Problem 1: e. Not enough information.. We do not know the stopping distance. If you have short hair and had your knees locked, then the stopping distance is short and the force is large. If you have lots of hair and your knees were bent, then the stopping distance is larger and the force is smaller.
Problem 2: a) p = mv so v = p/m = 2*10^4 kg m/s / 1000 kg = 20 m/s
b) KE = 1/2 mv^2 = 1/2 (1000 kg) (20 m/s)^2 = 2*10^5 J
c)
Impulse = delta p = Ft.--> F = delta p / t = 2*10^4 kg m/s / 5 s =
4*10^3 N
Problem 3: This problem was misprinted. The 1 kg glider is moving
at 2 m/s to the right (or it would never hit the other one).
p_after = p_before = p1 + p2 = m1*v1 + m2*v2 = (1 kg * 2 m/s) + (2
kg * 1 m/s) = 4 kg m/s
Problem 4: This is an energy problem. The kinetic energy of the car is reduced to zero by the work done by the brakes. If the car is going 3 times faster, then it has nine times as much kinetic energy (since KE = 1/2 mv^2). Therefore, the brakes need to do nine times more work. The force is the same. Therefore, since W = Fd, the distance must be 9 times farther. Therefore d = 9 * 10 m = 90 m.
Problem 5: The diver has potential energy but no kinetic energy when she is standing on the board. As she falls, her potential energy is converted to kinetic energy. She starts with PE = mgh = weight * h = 500 N * 10 m = 5000 J. Therefore, she ends up with 5000 J of kinetic energy just before she hits the water. Note that the problem gives the diver's weight, not her mass.
Problem 6: (b) Twice as much PE . PE = mgh. If you double h, then PE doubles.
Problem 7: (a) doubles. Its rotational speed does not change. Since the distance from the center doubles, its linear speed must double (linear speed is proportional to rotational speed * radius). Note that angular momentum is NOT conserved in this problem since it explicitly states that rotaional speed is kept constant (presumably by the motor on the phonograph).
Problem 8: (b) closer to the rim. Rotational inertia is largest when the mass is farthest from the axis of rotation.
Problem 9: (a) double. Torque = Force * lever arm. If you double the lever arm, then the torque will double.
Problem 10: (b) directed toward the center of the circle. You need a centripetal force to make the car go in a circle. If there is no force, the car will go at constant velocity (ie: in a straight line at constant speed).
Problem 11: (a) increases. Angular momentum is conserved. By moving her arms inward, she decreases her rotational inertia, therefore her rotational speed must increase to compensate.
Problem 12: (a) increase. Apparent weight = mv^2/R. If you increase v, the apparent weight will increase.
Problem 13: (e) other. If you increase the distance, the force will DECREASE. Since F ~ 1/R^2, if you double the distance, the force will decrease by a factor of 4.
Problem 14: (b) decrease to half your original weight. Your weight is the gravitational force between you and the Earth. This force F = GMm/R^2. If M decreases by a factor of 2, then so will F.
Problem 15: (c) do not have any normal force acting on them. Passengers in the space shuttle feel weightless because they are forever falling. The only force acting on them is gravity. There is no normal force. The normal force is what we perceive as weight.
Problem 16: (a) it has a large tangential velocity. The moon is in free fall. Its acceleration is always toward the Earth. Like all other objects in orbit around the Earth, it misses the Earth because it has a large tangential velocity.
Problem 17: (b) all parts of it are practically the same distance from the moon. There actually are tides, but they are so small they can be neglected. Remember that the strength of the tidal forces increases with the size of the object. Therefore, the Earth (size = 12000 km) experiences tides from the moon. Your swimming pool (size = 10 m) will experience tides 1 million times smaller than the Earth will.