Problem 7-2. A freight train moves due north with speed of 1.4 m/s. The mass of the train is 4.5 × 105 kg. How fast would an 1800-kg automobile have to be moving due north to have the same momentum as the train?
The momentum of the freight train is mv = (4.5 × 105 kg)(1.4 m/s) = 6.3 × 105 kg-m/s. In order to have a momentum of the same magnitude, an automobile of mass 1800 kg would need to have a speed of 6.3×105 kg-m/s ÷ 1800 kg = 350 m/s. (About 800 mph!)
Problem 7-10. A basketball (m = 0.60 kg) is dropped from rest. Just before striking the floor, the magnitude of the basketball's momentum is 3.1 kg-m/s. At what height was the basketball dropped?
(You could do this is several different ways) If the momentum of the basketball is 3.1 kg-m/s, its speed is 3.1 kg-m/s ÷ 0.60 kg = 5.17 m/s. In free fall, it would have this speed after falling a distance y = v2/2g = (5.17 m/s)2/(2×9.8 m/s2) = 1.36 m.
Problem 7-19. A 1550-kg car, traveling with a velocity of +12.0 m/s, plows into a 1220-kg stationary car. During the collision, the two cars lock bumpers and then move together as a unit. (a) What is their common velocity just after the impact? (b) What fraction of the initial kinetic energy remains after the collision?
This is what I call a sticking-together collision. The initial momentum of the 1550-kg car is 1550 kg × 12.0 m/s = 18600 kg-m/s. Since the other car is stationary, this is the total initial momentum. Momentum is conserved in the collision, so this is also the final momentum. Both cars, with a total mass of 1550 + 1220 = 2770 kg move at the same speed. In order for the momentum to have the correct value, the speed is 18600 kg-m/s ÷ 2770 kg = 6.71 m/s.
The initial kinetic energy (all due to the 1550-kg car) is ½(1550 kg)(12.0 m/s)2 = 111600 J. The final kinetic energy (use the total mass here) is ½(2770 kg)(6.71 m/s)2 = 62448 J. This is a fraction 62448/111600 = 0.559 of the initial energy.
Problem 7-30. A cue ball (mass = 0.165 kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick which applies an impulse of +1.50 N-s to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.
The velocity of the cue ball immediately after it is hit is found by setting the impulse of the force equal to the momentum of the ball (since it was initially at rest). Then its velocity is (1.50 N-s)/(0.165 kg) = 9.09 m/s. The collision is an elastic collision between two objects, one of which is initially at rest. This is the type of collision described by equations 7.8a and 7.8b. To find the velocity of the second ball in an unthinking way, use equation 7.8b
You should then remember that in a collision of this type between two objects of equal mass, all the momentum is transferred from the first object to the second.
Problem 7-41. The earth and moon are separated by a center-to-center distance of 3.85 × 108 m. The mass of the earth is 5.98 × 1024 kg and that of the moon is 7.35 × 1022 kg. How far does the center of mass lie from the center of the earth?
Choose to measure all distances from the center of the earth. Then
The radius of the earth is 6.38×106 m, so the center of mass is inside the earth.