Question

A bullet of mass m is fired at speed v_0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block.
The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction $\mu$. Which one of the following expressions determines how far the block slides before it comes to rest (the magnitude of displacement $\Delta x$ in the figure)?

• A. $\Delta x=\sqrt{\frac{m}{m+M}}\left(\frac{v_0^2}{\mu g}\right)$
• B. $\Delta x=\sqrt{\frac{m}{mM}}\left(\frac{v_0^2}{\mu g}\right)$
• C. $\Delta x=\left(\frac{v_0^2}{\mu g}\right)$
• D. $\Delta x=\frac{1}{2\mu g}{\left(\frac{m}{M+m}{V}_0\right)}^2$
  
  
  

Answer 1

The correct option is D

$\Delta x=\frac{1}{2\mu g}{\left(\frac{m}{M+m}{V}_0\right)}^2$

$p_0=m{v}_0;p_f=(M+m)v_f$

$p_0=p_f\Rightarrow v_f=\frac{m}{M+m}{v}_0$

So, retardation would be due to kinetic friction, and would be equal to $-\mu g\left(constant\right)$.

So, from equations of motion we can obtain the distance travelled by the block embedded with bullet before coming to rest. So, using

$v^2=u^2+2as,v_f=\frac{m}{M+m}{v}_0,0={\left(\frac{m}{M+m}{v}_0\right)}^2+2(-\mu g)s,s=\frac{1}{2\mu g}{\left(\frac{m}{M+m}{v}_0\right)}^2$