Question

A bullet fired horizontal with a speed of $400m/sec$. It strikes a wooden block of mass $5kg$ initially at rest placed on a horizontal floor as shown in the figure. It emerges with a speed of $200m/sec$ and the block slides a distance $20cm$ before coming to rest. If the coefficient of friction between block and the surface is $\frac{\lambda }{50}$ then find $\lambda$. Mass of bullet is $20gm$ (take $g=10m/s^2$)

Answer 1

Given mass of bullet is $m_b=20gm=0.02kg$ and mass of wooden block is $m_w=5kg$

Let speed of wooden block is $v_w$

Wooden block is at rest during collision so there is no external force acting then linear momentum will be conserved.

Linear momentum before collision = Linear momentum after collision

$\Rightarrow 0.02\times 400=0.02\times 200+5\times v_w$ $\Rightarrow v_w=0.8m/s$

Let the retardation of wooden block is $a$.

Using $v^2=u^2+2as$ $\Rightarrow 0=-{0.8}^2-2a\times 0.2$

$\Rightarrow a=1.6m/s^2$

$\Rightarrow a=\mu g$ , Where $\mu$ is coefficient of friction.

$\Rightarrow \mu =\frac{a}{g}=\frac{8}{50}$

Hence, $\lambda =8$