More physics fun................?

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spaniard

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A 68.0 kg person throws a 0.0480 kg snowball forward with a ground speed of 31.0 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 2.00 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?
catcher?
 
concept is conservation of momentum...

the first frame is where a person moving with a snowball at 2m/s

momentum before throwing = (68+0.0480)*2 = 136.096
momentum after throwing = momentum of ball + momentum of man
--------> = 0.048*31 + 68*x
68x = 136.096 - 1.488 = 134.608
--> x=1.98m/s -----> velocity of person1 after throwing


frame 2 : initial momentum = 60*0 + 0.048*31 = 1.488
momentum after catching = (60 + 0.048)*v = 1.488
---> = 1.488/ 60.048
--> v= 0.0247m/s.

thus velocity of second person is 0.0247m/s

Hope this helped you... :)
 
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