Question

An iron ball of mass $m=50g$ falls from a height of $h_1=5m$ and rises upto $h_2=2m$ after colliding with the horizontal surface. If the time of the glass half is $\Delta t=0.02s$, find the average contact force exerted on the ball by the horizontal surface.

Answer 1

The change in linear momentum during collision is

$\Delta \overrightarrow{P}=m\overrightarrow{v_2}-m\overrightarrow{v_1}$

$=m\left(\overrightarrow{v_2}-\overrightarrow{v_1}\right)$

$=m[v_2\hat{j}-(-v_i\hat{j})]=m[v_1+v_2]\hat{j}$

The average force during the collision is

$\overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}=\frac{m(v_1+v_2)\hat{j}}{\Delta t}$

We can calculate $v_1$ and $v_1$ by using kinematics.

using $v^2=u^2+2as$

Putting a=g and s=h, we get

$v_1=,\sqrt{2g{h}_1},v_2=\sqrt{2g{h}_2}$

Or $\overrightarrow{F}=\frac{m(\sqrt{2g{h}_2}+\sqrt{2g{h}_2})\hat{j}}{\Delta t}$

$=\left(\frac{50}{1000}\right)\frac{\sqrt{2\times 10\times 5}+\sqrt{2\times 10\times 3.2}}{0.02}\hat{j}=45\hat{j}N$

During collision, the horizontal surface pushes the ball up with an average force of $45N$.