Question

The track shown in figure is frictionless. The block $B$ of mass $2m$ is lying at rest and the block $A$ of mass $m$ is pushed along the track with some speed. The collision between $A$ and $B$ is perfectly elastic. With what velocity should the block $A$ be started to get the sleeping man awakened?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let the velocity of $A$ = $u_1$.

Let the final velocity when reaching at $B$ becomes collision = $v_1$.

$\therefore \left(\frac{1}{2}\right)m{v}_1^2-\left(\frac{1}{2}\right)m{u}_1^2$ = $mgh$

$\Rightarrow v_1^2-u_1^2$ = $2gh$ $\Rightarrow v_1$ = $\sqrt{2gh-u_1^2}$ ......(1)

When the block $B$ reached at the upper man's head, the velocity of $B$ is just zero.

For $B$, block

$\therefore \left(\frac{1}{2}\right)\times 2m\times 0^2-\left(\frac{1}{2}\right)\times 2m\times v^2$ = $mgh$ $\Rightarrow v$ = $\sqrt{2gh}$

$\therefore$ Before collision velocity of $A$ = $v_1$. $u_B$ = 0

After collision velocity of $v_A$ = $v$(say) $v_B$ = $\sqrt{2gh}$

Since it is an elastic collision the momentum and K.E. should be conserved.

$\therefore m\times v_1+2m\times 0$ = $m\times v+2m\times \sqrt{2gh}$

$\Rightarrow v_1-v$ = $2\sqrt{2gh}$ .........(1)

Also, $\left(\frac{1}{2}\right)\times m\times v_1^2+\left(\frac{1}{2}\right)2m\times 0^2$ = $\left(\frac{1}{2}\right)\times m\times v^2+2m\times (\sqrt{2gh}{)}^2$

$\Rightarrow v_1^2-v^2$ = $2\times \sqrt{2gh}\times \sqrt{2gh}$ .......(2)

Dividing (2) by (1)

$\frac{(v_1+v)(v_1-v)}{(v_1-v)}$ = $\frac{2\times \sqrt{2gh}\times \sqrt{2gh}}{2\times \sqrt{2gh}}$ $\Rightarrow v_1+v$ = $\sqrt{2gh}$ ........(3)

Adding (1) and (3)

$2v_1$ = $3\sqrt{2gh}$ $\Rightarrow v_1$ = $\left(\frac{3}{2}\right)\sqrt{2gh}$

But $v_1$ = $\sqrt{2gh+u^2}$ = $\left(\frac{3}{2}\right)\sqrt{2gh}$

$\Rightarrow 2gh+u^2$ = $\frac{9}{4}\times 2gh$

$\Rightarrow u^2$ = $\frac{5}{2}gh$

So the block will travel with a velocity greater than $\sqrt{2.5gh}$, to awake the man by $B$.