Question

A bullet of mass m travels at a very high velocity V (as shown in the figure) and gets embedded inside the block of mass M initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance s along the floor. Assuming $\mu$ to be the coefficient of kinetic friction between the block and the floor and g the acceleration due to gravity what is the velocity V of the bullet?

• A. $\frac{M+m}{m}\sqrt{2\mu gs}$
• B. $\frac{M-m}{m}\sqrt{2\mu gs}$
• C. $\frac{\mu (M+m)}{m}\sqrt{2\mu gs}$
• D. $\frac{M}{m}\sqrt{2\mu gs}$

Answer 1

The correct option is A $\frac{M+m}{m}\sqrt{2\mu gs}$

From conservation of linear momentum
mV= (M+m) U

$\therefore U=\frac{mV}{M+m}$
Kinetic friction,
${\overrightarrow{f}}_k=\mu N(-\hat{i})$
$(M+m)\overrightarrow{a}=\overrightarrow{f_k}$
${\overrightarrow{f}}_k=\mu (M+m)g(-\hat{i})$
$\overrightarrow{a}=\frac{{\overrightarrow{f}}_k}{(M+m)}=\frac{\mu (M+m)g(-\overrightarrow{i})}{(M+m)}$
$\overrightarrow{a}=\mu g(-\hat{i})$
Now $V_2^2=U^2-2\mu gs$
$\therefore 0={\left(\frac{mV}{M+m}\right)}_{-}^{2}2\mu gs$
$\therefore V=\left(\frac{M+V}{m}\right)\sqrt{2\mu gs}$