A ball of mass m1= 0.08 kg starts from rest and falls vertically downward from a height 3 m. After colliding with the ground, it bounces up to a height of 2 m. The collision takes place over a time interval of ∆t= 0.005 s. Calculate: 1. The momentum of the ball immediately before and immediately after the collision. 2. Average force exerted by the ground on the ball . 3. Impulse imparted to the ball.

Expert Answers

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We can use equations of motion to solve this problem. Using 

`v^2 = u^2 + 2as`

for the part where the ball drops from the height of 3 m, we can calculate the ball's velocity as it hits the ground. Here, u = 0, a = g = 9.81 m/s^2 and s = 3 m.

Thus, v^2 = 0^2 + 2 x 9.81 x 3

solving the equation, we get, v = 7.67 m/s.

Thus, momentum of the ball immediately before collision = mv

= 0.08 kg x 7.67 m/s = 0.61 kg m/s.

After the impact, the ball rises up to a height of 2 m. Using the same equation of motion, we know that v = 0 m/s, s = 2 m and a = -g = -9.81 m/s^2

we can find u by: u^2 = v^2 - 2as = 0 - 2 x (-9.81) x 2 

solving this, we get, u = 6.26 m/s.

Thus the momentum of the ball immediately after collision with the ground is 

= m x u = 0.08 kg x 6.26 m/s = 0.50 kg m/s.

Force can be calculated as the rate of change of momentum. 

Thus, the average force exerted = (0.61 - 0.5)/0.005 = 22 N.

The impulse is nothing but the change in momentum and is equal to 0.11 kg m/s.

Hope this helps. 

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