Chapter 6, Problem 27P

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# A 65.0-kg person throws a 0.045 0-kg snowball forward with a ground speed of 30.0 m/s. A second person, with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.50 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

To determine
The velocity of the thrower and catcher.

Explanation

Given Info: Mass of the thrower is mthrower=65.0Â kg , Mass of the snowball is mb=0.0450Â kg ,

Â Velocity of snowball is vb=30.0Â m/s and initial velocity of thrower is (vthrower)i=2.50Â m/s , mass of the catcher is mcatcher=60.0Â kg and initial velocity of catcher is (vcatcher)i=0

Explanation:

Apply conservation of momentum to ball -thrower system is

(mthrowervthrower+mbvb)i=[(mthrower+mb)vthrower]f

• mthrower is the mass of the first person throws a snowball.
• vthrower,i is the initial velocity of the first person before throws a snowball.
• mb is the mass of the snowball.
• vb is the initial velocity of the snowball.
• vthrower,f is the final velocity of the first person after throws a snowball.

Substitute 65.0â€‰kg for mthrower , 2.50â€‰m/s2 for (vthrower)i , 0.450â€‰kg for mb , 30.0â€‰m/s for vb in equation. (1)

(65.0â€‰kg)(2.50â€‰m/s)+(0.0450â€‰kg)(30.0â€‰m/s)=(65.0â€‰kg+0.0450â€‰kg)vthrower162.5â€‰kgâ‹…m/s+1.35â€‰kgâ‹…m/s=(65.045â€‰kg)vthrower163

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