Question

An iron ball of mass $m=50\text{gm}$ falls from a height of $h_1=5\text{m}$ and rises upto $h_2=3.2\text{m}$ after colliding with the horizontal surface. If the time of contact with the surface is $\Delta t=0.02\text{s},$ find the average contact force$\left(\text{Newtons}\right)$ exerted on the ball by the horizontal surface. Take $g=10{\text{m/s}}^2$.

• A. $25\text{N}$
• B. $60\text{N}$
• C. $45\text{N}$
• D. $95\text{N}$

Answer 1

The correct option is C $45\text{N}$

The change in linear momentum in vertical direction ($\text{y-direction}$) during collision is:


$\Delta \overrightarrow{p}=m\overrightarrow{{\text{v}}_2}-m\overrightarrow{{\text{v}}_1}$
$=m(\overrightarrow{{\text{v}}_2}-\overrightarrow{{\text{v}}_1})$
$=m[{\text{v}}_2\hat{j}-(-{\text{v}}_1\hat{j}\left)\right]=m({\text{v}}_1+{\text{v}}_2)\hat{j}$

$\Rightarrow$The average force during the collision is
$\overrightarrow{F}=\frac{\Delta \overrightarrow{p}}{\Delta t}=\frac{m({\text{v}}_1+{\text{v}}_2)\hat{j}}{\Delta t}..............i$

Using kinematic equation:
$v^2=u^2+2as$

Putting $a=-g$ and $s=+h$; where $s=$ Displacement of ball after colliding with surface, we get:

${\text{v}}_1=\sqrt{2g{h}_1},{\text{v}}_2=\sqrt{2g{h}_2}$

or $\overrightarrow{F}=\frac{m(\sqrt{2g{h}_1}+\sqrt{2g{h}_2})\hat{j}}{\Delta t}$

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

$\Rightarrow$During collision, the horizontal surface pushes the ball up with an average force of $45\text{N}$.