A block of mass M=5kg is moving on a horizontal table and the coefficient of friction is $\mu =0.4.$ A clay ball of mass m=1 kg is dropped on the block, hitting it with a vertical velocity of u=10 m/s. At the instant of hit, the block was having a horizontal velocity v=2 m/s. After an interval of $\Delta t,$ another similar clay ball hits the block and the system comes to rest immediately after the hit. Assume that the clay balls stick to the block and collision is momentary. Take $g=10m/s^2.$ Value of $\Delta t=\frac{2}{x}\sec .$ Value of ‘$x$’ is

 

$\Delta t=\frac{1}{12}s$

During the interaction period of the clay ball and the block, the vertical impulse (due to normal force) applied by ground is $J_v=\int Ndt=change in momentum is$

Vertical direction $=1\times 10=10kgm/s\mathrm{.............}\left(i\right)$

Horizontal impulse of ground friction during the same period is $J_H=\int \mu Ndt=0.4\times 10=4kgm/s\mathrm{..................}\left(ii\right)$

Velocity $\left(V_1\right)$ of the (block + ball) system just after the impact is given by $(M+m)V_1=MV-J_H$

$V_1=\frac{5\times 2-4}{6}=1m/s$

Let the velocity of (M+m) be reduced to $V_2$ [due to friction] in interval $\Delta t.$

At this point another clay ball hits the block.

$J_{v}and{J}_H$ given by (i) and (ii) remain same for the second impact

$\therefore 0=(M+m)V_2-J_H$

$\therefore V_2=\frac{4}{6}=\frac{2}{3}m/s$