Page 112 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 8] MOMENTUM 97

We now set equal the two expressions for v and solve for v :

2
1
−3kg·m/s − v kg

1
v + 6.3 m/s =
1
2kg
−3kg·m/s

v (1 + 0.5) = − 6.3 m/s =−7.8 m/s
1
2kg

v =−5.2 m/s
1

The minus sign means that the 1-kg ball moves off in the opposite direction from its initial one. To find v ,itis
2

simplest to use v = v + 6.3 m/s:
2
1
v =−5.2 m/s + 6.3 m/s = 1.1 m/s

2
The 2-kg ball also reverses its direction as a result of the collision.
Multiple-Choice Questions
8.1. A 300-g iron ball has the same diameter as a 105-g aluminum ball. The balls are dropped at the same time from a
cliff. Just before they reach the ground, they have the same
(a) acceleration (c) kinetic energy
(b) momentum (d) potential energy
8.2. A ball with the momentum p strikes a wall and bounces off. The change in the ball’s momentum is ideally
(a) 0 (c) 2p
(b) p (d) −2p
8.3. An elastic collision conserves

(a) momentum but not KE (c) both momentum and KE
(b) KE but not momentum (d) neither momentum nor KE

8.4. An inelastic collision conserves
(a) momentum but not KE (c) both momentum and KE
(b) KE but not momentum (d) neither momentum nor KE

8.5. A 60-kg skater pushes a 50-kg skater, who moves away at 2.0 m/s. As a result, the first skater moves backward at
(a) 0.6 m/s (c) 2.0 m/s
(b) 1.7 m/s (d) 2.4 m/s
8.6. A 60-g tennis ball moving at 8.0 m/s strikes a stationary tennis racket perpendicularly and bounces off at 6.0 m/s.
The impulse given to the racket is
(a) 0.12 N·s (c) 0.48 N·s
(b) 0.36 N·s (d) 0.84 N·s

8.7. During a serve, a tennis racket exerts an average force of 250 N on a 60-g tennis ball, initially at rest, for 5.0 ms
(0.0050 s). The ball’s KE afterward is
(a) 0.78 J (c) 13 J
(b) 1.25 J (d) 127 J

8.8. A 1500-kg truck whose velocity is 60 km/h overtakes a 4000-kg truck moving in the same direction at 35 km/h. The
trucks collide and stick together, and the initial velocity of the wreckage is
(a) 9.1 km/h (c) 48 km/h
(b) 42 km/h (d) 53 km/h