Question

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 20 m/s. It is hit straight back at the pitcher with a final speed of 25 m/s. Assume the direction of the initial motion of the baseball to be positive.(a) What is the impulse delivered to the ball? (b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0?

Answer 1

a.-6.75 kgm/s

b.
`3375 N`

Step-by-step explanation:

We are given that

Mass of baseball=0.15 kg

Initial speed=
`u=20m/s`

Final speed=
`v=-25m/s`

a.We know that

Impulse=Change in momentum=
`\Delta p=mv-mu=m(v-u)`

Momentum=
`mass* velocity`

Using the formula

Impulse=
`0.15(-25-20)=-6.75 kgm/s`

b.Time=
`2* 10^(-3) s`

Force=
`(Impulse)/(time)`

Using the formula

Average force exerted by the bat on the ball=
`(-6.75)/(2* 10^(-3))` N

Average force exerted by the bat on the ball=
`3375N`