Velocity Question?

Question

A skater stands at rest in a skating rink, and throws a baseball (m = 0.15 kg) at a speed of 35 meters per second. If the their mass is 50 kg, and if you neglect friction, what recoil velocity will the skater have?

Answer 1

Let 'm' be the mass of the ball
..."..'v' ............velocity of the ball
...'...'M'.....".....mass of the skater &
...",,'V' ....".....velocity of the skater
By law of conservation of momentum,
Sum of the initial momentum = 0 (velocities are zero)
Sum of the final momentum = 0 (Velocities are not zero)
ie; mv + MV =0
or, mv = -MV
mv = M(-V)
0.15 kg x 35 m = 50 kg x (-V)
(-V) = (0.15 x 35) / 50
(-V) = .105 m/s
V = - 0.105 m/s
ie; the skater moves backwards with a velocity 0.105 m/s
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Answer 2

Use conservation of momentum.

Initial momentum of (skater + ball) is zero. Therefore:
Final momentum of (skater + ball) is also zero.

This means the skater must move with a backwards momentum that exactly cancels the "forward" momentum of the ball. That is:

(skater's momentum) = (ball's momentum)

The formula for momentum is mv. So:

(m_skater)(v_skater) = (m_ball)(v_ball)

They give you the values of m_skater, m_ball and v_ball. Solve for v_skater.

Answer 3

m1*v1 + m2*v2 = m1*v1' + m2*v2'
the first half of the equation is 0 (both velocities are zero)
so then the equation becomes:
m1*v1' = -m2v2'
solving for v2'
v2' = (m1*v1')/m2 = (.15*35)/50 = .105m/s

we can neglect the negative sign because it is going the opposite direction as the ball