Question

A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)

Answer 1

Given:

Initial velocity, vi = 41.0 m/s

Mass of ball, m = 195 g = 0.195 kg

Final velocity, vf = 37.0 m/s

Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:

• (A). What is the impulse delivered to the ball by the bat?

To find the impulse, apply the change in momentum formula:

`\Delta p=p_f-p_i`

Where:

pi is the initial momentum = -mvi

pf is the final momentum = mvf

Thus, we have:

`\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}`

Impulse can be said to equal change in momentum.

Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.

• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?

Apply the formula:

`\text{ Impulse = Force }\ast\text{ time}`

Rewrite the formula for force:

`\text{ Force=}(impulse)/(time)`

Where:

time = 3.00 m/s

impulse = 15.21 kg.m/s

Hence, we have:

`\begin{gathered} \text{ F=}(15.21)/(3) \\ \\ F\text{ = 5.07 kN} \end{gathered}`

Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.

ANSWER:

(A). 15.21 kg.m/s away from the bat

(B). 5.07 kN.