Question

In the figure shown above, a disc A of mass m and radius r is placed on a smooth horizontal (x-y) plane. The coordinates of the centre of the disk A are $(\frac{r}{\sqrt{2}},\frac{r}{\sqrt{2}})$ another identical disc B having mass and radius same as that of A moving along the line $x=\frac{-r}{\sqrt{2}}$ with its plane horizontal to x-y plane with speed $v_0$, makes elastic impact with A. The time of impact is $△t$. In the elastic impact, the kinetic energy of the system is conserved. The velocity of the disc B after collision is :

• A. $\frac{v_0}{\sqrt{2}}(-\hat{i}+\hat{j})$
• B. $\frac{v_0}{\sqrt{2}}\hat{i}$
• C. $v_0\hat{i}$
• D. $\frac{v_0}{2}(-\hat{i}+\hat{j})$

Answer 1

The correct option is A $\frac{v_0}{\sqrt{2}}(-\hat{i}+\hat{j})$

A disc A, of mass $=m$

Radius $=r$

The coordinates of the centre of the disc $(\frac{r}{\sqrt{2}},\frac{r}{\sqrt{2}})$

Then,

another identicak B,

Mass and radius are

Same as A

B lying along the line

$x=\frac{-n}{\sqrt{2}}$

Speed$=v_0$

Makes the elastic impact with A

Time of impact $=\Delta t$

Find the velocity of B after collision.

Since, this is along $x-axis$

$Y=0$

Hence the co-ordinate of the velocity of B after collision $=\frac{1}{\sqrt{2}}(-i+j)$

where $r=1$

Now, $v_0$ be the speed of A

$\therefore Hence=\frac{v_0}{\sqrt{2}}(-i+j)$