Question

A bullet of mass $m_{1}$ is moving with speed $v_{0}$ hits a sand bag of mass $m_{2}$. If the speedof the bullet after passing the sand bag is $\frac{v_{0}}{3}$, then the height $h$ up to which the bag rises is
(Assume $g =$ acceleration due to gravity)

• A. $h=\frac{1}{2 g}\left(\frac{2 m_{1} v_{0}}{3 m_{2}}\right)^{2}$
• B. $h=\frac{2 m_{1} v_{0}}{3 m_{2}}$
• C. $h=\frac{1}{2 g}$
• D. $h=\left(\frac{2 m_{1} v_{0}}{3 m_{2}}\right)^{2}$

Answer 1

Answer: (A)

The given situation is as shown in figure,

As, momentum is conserved in collision.
$\therefore p_{i}=p_{f} $
$\Rightarrow m_{1} v_{0}=m_{2} v_{2}+m_{1} \frac{v_{0}}{3} $
$\Rightarrow v_{2}=\frac{2 m_{1} v_{0}}{3 n_{2}} \ldots $ (i)
Given that, the bag rises upto height $h$.
So, at this height the kinetic energy is equal to potential energy.
i.e., $K E=P E $
$\Rightarrow \frac{1}{2} m_{2} v_{2}^{2}=m_{2} g h $
$\Rightarrow h=\frac{1}{2 g} v_{2}^{2}=\frac{1}{2 g}\left(\frac{2 m_{1} v_{0}}{3 m_{2}}\right)^{2}$