Question

If the velocity of a pitched ball has a magnitude of 43.5 m/s and the batted ball's velocity is 57.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. Express your answer to three significant figures and include the appropriate units.

Answer 1

Incomplete question as the mass of ball is missing.So I assume the mass of ball is 0.35 kg.The complete question is here

A baseball has mass 0.35 kg. (a) If the velocity of a pitched ball has a magnitude of 43.5 m/s and the batted ball’s velocity is 57.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.Express your answer to three significant figures and include the appropriate units.

Answer:

J=4.90 kg.m/s

Step-by-step explanation:

First, we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction So we have:

`v_(i)=43.5m/s\\ v_(f)=57.5m/s`

We plug our values for m and vi.So we get the initial momentum of ball:

`p_(1)=(0.35kg)(43.5m/s)\\p_(1)=15.225kg.m/s\\`

We plug our values for m and vf.So we get the final momentum of ball:

`p_(2)=(0.35kg)(57.5m/s)\\p_(2)=20.125kg.m/s\\`

So the change in momentum is:

Δp=p₂ - p₁

Δp=(20.125 kg.m/s) - (15.225 kg.m/s)

Δp=4.90 kg.m/s

We get the impulse applied to the ball by bat as follow

J=Δp

J=4.90 kg.m/s