A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of \(2\text{ms}^{-1}\). Determine the magnitude of the resulting impulse

Assessment: JAMB UTME - Physics - 2018 Subject: Physics

Question 1 Report

A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of \(2\text{ms}^{-1}\). Determine the magnitude of the resulting impulse

Answer Details
The magnitude of the resulting impulse can be calculated using the formula impulse = change in momentum. In this scenario, the ball experiences a change in velocity (speed) as it hits the wall. The ball's initial momentum is equal to its mass times its velocity, and its final momentum is zero since it comes to a stop after hitting the wall. The change in momentum is equal to the final momentum minus the initial momentum, which is equal to the negative of the initial momentum. Since the ball has a mass of 5.0 kg and a velocity of 2 m/s, its initial momentum is 5.0 kg * 2 m/s = 10.0 kg m/s. Therefore, the change in momentum is -10.0 kg m/s and the magnitude of the resulting impulse is 10.0 kg m/s, which is equal to 10.0 Ns. So, the correct answer is 10.0kgms−1.

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