A bullet of mass m travelling at speed v stri | vedclass

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Question

(A) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum.$mv + M(0) = (M + m)V_{\

Vedclass · Accepted Answer

$\frac{1}{2}mv^2 	imes \left( \frac{m}{m + M} ight)$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum.$mv + M(0) = (M + m)V_{	ext{common}}$$	herefore V_{	ext{common}} = \frac{mv}{M + m}$Now,the final kinetic energy $(KE_{	ext{final}})$ of the combined system is given by:$KE_{	ext{final}} = \frac{1}{2}(M + m)V_{	ext{common}}^2$Substituting the value of $V_{	ext{common}}$:$KE_{	ext{final}} = \frac{1}{2}(M + m) \left( \frac{mv}{M + m} ight)^2$$KE_{	ext{final}} = \frac{1}{2}(M + m) \frac{m^2v^2}{(M + m)^2}$$KE_{	ext{final}} = \frac{1}{2}mv^2 	imes \left( \frac{m}{M + m} ight)$

$A$ bullet of mass $m$ travelling at speed $v$ strikes a block of mass $M$ at rest and gets embedded in it. Finally,both will have a combined kinetic energy of:

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