A bullet of mass m_1 is moving with speed v_0 | vedclass

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Question

(A) The given situation involves a collision where linear momentum is conserved.Let $v_2$ be the velocity of the sand bag immediately after the bullet

Vedclass · Accepted Answer

$h=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2$

Vedclass · Answer

(A) The given situation involves a collision where linear momentum is conserved.Let $v_2$ be the velocity of the sand bag immediately after the bullet passes through it.According to the law of conservation of linear momentum:$p_i = p_f$$m_1 v_0 = m_2 v_2 + m_1 \left(\frac{v_0}{3}\right)$$m_2 v_2 = m_1 v_0 - \frac{m_1 v_0}{3} = \frac{2 m_1 v_0}{3}$$v_2 = \frac{2 m_1 v_0}{3 m_2}$Now,the sand bag rises to a height $h$. By the law of conservation of mechanical energy,the kinetic energy of the bag at the bottom is converted into potential energy at the maximum height $h$:$KE = PE$$\frac{1}{2} m_2 v_2^2 = m_2 g h$$h = \frac{v_2^2}{2 g}$Substituting the value of $v_2$:$h = \frac{1}{2 g} \left(\frac{2 m_1 v_0}{3 m_2}\right)^2$

$A$ bullet of mass $m_1$ is moving with speed $v_0$ and hits a sand bag of mass $m_2$. If the speed of the bullet after passing through 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).

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