Question

Two blocks of equal masses are placed on a horizontal surface. The surface of $A$ is smooth but that of $B$ has a friction coefficient of $0.2$ with the floor. Block $A$ is given a speed $5\text{m/s}$, towards $B$ which is kept at rest. Find the distance travelled by $B$, if the collision is perfectly elastic.
(take $g=10{\text{m/s}}^2$ )

• A. $\frac{5}{4}\text{m}$
• B. $5\text{m}$
• C. $\frac{25}{4}\text{m}$
• D. $\frac{5}{2}\text{m}$

Answer 1

The correct option is C $\frac{25}{4}\text{m}$

Before collision

Given:
Initial speed of block $A,u_1=5\text{m/s}$
Initial speed of block $B,u_2=0\text{m/s}$
Also $m_1=m_2=m$

After collision:



Let,

$v_1=$ Speed of Block $A$ towards right
$v_2=$ Speed of Block $B$ towards right

By momentum conservation along the motion of the blocks (because no external force is present along that direction)

$\Rightarrow mu+0=m{v}_1+m{v}_2$

$\Rightarrow u=v_1+v_2.....\left(1\right)$

Also we know that,

$e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$

$\Rightarrow e=\frac{v_2-v_1}{u-0}=1$
(Because we have elastic collision)

$\Rightarrow v_2-v_1=u.....\left(2\right)$

From eq $\left(1\right)$ and $\left(2\right)$ we have

$v_2=u,v_1=0$

Again, a frictional force is acting on Block $B$, so, the situation can be shown below


From the equation of motion we have

$v^2-u^2=2as$

Here, final speed $v=0$ and $u=5\text{m/s}$

also, $a=-\frac{\mu mg}{m}=-\mu g=-0.2\times 10=-2{\text{m/s}}^2$

Putting the values in the above equation we have

$0-5^2=2\times (-2)\times s$

$\Rightarrow s=\frac{25}{4}\text{m}.$

Hence, the distance travelled by $B$ is $\frac{25}{4}\text{m}.$