A 70.0-kg person throws a 0.0400-kg snowball forward with a ground speed of 33.0

Physics
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Jun 10th, 2015

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Jun 10th, 2015

A 70.0-kg person throws a 0.0400-kg snowball forward with a ground speed of 33.0 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 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.

needs 

thrower.....

catcher......


Jun 10th, 2015

for the thrower:

(m1 + ms)v1 = m1v2 + msvs 

m1 = mass of the thrower (70 kg) 
ms = mass of the snowball (.04 kg) 
v1 = initial velocity of the thrower + snowball (2.2 m/s) 
v2 = final velocity of the thrower 
vs = final velocity of the snowball (33 m/s) 

thus v2=(70.04*2.2-0.04*33)/70=2.18m/s

for the catcher:

msvs + m2v3 = (ms + m2)v4 

where 
ms = mass of the snowball (.04 kg) 
vs = initial velocity of the snowball (33 m/s) 
m2 = mass of the cather (60 kg) 
v3 = initial velocity of the catcher (0 m/s) 
v4 = final velocity of the snowball and catcher 

thus v4=(0.04*33+60*0)/(0.04+60)=0.022m/s

Jun 10th, 2015

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