Question

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

Answer 1

Given:
Mass of the block, A = m
Mass of the block, B = 2m 

Let the initial velocity of block A be u₁ and the final velocity of block A,when it reaches the block B be v₁.

Using the work-energy theorem for block A, we can write:
Gain in kinetic energy = Loss in potential energy

`∴ (1/2)mv_1^2 - (1/2) m u_1^2 = mgh` 

`Rightarrow v_1^2 - u_1^2 = 2gh`

`Rightarrow v_1= sqrt(2gh  + u_1^2)` .......(1)

Let the block B just manages to reach the man's head.
i.e. the velocity of block B is zero at that point.

Again, applying the work-energy theorem for block B, we get:

`(1/2) xx 2m xx (0)^2 - (1/2 ) xx 2m xx v^2 = mgh`

`Rightarrow v = sqrt(2 gh)`

\[\text{ Therefore, before the collision }: \]

\[ \text{ Velocity of A, u}_A = v_1 \]

\[ \text{ Velocity of B, u}_B = 0\]

\[\text{ After the collision: }\]

\[ \text{ Velocity of A, v}_A = v (\text{ say })\]

\[\text{ Velocity of B, v}_B = \sqrt{2gh}\]

As the collision is elastic, K.E. and momentum are conserved.

\[m v_1 + 2m \times 0 = mv + 2m\sqrt{2gh}\]

\[ \Rightarrow v_1 - v = 2\sqrt{2gh} . . . \left( 2 \right)\]

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

\[ \Rightarrow v_1^2 - v^2 = 2 \times \sqrt{2gh} \times \sqrt{2gh} . . . \left( 3 \right)\]

\[\text{  Dividing equation (3) by equation (2), we get: } \]

\[ v_1 + v = \sqrt{2gh} . . . \left( 4 \right)\]

\[\text{ Adding the equations } \left( 4 \right) \text{ and } \left( 2 \right), \text{ we get: } \]

\[2 v_1 = 3\sqrt{2gh}\]

\[\text{ Now using equation } \left( 1 \right) \text{ to substitue the value of v}_1 , \text{ we get: }\]

\[\sqrt{2gh + u^2} = \left( \frac{3}{2} \right)\sqrt{2gh}\]

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

\[ \Rightarrow u = \sqrt{2 . 5 gh}\] 

Block a should be started with a minimum velocity of \[\sqrt{2 . 5gh}\]

 to get the sleeping man awakened.