Question

A 0.135-kg baseball pitched horizontally at 30.5 m/s strikes a bat and is popped straight up to a height of 36.5 m.

Part A:

If the contact time between bat and ball is 3.0 ms , calculate the magnitude of the average force between the ball and bat during contact.

Favg = ____ N

Part B:

Find the direction of the average force on the ball.

Theta = __ degrees

Answer 1

(a) Contact time between bat and ball, t = 3.0 ms = 0.003 s

In this problem, direction is a big factor, so we will use arrows to represent the direction.

If the ball reaches a height of 35.5 m, then the vertical component of velocity (u) when the ball leaves the bat can be calculated:
v↑² = u↑² + 2gs
0² = u↑² + 2 (-9.8) (35.5)
u↑² = 695.8
u↑ = 26.4 m/s (vertical)

Impulse equation:
Ft = change of momentum = ∆p = m∆v
where:
∆v = final velocity – initial velocity = final velocity + ( - initial velocity)
∆v = (v↑) – (←u) = (26.4↑) + (- ←30.5)

=> ∆v = (26.4↑) + (→30.5)

=> ∆v = √ (26.4² + 30.5²) = 40.3 m/s

So -

Ft = m∆v
F = m∆v / t = (0.135 x 40.3) / 0.003

=> F = 1813.5 N

So, magnitude of average force between the ball and the bat = Favg = 1813.5 N

(b) Direction of the average force is given as -

tan θ = 26.4 / 30.5 = 0.866

=> θ = tan^-1(0.866) = 40.88 ° (Answer)