# Sample Midterm I Key

Physics 135-1 Midterm, Spring 2014
Show all work for partial credit
Name ________________________________
1) (10 points) Suppose a ferris wheel with a radius of
3.5 meters is rotating clockwise at a rate of 5 revolutions
per minute. The bottom of the wheel is elevated 1.75 m
above the ground, as shown. Then, just as she reaches the
very top of the wheel, a lady accidently drops her teddy
bear. At what horizontal distance from the center of the
wheel will the teddy bear hit the ground?
Solution
The ferris wheel is rotating at 5 / 60 = 0.08333 Hz, or
 = 2f = (2)(3.14159)(0.08333) = 0.5236 rad/s. This
means that the teddy bear will leave the ferris wheel with a
tangential (horizontal) velocity of v = R = (0.5236)(3.5) = 1.83 m/s.
The teddy bear will fall (2)(3.5) + 1.75 = 8.75 m to the ground, and we can find out how long it falls
from d = ½ gt2. That is, t = (2d/g)1/2 = [(2)(8.75) / 9.8]1/2 = 1.34 s. Therefore the teddy bear will
drift (1.83)(1.34) = 2.45 m to the right from the center of the ferris wheel.
2) Two people are pulling on a barge in a river with ropes.
One is pulling at 34° south of horizontal, and the other is
pulling at 45° north of horizontal. The barge itself is
moving only along the x-axis (horizontal). It has a mass of
200 kg.
2a) (5 points) If the barge is resisting their pull with
180 N of frictional force, and is gently accelerating at
0.2 m/s2, what is the tension in the north and south ropes?
2b) (5 points) Assuming the barge starts at rest, how much work (in joules) will the two people
have expended after pulling on it for one minute in this way?
Solutions
2a) The y-components of the tensions in the two ropes must balance, because the barge is not
moving along the y-axis. Hence TN sin(45°) = TS sin(34°). The combined force of the two tensions
along the x-axis must equal the friction plus “ma”, or 180 N + (200 kg)(0.2 m/s2) = 220 N. Hence
we have TN cos(45°) + TS cos(34°) = 220 N. There are several ways to do the algebra from this
point; we will set TN = TS sin(34°) / sin(45°) = 0.7908 TS to start, then substitute back into the
second equation: 0.7908 TS (0.7071) + TS (0.8290) = 220, or TS = 158.5 N. This then yields
TN = (0.7908)(158.5 N) = 125.3 N.
2b) The barge is accelerating at 0.2 m/s2, so after one minute it will have a velocity of v = at =
(0.2)(60) = 12 m/s. This corresponds to a kinetic energy of E = ½ (200 kg)(12 m/s)2 = 14,400 J. In
that same time it will have moved d = ½ at2 = ½ (0.2)(60)2 = 360 m. This corresponds to a work
expended of W = Fd = (180 N)(360 m) = 64,800 J. So, the total work done is 79,200 J.
3a) (6 points) A rock A of mass 1 kg is
moving at 40 m/s along the positive x-axis.
Then it collides with another rock B of equal
mass and ricochets away at a positive 30°
relative to the x-axis. The second rock B
moves away at a negative 45° relative to the
x-axis. How fast is each rock moving after the
collision?
3b) (4 points) How much heat is generated by this collision?
Solutions
3a) Conservation of momentum along the y-axis means: (1 kg) vA sin(30°) = (1 kg) vB sin(45°),
and similarly along the x-axis means: (1 kg)(40 m/s) = (1 kg) vA cos(30°) + (1 kg) vB cos(45°).
The y-equation means that vA = 1.414 vB, and substitution of this into the x-equation gives
40 = 1.414 vB (0.866) + vB (0.707), or vB = 20.7 m/s. Thus vA = 29.3 m/s.
3b) The kinetic energy of rock A before the collision is E = ½ (1 kg)(40 m/s)2 = 800 J. The
energies of the rocks after the collision are EA = ½ (1 kg)(29.3 m/s)2 = 429 J and
EB = ½ (1 kg)(20.7 m/s)2 = 214 J, so by conservation of energy the heat released must be
800 – 429 – 214 = 157 J.
4) (10 points) A sphere of mass M = 5 kg and
radius R = 40 cm has a cord tied around its
equator, which is also attached to a small mass
of m = 1.5 kg. The system is originally at rest.
Then the small mass is released and allowed to
fall. After it has fallen 50 cm, how fast (in
radians per second) will the sphere be rotating?
There is no friction in the system, and the
pulley has no mass.
Solution
The gravitational potential released by the falling weight must equal the kinetic energies of the
falling mass and the rotating sphere, so mgh = ½ mv2 + ½ I 2. We can relate v to  with the
formula v = R, and I for a sphere is 2/5 MR2. Thus mgh = ½ m2R2 + ½ (2/5 MR2)2. Inserting
numbers: (1.5)(9.8)(0.5) = (0.5)(1.5)(0.4)22 + (0.2)(5)(0.4)22, or  = (7.35 / 0.28)1/2 = 5.12 rad/s.
5) Assume that a frictionless cart of mass 4m is at the
origin on a one-dimensional track. Assume that
another cart of mass 2m is at x = 1 m and a third cart
of mass m is at x = 2 m. The first cart is moving at
v0 = 3 m/s; the other two carts are initially motionless.
5a) (5 points) Where is the center of mass of this system at t = 0?
5b) (5 points) Assume that the first cart collides with and sticks to the second cart, and then the
second cart collides with and sticks to the third cart. Where is the center of mass of this system at
t = 5 sec?
Solutions
5a) Using the center of mass formula: xCM = [4m(0) + 2m(1) + m(2)] / (7m) = 0.571 meter
5b) After all the carts have collided and are stuck together, then conservation of momentum tells us
that they must be moving at: (4m)(3 m/s) = (7m)v, or v = 1.714 m/s. That is, their CM must also
be moving at 1.714 m/s. But, the motion of the CM cannot be affected by collisions between the
different parts of the system, so this means that the CM was moving at 1.714 m/s before any of the
collisions even happened and indeed, must continue to move at this speed regardless of whether the
carts continue to stick together in the future or not! (Don’t you love physics?)
So, at t = 5 s the CM must be at 0.571 m + (1.714 m/s)(5 s) = 9.14 m.