A bullet of mass m moving with velocity v str | vedclass

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(A) Initial momentum of the system,$p_i = mv$.Let $V$ be the velocity of the composite system after the collision.Final momentum of the system,$p_f =

Vedclass · Accepted Answer

$\frac{1}{2}mv^2 	imes \frac{m}{m+M}$

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(A) Initial momentum of the system,$p_i = mv$.Let $V$ be the velocity of the composite system after the collision.Final momentum of the system,$p_f = (m+M)V$.By the law of conservation of linear momentum,$p_i = p_f$,so $mv = (m+M)V$.Therefore,the velocity of the composite block is $V = \frac{mv}{m+M}$.The kinetic energy of the composite block is $K = \frac{1}{2}(m+M)V^2$.Substituting the value of $V$,we get $K = \frac{1}{2}(m+M) \left( \frac{mv}{m+M} ight)^2$.Simplifying this expression,$K = \frac{1}{2}(m+M) \frac{m^2v^2}{(m+M)^2} = \frac{1}{2}mv^2 	imes \frac{m}{m+M}$.

$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

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