Question

A small block of mass $M$ moves with velocity $5m{s}^{-1}$ towards an 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. $3m$
• B. $4m$
• C. $2m$
• D. $1.5m$

Answer 1

Answer: (A)

$3m$

Given,

$Mass=m,Velocity=5m/s,d=2m,\mu =0.25,g=10m/s^2$

So for the mass 1,

$V_{in}=5m/s$ Let the final velocity that is imparted to the other mass before the collision be $V_f$

So Acceleration $=\mu \times g=0.25\times 10=2.5\Rightarrow a=-2.5m/s^2$(negative sign shows friction acts opposite to the direction of motion)

Now, $V_f^2=V_{in}^2+2as\Rightarrow V_f^2=25-2\times 2.5\times 2=15\Rightarrow V_f=\sqrt{15}m/s$

At the time of collision, momentum will be conserved. So the velocity of both mass will exchange.

For the second mass,

Velocity gained $V_f=\sqrt{15}m.s=$ to the initial velocity of the second mass= $u$

Now, A/C to the question the mass cofinal velocity of the second mass to rest so$v=0$

Thus, By the kinematic equations

$V^2=U^2+2as\Rightarrow ==(\sqrt{15}{)}^2+2\times (-2.5)\times s\Rightarrow s=\frac{15}{5}=3m$