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

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Question

Velocity of the system just after the collision

$\mathrm{mv}_{0}=(\mathrm{m}+\mathrm{M}) \mathrm{V'} \quad \Rightarrow \quad \mathrm{V}^{\prime

Vedclass · Accepted Answer

${\left( {\frac{{{m^2}v_0^2}}{{(M + m)k}}} \right)^{1/2}}$

Vedclass · Answer

Velocity of the system just after the collision

$\mathrm{mv}_{0}=(\mathrm{m}+\mathrm{M}) \mathrm{V'} \quad \Rightarrow \quad \mathrm{V}^{\prime}=\frac{\mathrm{mv}_{0}}{(\mathrm{m}+\mathrm{M})}$

Using work energy theorem

$\Delta \mathrm{K}=\mathrm{W}_{\mathrm{All}}=\mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{N}}+\mathrm{W}_{\mathrm{S}}$

(Assume friction force is absent)

$0-\frac{1}{2}(\mathrm{m}+\mathrm{M}) \mathrm{V}^{2}=0+0-\frac{1}{2} \mathrm{K} \mathrm{X}_{\max }^{2}$

$\frac{\mathrm{m}_{0}^{2} \mathrm{v}_{0}^{2}}{(\mathrm{m}+\mathrm{M})}=\mathrm{KX}_{\max }^{2} \Rightarrow \mathrm{X}_{\max }=\frac{\mathrm{m}_{0} \mathrm{v}_{0}}{\sqrt{\mathrm{K}(\mathrm{M}+\mathrm{m})}}=\sqrt{\frac{\mathrm{m}_{0}^{2} \mathrm{v}_{0}^{2}}{\mathrm{K}(\mathrm{M}+\mathrm{m})}}$

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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