Search any question & find its solution
Question:
Answered & Verified by Expert
A ball of 0.5 kg collided with wall at $30^{\circ}$ and bounced back elastically. The speed of ball was $12 \mathrm{~m} / \mathrm{s}$. The contact remained for 1 s . What is the force applied by wall on ball?
Options:
Solution:
2091 Upvotes
Verified Answer
The correct answer is:
$6 \sqrt{3} N$
Given, $\mathrm{m}=0.5 \mathrm{~kg}, \mathrm{v}=12 \mathrm{~m} / \mathrm{s}, \Delta \mathrm{t}=1 \mathrm{~s}, \theta=30^{\circ}$
Force applied by wall on ball, $\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
or $\quad \mathrm{F}=\frac{\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}-\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}}{\Delta \mathrm{t}}$
In this elastic collision, final and initial velocity will be same but direction will changed.
$\therefore$ Horizontal component, $\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}=\mathrm{mv} \cos \theta$ and $\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}=-\mathrm{mv} \cos \theta$
$\begin{aligned} & \therefore \mathrm{F}=\frac{\mathrm{mv} \cos \theta-(-\mathrm{mV} \cos \theta)}{\Delta \mathrm{t}}=\frac{2 \mathrm{mV} \cos \theta}{\Delta \mathrm{t}} \\ & =\frac{2 \times 0.5 \times 12 \times \cos 30^{\circ}}{1}=6 \sqrt{3} \mathrm{~N} \\ & \end{aligned}$
Force applied by wall on ball, $\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
or $\quad \mathrm{F}=\frac{\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}-\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}}{\Delta \mathrm{t}}$
In this elastic collision, final and initial velocity will be same but direction will changed.
$\therefore$ Horizontal component, $\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}=\mathrm{mv} \cos \theta$ and $\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}=-\mathrm{mv} \cos \theta$
$\begin{aligned} & \therefore \mathrm{F}=\frac{\mathrm{mv} \cos \theta-(-\mathrm{mV} \cos \theta)}{\Delta \mathrm{t}}=\frac{2 \mathrm{mV} \cos \theta}{\Delta \mathrm{t}} \\ & =\frac{2 \times 0.5 \times 12 \times \cos 30^{\circ}}{1}=6 \sqrt{3} \mathrm{~N} \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.