Question

A small sphere of mass $m=1kg$ is moving with a velocity $(4\hat{i}-\hat{j})m/s$. It hits fixed smooth wall and rebounds with velocity $(\hat{i}+3\hat{j})m/s$. The coefficient of restitution between the sphere and the wall is $n/16$. Find the value of $n$.

Answer 1

During the collision, the component of the velocity $\parallel$ to the wall remains unchanged.

$\Rightarrow$ Let $\stackrel{⟶}{u}$ be the velocity before collision and $\stackrel{⟶}{v}$ be the velocity after collision.

The difference between $\stackrel{⟶}{u}$ and $\stackrel{⟶}{v}$ will always remain $\perp$ normal to the wall canceling the components along the wall.

Let $\stackrel{\wedge }{n}=\alpha \stackrel{\wedge }{i}+\beta \stackrel{\wedge }{j},({\alpha }^2+{\beta }^2=1)$, be the unit vector

Also, $\stackrel{⟶}{v}-\stackrel{⟶}{u}=\left(\right(\stackrel{⟶}{v}-\stackrel{⟶}{u}\left)\stackrel{\wedge }{n}\right)\stackrel{\wedge }{n}$

$\stackrel{⟶}{v}=4\stackrel{\wedge }{i}-\stackrel{\wedge }{j}$ $\stackrel{⟶}{u}=\stackrel{\wedge }{i}+3\stackrel{\wedge }{j}$

$\therefore \alpha =\frac{-3}{5},\beta =\frac{4}{5}[\because {\alpha }^2+{\beta }^2=1]$

So, we get

$\stackrel{\wedge }{n}=\frac{-3}{5}\stackrel{\wedge }{i}+\frac{4}{5}\stackrel{\wedge }{j}$

We know that,

Coefficient of restitution,

$e=-\frac{\stackrel{⟶}{v}\stackrel{\wedge }{n}}{\stackrel{⟶}{u}\stackrel{\wedge }{n}}=-\frac{(-3)+12}{-12-4}=\frac{9}{16}$

$\therefore \frac{n}{16}=\frac{9}{16}$

$\Rightarrow n=9$.