Question

The track shown in figure is frictionless. The block B 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 ?

Answer 1

Let the velocity of A is $u_1$ and the final velocity when reaching at B is $v_1$.

The final speed is given as,

$\frac{1}{2}m{v_1}^2-\frac{1}{2}m{u_1}^2=mgh$

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

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

It can be written as,

$\frac{1}{2}2m{\left(0\right)}^2-\frac{1}{2}2m{v_B}^2=mgh$

$v_B=\sqrt{2gh}$

Let after the collision the velocity of A is $v$ and before the collision is $v_1$.

$m{v}_1+2m\times 0=m\times v+2m\sqrt{2gh}$

$v_1-v=2\sqrt{2gh}$

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

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

From equation (1) and (2), we get

$v_1+v=\sqrt{2gh}$ (3)

From equations (1) and (3), we get

$2v_1=3\sqrt{2gh}$

$v_1=1.5\sqrt{2gh}$

$u=2.5\sqrt{2gh}$

Thus, the velocity of block A is $2.5\sqrt{2gh}$.