Question

A small block of mass M moves with velocity 5m/s towards another block of same mass M placed at a distance of 2m on a rough horizontal surface. Coefficient of friction between the blocks and ground is 0.25. Collision between the two blocks is elastic, the separation between the blocks, when both of them come to rest, is (g = 10m/s$^2$)
A. 3 m
B. 4 m
C. 2 m
D. 1.5 m

Answer 1

The acceleration possessed by the blocks are due to the frictional force. Final velocity of the first block is equal to the initial velocity of the second block when the first block collides with the second block.

Complete step by step answer:
A small block of mass M is moving with an initial velocity(u$_1$) of 5m/s towards an another block whose mass is same i.e. M. The distance (s) between them is 2m. As the blocks are moving on a rough surface. So, the blocks will experience frictional force that opposes the motion and the acceleration due to this force(a)= $\mu g = 0.25 \times 10 = 2.5$m/s$^2$ where μ is coefficient of acceleration between the blocks and ground. Let the final velocity of the first block be v$_1$, initial velocity of the second block be u$_2$ and final velocity be v$_2$.
From the kinematic equation, ${v_1}^2 = {u_1}^2 + 2as$
${v_1}^2 = {5^2} + 2 \times \left( { - 2.5} \right) \times 2$ [The negative sign of acceleration denotes the acceleration due to friction is in opposite direction of motion]
${v_1}^2 = 25 - 10 = 15$
${v_1}^2 = 15$
${v_1} = \sqrt {15}$ m/s
In an elastic collision, momentum (product of mass and velocity) is conserved. So, the velocity of the first block after the collision is equal to the velocity of the second block before the collision as masses are the same i.e. final velocity of the first block is imparted to the second block v$_1$ = u$_2$.
Now, ${v_2}^2 = {u_2}^2 + 2a{s_1}$ where s$_1$ is the separation between the blocks when both of them come to rest.
${0^2} = {\left( {\sqrt {15} } \right)^2} + 2 \times \left( { - 2.5} \right) \times {s_2}$
$0 = 15 - 5{s_1}$ as final velocity of the second block is zero.
$5{s_1} = 15$
$\therefore {s_1} = \dfrac{{15}}{5} = 3m$

So, the correct answer is “Option A”.

Note:
In an elastic collision, both kinetic energy and momentum are conserved. So, the final velocity of the first block is transferred to the second block when the first block collides with the second block due to which it moves to a certain distance before it comes to rest.