Question

. A 0.145 kg baseball pitched at 35.0 m/s is hit on a horizontal line drive straight back at the pitcher at 56.0 m/s. If the contact time between the bat and ball is 5.00 × 10−3 ????, calculate the force (assumed to be constant) between the ball and bat.

Answer 1

2639 N

Step-by-step explanation:

The impulse on the baseball is given by:

`I=F \Delta t` (1)

where F is the force exerted on the ball and
`\Delta t` the contact time between the bat and the ball.

The impulse is also equal to the change in momentum of the ball:

`I=\Delta p = m (v-u)` (2)

where m is the ball's mass, v is its final velocity, u is its initial velocity.

By using (1) and (2) simultaneously we can write an expression for F, the force exerted on the ball:

`F=(m(v-u))/(\Delta t)`

where:

m = 0.145 kg

u = 35.0 m/s

v = -56.0 m/s

`\Delta t=5.00 \cdot 10^(-3)s`

Substituting,

`F=((0.145 kg)((-56.0 m/s)-35.0 m/s))/(5.00\cdot 10^(-3) s)=-2639 N`

and the negative sign means that the force is simply in the opposite direction to the ball's initial direction.