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

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Question

(A) Step $1$: Apply the Conservation of Linear Momentum $(COLM)$ during the perfectly inelastic collision between the bullet and the block.$mv_0 + M(0

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Step $1$: Apply the Conservation of Linear Momentum $(COLM)$ during the perfectly inelastic collision between the bullet and the block.$mv_0 + M(0) = (m + M)V$$V = \frac{m}{m + M}v_0$  --- $(1)$Step $2$: Apply the Conservation of Mechanical Energy $(CONE)$ for the system (block + bullet + spring) from the moment after the collision until the spring reaches maximum compression $(x)$. At maximum compression,the velocity of the combined mass becomes zero.$\frac{1}{2}(m + M)V^2 = \frac{1}{2}kx^2$  --- $(2)$Step $3$: Substitute the value of $V$ from equation $(1)$ into equation $(2)$:$\frac{1}{2}(m + M) \left( \frac{m}{m + M}v_0 \right)^2 = \frac{1}{2}kx^2$$(m + M) \frac{m^2}{(m + M)^2}v_0^2 = kx^2$$\frac{m^2 v_0^2}{(m + M)} = kx^2$$x^2 = \frac{m^2 v_0^2}{(m + M)k}$$x = \sqrt{\frac{m^2 v_0^2}{(m + M)k}} = \left( \frac{m^2 v_0^2}{(m + M)k} \right)^{1/2}$Thus,the correct option is $A$.

$A$ 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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