Question

E-4 A steel ball of mass 0.5 Kg is dropped from a height of 4 m on to a horizontal heavy steel slab.The ball strike the slab and rebounds to its original height $(takeg=10m/se{c}^2)$
(a) calculate the impulse delivered to the ball during impact.
(b) if the ball is in contact with the slab for 0.002s,find the average reaction force on the ball during impact.

Answer 1

Given,

Mass of ball, $m=0.5kg$

Initial velocity is $u=0$.

Apply kinematic velocity, $v^2-u^2=2as$

Striking velocity, $v=\sqrt{2gh}=\sqrt{2\times 10\times 4}=4\sqrt{5}m{s}^{-1}$

Ball reach to same height, hence rebound velocity magnitude is equal to striking velocity

Rebound velocity, $v_1=-4\sqrt{5}m{s}^{-1}$

Impulse, $I=m\Delta V=m[v-(-v_1\left)\right]=0.5(2\times 4\sqrt{5})=4\sqrt{5}N{s}^{-1}$

Force, $F=\frac{m\Delta V}{\Delta t}=\frac{4\sqrt{5}}{0.002}=4472.13N$

Hence, impulse is $4\sqrt{5}N{s}^{-1}$ and Force is $4472.13N$