Force/Momentum problems

Question

A 0.145 kg baseball is pitched at 35.0 m/s is hit on a horizontal line drive straight back at the pitcher at 50 m/s. If the contact tiem between the bat and ball is 0.02 seconds, then

a) what is the change in momentum of the ball

b) what is the average force between th ebat and the ball during contact

Answer 1

Delta momentum is the mass times the change in velocity.

Mass = 0.145 Kg

Change in velocity = 35 m/s - (-50 m;s} = 85 m/s

(velocity is a vector quantity, note direction reversal here)

So, delta momentum = 0.145 * 85 = 12.3 Kg m/s

Average force = delta momentum / delta time

= 12.3 / 0.02 = 616 Kg m/sec/sec = 616 Nt

Answer 2

Change in momentum = mass × change in velocity

Given: mass= 0.145kg, initial velocity = 35.0 m/s, final velocity = 50m/s, time= 0.02 seconds

Therefore,

Change in momentum = 0.145kg × (50 - 35)m/s = 2.175 Ns

The average force = change in momentum / time

Therefore,

Average force = 2.175 Ns/ 0.02sec = 108.75 N