A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m. (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 2.00 ms, find the average force on the ball during impact.
Solution 67P To solve this question, we shall have to take into account the equation of motion v = u + 2gh…..(1) , where v is the final velocity, u is the initial velocity and h is the height from which the ball is dropped. Moreover, we shall have to use the relation impulse = change in momentum and impulse = force × time. Given, h = 2.00 m Initial speed u = 0 Therefore, from equation 1, v = 0 + 2 × g × 2.00 m v = 2 × 9.80 m/s × 2.00 m v = 6.3 m/s The ball rebounds to a height of 1.60 m. So, final velocity at the maximum height is 0. Let u be1the initial speed. 2 2 Therefore, 0 = u 1 2 × 9.80 m/s × 1.60 m u 1 5.6 m/s Since the ball rebound upward let the velocity u be negative, so u = 5.6 m/s 1 1 (a) Therefore, the change in velocity of the ball is = 6.3 m/s ( 5.6) m/s = 11.9 m/s Mass of the ball is m = 40.0 g = 0.04 kg The change in momentum of the ball is P = mass × change in velocity P = 0.04 kg × 11.9 m/s = 0.476 kg.m/s We know that the change in momentum is impulse. Therefore, the impulse delivered to the ball is 0.476 kg.m/s. 3 (b) Given, time of contact = 2.00 ms = 2.00 × 10 s We know that, impulse = force × time force = impulse= 0.476 k3m/s time 2.00×10 s force = 238 N Therefore, the average force on the ball during the impact is 238 N.