Question

The track shown in figure (9-E16) 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 ?

Answer 1

Let the velocity of $A=\mu$

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

$\therefore \left(\frac{1}{2}\right)m_1^2-\left(\frac{1}{2}\right)m{v}_1^2=mgh$

$\Rightarrow v_1^2-u_1^2=2gh$

$\Rightarrow v_1=\sqrt{2gh+u_1^2}...\left(i\right)$

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 $u_A=v_1$

$u_B=0$

After collision velocity of $v_A=V$ (say)

$u_B=\sqrt{2gh}$

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

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

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

Also $\left(\frac{1}{2}\right)\times m\times v_1^2+\left(\frac{1}{2}\right)\times 2m\times (0{)}^2$

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

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

Dividing (i) and (ii),

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

$\Rightarrow v_1+v=\sqrt{2gh}$

Dividing (i) and (iii),

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

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

But $v_1=\sqrt{2gh+u^2}$

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

$\Rightarrow 2gh+u^2=\left(\frac{9}{4}\right)\left(2gh\right)$

$\Rightarrow u=\sqrt{2.5gh}$

So the block will travel with a velocity greater than $\sqrt{2.5gh}$ i.e. man must awake.