50:750:203 General Physics Name___________________
Test 2. July 1999
Attempt all questions. You must show some working. Credit will not be given for an answer without working.
1. A 120-kg crate rests on the flatbed of a truck that moves at a speed of 18.0 m/s around an unbanked curve whose radius is 66.0m. The crate does not slip relative to the truck. Obtain the magnitude of the static frictional force that the truck bed exerts on the crate
The centripetal force on the truck is
FC = mv2/r = (120 kg)(18.0 m/s)2/(66.0 m)=589 N
The only horizontal force acting on the crate is the frictional force, so this is its magnitude.
2. (Conceptual question) A person is riding on a Ferris wheel. When the wheel makes one complete turn, is the net work done by the gravitational force positive, negative, or zero? Justify your answer.
The work done by the gravitational force depends only on the difference between the initial and final heights. In a complete turn, the person returns to the initial position, so the net work done is zero. Gravity does work during one half of the path, and work is done against gravity (by the drive system) during the other half.
3. A cyclist approaches the bottom of a gradual hill at a speed of 12 m/s. The hill is 4.5 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Ignoring air resistance and friction, find the speed at which the cyclist crests the hill.
Use conservation of mechanical energy. The cyclist has initial kinetic energy (½)mv02. At the top of the hill, she has gravitational potential energy, mgh, and the remaining kinetic energy is (½)mv2. So
(½)mv02 = (½)mv2 + mgh
Notice that the "m" cancels. Then
(½)(12 m/s)2 = (½)v2 + (9.8)(4.5 m)
From this, v is found to be 7.47 m/s.
4. With the engines off, a spaceship is coasting at a velocity of +210 m/s through outer space. It fires a rocket straight ahead at an enemy vessel. The mass of the rocket is 1200 kg, and the mass of the spaceship (not including the rocket) is 4.0×106 kg. The firing of the rocket brings the spaceship to a halt. What is the velocity of the rocket?
You must use conservation of momentum for this. Mechanical energy is not conserved because of the burning of fuel, which converts chemical energy into mechanical. Momentum conservation is valid even when energy conservation is not.
Initially, the spaceship and rocket together have momentum
p = (4.0×106 + 1200 kg)(210 m/s) = 8.40×108 kg-m/s
(Notice that the mass of the rocket is pretty much negligible in this part)
Finally, the spaceship is at rest, so this momentum is all carried by the rocket. Its velocity is therefore
v = (8.40×108 kg-m/s)/(1200 kg) = 7.00×105 m/s
5. A 55.0-kg person, running horizontally with a velocity of +4.0m/s, jumps onto a 12.0-kg sled that is initially at rest.
(a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 40.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
(a) This is an example of a sticking-together collision, highly inelastic. You must use momentum conservation. The 55.0-kg person has initial speed of 4.0 m/s. The final combination of person and sled has a mass of (55.0 + 12.0) kg. The speed of the combination is v, where
(55.0 kg)(4.0 m/s) = (55.0 kg + 12.0 kg)v
v = 55.0×4.0 / 67.0 = 3.28 m/s
(b) There are several ways to do this. The person-sled combination has kinetic energy
(½)mv2 = (½)(67.0 kg)(3.28 m/s)2 = 361 J
In bringing the sled to rest, the work done against the frictional force must equal this. Since the distance the sled travels is 40.0 m, the frictional force must be
(361 J)/(40.0 m) = 9.03 N
The frictional force is
f = µFN = µmg, so µ = (9.03 N)/(9.8 × 67.0 kg) = 0.0138
6. The angular velocity of a fan in an air conditioning unit increases to +240 rad/s from +160 rad/s in 12 revolutions. What is the angular acceleration in rad/s2.
Use the rotational equation analogous to v2 = v02 + 2ax:
w2 = w02 + 2aq
The final angular velocity, w , is 240 rad/s. The initial value is 160 rad/s. The angle turned through is 12 revolutions. You must convert this into the same units for angle. One revolution is 2p radians, so the total angle is 12×2 p. Then
a = ((240)2 - (160)2)/(2×24p) = 212 rad/s2
7. Three objects are situated on the x-axis. Their weights and positions are: 45.0 N at x = +2.40 m, 19.0 N at x = -0.800 m, and 29.5 N at x = -1.80 m. Where on the x-axis is (a) the center of gravity and (b) the center of mass of this group of objects?
The position of the center of gravity is given by
xCG = (w1x1 + w2x2 + w3x3)/(w1 + w2 + w3)
=(45.0×(+2.40)+19.0×(-0.800)+29.5×(-1.80))/(45.0+19.0+29.5)
= +0.425 m.
(There is no need to move the origin of x. Just keep track of the signs as they are given.)
(b) While you could calculate the masses, and then find the center of mass in the same way, the answer must be the same.
8. A baseball is thrown such that the translational speed of its center of mass is 33 m/s, and its angular speed about the center of mass is 160 rad/s. Treat the baseball as if it were a uniform solid sphere of radius 3.7 cm. What fraction of the ball's total kinetic energy is rotational kinetic energy? (For a solid sphere, I = (2/5)MR2.)
You are not given the mass of the ball, so call it "m" and expect it to cancel. The value 160 rad/s is the rotational speed of the ball, . 3.7 cm is .037 m.
Translational kinetic energy = ½mv2 = ½ m (33)2 = 544.5m
Rotational kinetic energy = ½I2 = (½)(2/5)m(.037)2(160)2 = 7.01m
As a fraction of the total, the rotational energy is
7.01m / (544.5m + 7.01m) = .0127 (and the m does cancel).