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 of $5m/s$, towards $B$ which is kept at rest. If the distance travelled by $B$ is $\frac{5}{4}\lambda$ then find $\lambda$, if the collision is perfectly elastic. Take $g=10m/s^2$.

Answer 1

Given , Collision is perfectly elastic.

Let speed of block $A$ and block $B$ is $v_A$ and $v_B$ respectively after collision and mass of block is $m$

Linear momentum before collision= Linear momentum after collision

$\Rightarrow m\times 5=m{v}_A+m{v}_B$ $\Rightarrow v_A+v_B=5$ .....1

Coefficient of restitution $(e=1)=-\frac{v_1-v_2}{u_1-u_2}$

$\Rightarrow 1=-\frac{v_A-v_B}{5}$ $\Rightarrow -v_A+v_B=5$ ....2

By adding the equation 1 and 2, we get, $v_B=5ms$

Now, retardation of block $B$, $a_B=\mu =0.2\times 10=2m/s^2$

Using $v^2=u^2+2as$ $\Rightarrow 0=5^2-2\times 2\times \frac{5\lambda }{4}$ $\Rightarrow \lambda =5$