Question

If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 50.5 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.

Answer 1

The magnitude of change in momentum of the ball is
`97.5 m` and impulse is also
`97.5 m`

Step-by-step explanation:

Given:

Velocity of a pitched ball
`v _(i) = 47`
`(m)/(s)`

Velocity of ball after impact
`v_(f) = -50.5`
`(m)/(s)`

From the formula of change in momentum,

`\Delta P = m (v_(f) -v_(i) )`

Here mass is not given in question,

Mass of ball is
`m`

Change in momentum is given by,

`\Delta P = m (-50.5 -47)`

`\Delta P = -97.5 m`

Magnitude of change in momentum is

`\Delta P = 97.5 m`

And impulse is given by

`J = \Delta P`

`J = -97.5 m`

So impulse and

Therefore, the magnitude of change in momentum of the ball is
`97.5 m` and impulse is also
`-97.5 m`