Alex throws a 0.10-kg rubber ball down onto the floor. The ball's speed just before impact is 5.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for 0.02 s, what is the magnitude of the average force applied by the floor on the ball?Select one:a.45 Nb.27.5 Nc.17.5 Nd.0.9 Ne.10 N
Question
Alex throws a 0.10-kg rubber ball down onto the floor. The ball's speed just before impact is 5.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for 0.02 s, what is the magnitude of the average force applied by the floor on the ball?Select one:a.45 Nb.27.5 Nc.17.5 Nd.0.9 Ne.10 N
Solution
To solve this problem, we need to use the formula for force, which is the change in momentum divided by the change in time.
Momentum is mass times velocity, so the change in momentum (Δp) is the mass times the change in velocity (Δv).
The change in velocity is the final velocity minus the initial velocity. In this case, the ball is bouncing back, so its final velocity is -3.5 m/s (the negative sign indicates that the direction of the velocity has reversed). The initial velocity is -5.5 m/s (negative because the ball was thrown downwards).
So, Δv = -3.5 m/s - (-5.5 m/s) = 2 m/s.
The mass of the ball (m) is 0.10 kg, so Δp = m * Δv = 0.10 kg * 2 m/s = 0.2 kg*m/s.
The time (t) the ball is in contact with the floor is 0.02 s, so the average force (F) is Δp / t = 0.2 kg*m/s / 0.02 s = 10 N.
So, the magnitude of the average force applied by the floor on the ball is 10 N. The correct answer is e. 10 N.
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