Question

A 438 gram ball is traveling east at 39 m/s when it is hit by a 2.4 kg softball bat. After being in contact with the bat for 302 ms, the ball rebounds at a speed of 28 m/s.
Determine the impulse vector (magnitude and direction) in kg*m's felt by the bat. Treat east as the positive direction.

Answer 1

Answer:

The magnitude of the impulse vector of the bat is 29.346 kg.m/s

The direction of the impulse vector of the bat is in the initial direction of the ball before impact

Explanation:

The given parameters are;

The mass of the ball, m₁ = 438 g

The speed of the ball = 39 m/s

The mass of the softball bat = 2.4 kg

The time of contact = 302 ms = 0.302 seconds

The speed of rebound = 28 m/s

The impulse = The change in momentum = Δp = F × Δt = m × Δv

The impulse on the ball = m₁ × Δv = 0.438 × (39 - (-28)) = 29.346 kg·m/s

Given that force of reaction of the bat is in opposite direction but equal to the force of the ball, and the time and the duration of contact with the ball is the same for both the ball and the bat, the impulse vector ore equal and opposite

Therefore, the magnitude of the impulse vector of the bat = 29.346 kg.m/s

The direction of the impulse vector of the bat = The initial direction of the ball before impact.