Question

A 69.0-kgperson throws a 0.0500-kgsnowball forward with a ground speed of 32.5m/s.A second person, with a mass of 55.5kg,catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.90m/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. (Take the direction the snowball is thrown to be the positive direction. Indicate the direction with the sign of your answer.)

thrower	m/s (Give your answer to at least three decimal places.)
catcher	m/s

Answer 1

Conerving momentum of the system :

m1*u1 + m2*u2 = (m3 + m2)*v + m1*v1

here, m1 = 69 kg

m2 = 0.05 kg

m3 = 55.5 kg

u1 = 3.9 m/s

u2 = 32.5 m/s

v = final speed of the 2nd person after the exchange

v1 = final speed of the 1st person

So, 69*3.9 + 0.05*32.5 = (55.5 +0.05)*v + 69*v1 ------------- (1)

Conserving energy of the system :

m1*u1^2 + m2*u2^2 = (m3 + m2)*v^2 + m1*v1^2

69*3.9^2 + 0.05*32.5^2 = (55.5 +0.05)*v^2 + 69*v1^2 ------------ (2)

Solving the two equations simultaneously :

v1 = -0.09 m/s <------- thrower

v = 4 m/s <------ catcher