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posted by  Dee87 on 10/18/2007 4:14:39 PM  |  status: Live  

6.022

Course Textbook Chapter Problem
Algebra Based Physics College Physics (7th) by Serway, Faughn, Bennett 6 15P
Question Details:
A 68.0 kg person throws a 0.0500 kg snowball forward with a ground speed of 34.0 m/s. A second person, with a mass of 57.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.00 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.
thrower m/s
catcher m/s
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posted by oglybogly on 10/18/2007 4:40:54 PM  |  status: Live
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Response Details:
Assuming no friction:
Scholar
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posted by physics is phun! on 10/18/2007 4:42:39 PM  |  status: Live
Asker's Rating: Lifesaver   
Dee87's comment:
"thanx..."
Response Details:
Use the formula:
m1 = 68 kg
m2 = .05 kg (snowball)
v3 = 2 m/s
v2 = 34 m/s
v1 = ?
PLUG INTO THE FORMULA TO FIND V1:
You should get: 1.17733564 m/s.  This is the new velocity of the thrower.
 
Now use the same formula again just with different information.
m1 = .05 kg
m2 = 57 kg (mass of the catcher)
v2 = 0 m/s
v1 = 34 m/s
Plug these into the formula and you should get:
.0297984224 m/s = velocity of the catcher.
 
HOPE THIS HELPS.    PLEASE LET ME KNOW.
 
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