Question

The blocks shown in figure have equal masses The surface of $A$ is smooth but that of $B$ has a friction coefficient of $0.10$ with the floor. Block $A$ is moving at a speed of $10m/s$ towards $B$ which is kept at rest. Find the distance travelled by $B$ if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. Take $g=10m/s^2$.

Answer 1

$v_1=10m/s..................v_2=0$

$v_1$ and $v_2$ be the velocity of A and B respectively after the collision.

(a) The collision is perfectly elastic

$m_1+m{v}_2=v_1+v_2$

$10+0=v_1+v_2$

$v_1+v_2=10$ [Equation 1]

Now, $v_1-v_2=-(v_1-v_2)$

$v_1-v_2=-(10-0)$

$v_1-v_2=-10$

Subtracting Equation 2 from Equation 1

$2v_2=20$

$v_2=10m/s$

Deceleration of $B=\mu g$

Now, work energy principle,

$0.5\times m\times 0^2-\left(0.5\right)\times m\times v_2=-\mu \times mg\times d$[d is the distance travelled by B]

$-\left(0.5\right)\times {10}^2=-\mu gd$

$d=\frac{100}{(2\times 0.1\times 10)}$

$d=50m$

(b) The collision is perfectly inelastic

$m\times u_1+m\times u_2=(m+m)\times v$

$m\times 10+m\times 0=2m\times v$

$v=\frac{10}{2}\Rightarrow v=5m/s$

Two blocks will move together sticking to each other.

Now , work energy prinicple,

$\left(0.5\right)\times 2mx{0}^2-\left(0.5\right)\times 2m\times v^2=2m\times \mu g\times d_1$ [ $d_1$ = distance travelled by B]

$d_1=5^2(0.1\times 10\times 2)\Rightarrow d_1=12.5m$