A bullet of mass m strikes a block of mass M | vedclass

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(A) Step $1$: Apply the principle of conservation of linear momentum during the collision.Since the bullet gets embedded in the block,the collision is

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${\left( {\frac{{{m^2}V_0^2}}{{(M + m)k}}} \right)^{1/2}}$

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(A) Step $1$: Apply the principle of conservation of linear momentum during the collision.Since the bullet gets embedded in the block,the collision is perfectly inelastic.Let $V'$ be the velocity of the combined system $(M+m)$ immediately after the collision.$m V_0 = (M + m) V'$$V' = \frac{m V_0}{M + m}$Step $2$: Apply the law of conservation of energy for the system after the collision.The kinetic energy of the system is converted into the potential energy of the spring at maximum compression $X_{\max}$.$\frac{1}{2} (M + m) (V')^2 = \frac{1}{2} k X_{\max}^2$$(M + m) \left( \frac{m V_0}{M + m} \right)^2 = k X_{\max}^2$$\frac{m^2 V_0^2}{M + m} = k X_{\max}^2$$X_{\max}^2 = \frac{m^2 V_0^2}{(M + m)k}$$X_{\max} = \sqrt{\frac{m^2 V_0^2}{(M + m)k}} = \left( \frac{m^2 V_0^2}{(M + m)k} \right)^{1/2}$

$A$ bullet of mass $m$ strikes a block of mass $M$ connected to a light spring of stiffness $k$,with a speed $V_0$. If the bullet gets embedded in the block,then the maximum compression in the spring is

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