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.
