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A bullet of mass $m_1$ is moving with speed $v_0$ hits a sand bag of mass $m_2$. If the speed of the bullet after passing the sand bag is $\frac{v_0}{3}$, then the height $h$ upto which the bag rises is (assume, $g=$ acceleration due to gravity)
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The correct answer is:
$h=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2$
The given situation is as shown in figure,
As, momentum is conserved in collision,
$\therefore \quad p_i=p_f$
$\begin{array}{ll}\Rightarrow & m_1 v_0=m_2 v_2+m_1 \frac{v_0}{3} \\ \Rightarrow & v_2=\frac{2 m_1 v_0}{3 m_2}\end{array}$
Given that, the bag rises upto height $h$.
So, at this height the kinetic energy is equal to potential energy.
$\begin{array}{ll}\text { i.e., } & \mathrm{KE}=\mathrm{PE} \Rightarrow \frac{1}{2} m_2 v_2^2=m_2 g h \\ \Rightarrow & h=\frac{1}{2 g} v_2^2=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2\end{array}$
As, momentum is conserved in collision,
$\therefore \quad p_i=p_f$
$\begin{array}{ll}\Rightarrow & m_1 v_0=m_2 v_2+m_1 \frac{v_0}{3} \\ \Rightarrow & v_2=\frac{2 m_1 v_0}{3 m_2}\end{array}$
Given that, the bag rises upto height $h$.
So, at this height the kinetic energy is equal to potential energy.
$\begin{array}{ll}\text { i.e., } & \mathrm{KE}=\mathrm{PE} \Rightarrow \frac{1}{2} m_2 v_2^2=m_2 g h \\ \Rightarrow & h=\frac{1}{2 g} v_2^2=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2\end{array}$
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