Question

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.

Answer 1

Answer:

impulse = 0.476 kg.m/s

force = 238 N

Explanation:

from the question we are given the following

mass (m) = 40 g = 0.04 g

time (t) = 2 ms = 0.002 s

height of drop (H) = 2 m

rebound height (h) = 1.6 m

acceleration due to gravity (g) = 9.8 m/s

(a) impulse = force x time = mass x change in velocity (velocity when it hits the groung and when it rebounds)

Since the ball was at rest before it was dropped from the 2 meters its initial velocity would be 0 m/s and we need to find the final velocity when it hits the ground.

• While rebounding, it gets to a height of 1.6 m and stops, so its final velocity is 0 m/s and we need to find the initial velocity.
• We can apply the formula below

           (final velocity (Vf))^{2} = (initial velocity (Vi))^{2} - (2gh)

therefore

final velocity when the ball hits the ground:

Vf = $\sqrt{0 - 2gh}$

Vf = $\sqrt{2 x 9.8 x 2}$

Vf = 6.3 m/s = -6.3 m/s since it is moving downwards

initial velocity when the ball rebounds :

Vi =  $\sqrt{0 + 2gh}$

Vi =  $\sqrt{2 x 9.8 x 1.6}$

Vi = 5.6 m/s

recall that impulse = force x time = mass x change in velocity

impulse = mass x change in velocity

impulse = 0.04 x (5.6 - (-6.3)) =  0.476 kg.m/s

(b) Recall that  impulse = force x time = mass x change in velocity

 therefore

force x time = mass x change in velocity

f x 0.002 = 0.476

f = 238 N