A ball of mass 0.5 kg hits a vertical wall at a speed of 12 m s –1. It bounces back along its originalpath with a speed of 8 m s –1. The collision lasts for 0.10 s.12 m s–18 m s–1What is the average force on the ball due to the collision?A 0.2 N B 1 N C 20 N D 100 N
Question
A ball of mass 0.5 kg hits a vertical wall at a speed of 12 m s –1. It bounces back along its originalpath with a speed of 8 m s –1. The collision lasts for 0.10 s.12 m s–18 m s–1What is the average force on the ball due to the collision?A 0.2 N B 1 N C 20 N D 100 N
Solution
To solve this problem, we first need to find the change in velocity of the ball. The ball's initial velocity is 12 m/s towards the wall, and its final velocity is 8 m/s away from the wall. Because the directions are opposite, we consider one of the velocities as negative. So, the change in velocity (Δv) is:
Δv = final velocity - initial velocity Δv = -8 m/s - 12 m/s = -20 m/s
The negative sign indicates that the direction of the velocity has changed, which is expected as the ball bounced back.
Next, we use the formula for force, which is the change in momentum (mass × change in velocity) divided by the change in time. The mass of the ball (m) is 0.5 kg, and the change in time (Δt) is 0.10 s. So, the force (F) is:
F = mΔv / Δt F = 0.5 kg * -20 m/s / 0.10 s = -100 N
The negative sign indicates that the force is exerted in the opposite direction of the initial motion of the ball, which makes sense as the wall pushed the ball back. So, the magnitude of the force is 100 N. Therefore, the correct answer is D) 100 N.
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