the force in the rod, N1 and N2 are the normal
forces on boxes 1 and 2, f1 and f2 are the kinetic friction forces on boxes 1 and 2.
(b) For each block we take +x down the incline and +y in the direction of the normal force.
Applying Newton's second law to the x and y directions we obtain two equations for each box:
m1 g sin − f 1  T = m1 a 1
N 1 − m1 g cos  = 0 2
m2 g sin  − f 2 − T = m2 a 3
N 2 − m2 g cos = 0 4
Dividing Eq. (1) by m1 and dividing Eq. (3) by m2 gives
g sin − f 1/m1  T /m1 = a and g sin  − f 2 /m2 − T /m2 = a
Subtracting gives
−
f 1
m1

f 2
m2

T
m1

T
m2
= 0 ⇒ T
m1  m2
m1 m2
 =
m2 f 1 − m1 f 2
m1 m2
⇒ T =
m2 f 1 − m1 f 2
m1  m2
Using f 1 = 1 N 1 = 1 m1 g cos and f 2 = 2 N 2 = 2 m2 g cos we get
T =
m2 1 m1 g cos − m1 2 m2 g cos 
m1  m2
=
m1 m2 g cos1 − 2
m1  m2
=
1.65 kg3.30 kg9.8 m/s
2
cos 30
°
0.26 − 0.113
1.65 kg  3.30 kg
T = 1.05 N
(c) Substituting for f1 in Eq. (1) we ge
a = g sin − 1 g cos  
T
m1
= 9.8 m/s2
sin30°
− 0.2269.8 m/s2
cos 30°

1.05 N
1.65 kg
a = 3.62 m/s2
1

2. Physics 40A - Final Exam Solutions - Winter 2013
Problem 2 [25 points]
A pulley of rotational inertia 0.385 kg·m2 and radius 33 cm is mounted on a horizontal axle. A
bucket of mass 1.53 kg hangs from a massless cord which is wrapped around the rim of the
pulley. A frictional torque of 1.1 N·m acts on the pulley at the axle. Assume that the cord does
not slip. (a) Calculate the angular acceleration of the pulley and the linear acceleration of the
falling bucket. (b) Determine the angular velocity of the pulley and the linear velocity of the
bucket at time t = 3.0 s if the pulley (and bucket) start from rest at t = 0.
Solution:
(a) Suppose the pulley rotates counterclockwise due to the torque caused by the tension in the
cord. Then the frictional torque would tend to oppose the motion, i.e. it would tend to make it
rotate clockwise. Therefore, Newton's second law applied to the rotating pulley gives
T R − f = I 
Newton's second law applied to the bucket gives
T − m g = m−a
⇒ T = m g − a
Because the cord does not slip, the tangential component of the acceleration of a point P at the
rim of the pulley is equal to the acceleration of the bucket. Therefore, since P moves in a circle,
a = R
Now we have
I  = T R − f = mg − a R − f = mg − R R − f
Solving for  gives
 =
m g R − f
I  mR2
 =
1.53 kg9.8 m/s2
0.33 m − 1.1 N⋅m
0.385 kg⋅m2
 1.53 kg0.33 m2
= 6.98 rad/s2
Now we obtain the linear acceleration
a = R = 0.33 m6.98 rad/s2
 = 2.30 m/s2
(b) Since the angular acceleration is constant, after 3.0 s
 = 0   t = 0  6.98 rad/s3.0 s = 20.9 rad/s
The velocity of the bucket is the same as that of point P on the rim of the pulley:
v =  R = 20.9 rad/s0.33 m = 6.90 m
2

3. Physics 40A - Final Exam Solutions - Winter 2013
Problem 3 [25 points]
In the Figure below, a 2.5 kg block slides head on into a spring with a spring constant of 320
N/m. When the block stops it has compressed the spring by 7.5 cm. The coefficient of kinetic
friction between the block and the horizontal surface is 0.25. While the block is in contact with
the spring and being brought to rest, what are (a) the work done by the spring force, and (b) the
increase in thermal energy of the block-floor system? (c) What is the block's speed just as the
block reaches the spring?
Solution:
(a) The work done by the spring force is
W = ∫xi
xf
F dx = ∫xi
xf
−kxdx =
[−
1
2
k x2
]xi
xf
=
1
2
k xi
2
− x f
2

With xi = 0 and x f = 7.5 cm = 0.075 m we find
W =
1
2
320 N/m0 − 0.075 m2

W = −0.9 J
(b) The increase in thermal energy of the block-floor system is
 Eth = f k d
where fk is the frictional force and d = 7.5 cm is the distance moved during the time the block is
in contact with the spring and being brought to rest.. The frictional force on the block from the
floor is
f k = k N
and Newton's second law applied to the block in the vertical direction is
N − mg = 0
Therefore,
f k = k mg
and we find
 Eth = f k d = k m g d = 0.252.5 kg 9.8 m/s2
0.075 m
 Eth = 0.46 J
(c) Consider the motion of the block from the moment it makes contact with the spring (when its
speed is v) to the moment when it compresses the spring and comes to rest. There are no external
forces acting on the spring-block-floor system in the horizontal direction, so from conservation of
energy, we find
 Emec   Eth =  K  U   Eth = 0 ⇒  K = −U − Eth
0 −
1
2
mv2
= −
1
2
k d 2
− 0 − 0.46 J
1
2
mv2
=
1
2
k d 2
 0.46 J =
1
2
320 N/m0.075 m2
 0.46 J = 1.36 J
v =
21.36 J 
2.5 kg
v = 1.04 m/s
3

4. Physics 40A - Final Exam Solutions - Winter 2013
Problem 4 [25 points]
A steel ball of mass 0.5 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The
ball is then released when the cord is horizontal (see Figure). At the bottom of its path the ball
strikes a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find
(a) the velocity of the ball and (b) the velocity of the block both just after the collision.
Solution:
(a) First we find the speed of the ball (of mass m1) right before it hits the block. When the cord is
horizontal the ball is at height h and has zero velocity. Juts before the ball hits the block it has
zero height and has speed v. From conservation of mechanical energy we find
m1 g h =
l
2
m1 v2
⇒ v = 2gh = 29.8 m/s
2
0.7 m = 3.7 m/s
We now treat the elastic collision and find the velocity of the ball just after the collision using
v1f =
m1 − m2
m1  m2
v =
0.5 kg − 2.5 kg
0.5 kg  2.5 kg
3.7 m/s
v1f = −2.47 m/s
i.e. the ball moves to the left with speed 2.47 m/s.
(b) The velocity of the block just after the collision is given by
v2f =
2m1
m1  m2
v =
20.5 kg
0.5 kg  2.5 kg
3.7 m/s
v2f = 1.23 m/s
i.e. the block moves to the right with a speed of 1.23 m/s.
4