A tennis ball of mass 50g is hit against a wall. It hits the wall going right at 5 ms-1 and rebounds to the left at 4 ms-1, changing its momentum by 0.45 kg m/s left.If the ball is in contact with the wall for 0.2 s what is the average force applied by the wall on the ball (your answer is a vector)?
Question
A tennis ball of mass 50g is hit against a wall. It hits the wall going right at 5 ms-1 and rebounds to the left at 4 ms-1, changing its momentum by 0.45 kg m/s left.If the ball is in contact with the wall for 0.2 s what is the average force applied by the wall on the ball (your answer is a vector)?
Solution
To solve this problem, we first need to understand the concept of momentum and how it changes. Momentum (p) is the product of an object's mass (m) and its velocity (v). It's a vector quantity, which means it has both magnitude and direction.
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Calculate the initial momentum of the ball before it hits the wall. The initial momentum (p_initial) is the product of the ball's mass and its initial velocity.
p_initial = m * v_initial p_initial = 0.05 kg * 5 m/s p_initial = 0.25 kg m/s to the right
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Calculate the final momentum of the ball after it rebounds. The final momentum (p_final) is the product of the ball's mass and its final velocity.
p_final = m * v_final p_final = 0.05 kg * 4 m/s p_final = 0.20 kg m/s to the left
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Calculate the change in momentum (Δp). The change in momentum is the final momentum minus the initial momentum. Since the directions are opposite, we consider one of them as negative.
Δp = p_final - p_initial Δp = -0.20 kg m/s - 0.25 kg m/s Δp = -0.45 kg m/s to the left
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Calculate the average force (F) applied by the wall on the ball. The average force is the change in momentum divided by the time it takes for that change to occur (t), according to Newton's second law (F = Δp/t).
F = Δp / t F = -0.45 kg m/s / 0.2 s F = -2.25 N
So, the average force applied by the wall on the ball is 2.25 N to the left.
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