(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 =

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

(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 $K = \frac{1}{2}(m+M) \left( \frac{mv}{m+M} \right)^2$.Simplifying this expression,$K = \frac{1}{2}(m+M) \frac{m^2v^2}{(m+M)^2} = \frac{1}{2}mv^2 \times \frac{m}{m+M}$.

Class 11 Physics Work, Energy, Power and Collision Perfectly Inelastic Collision

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$A$ 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

A
$\frac{1}{2}mv^2 \times \frac{m}{m+M}$
B
$\frac{1}{2}mv^2 \times \frac{M}{m+M}$
C
$\frac{1}{2}mv^2 \times \frac{M+m}{M}$
D
$\frac{1}{2}mv^2 \times \frac{2m}{M+m}$

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Perfectly Inelastic Collision

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MediumJEE MAIN 2020

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$A$ body of mass $5 \times 10^{3} \, kg$ moving with speed $2 \, m/s$ collides with a body of mass $15 \times 10^{3} \, kg$ inelastically and sticks to it. The loss in $K.E.$ of the system will be (in $kJ$):

DifficultAIIMS 2019

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$A$ body of mass $m$ moving with a constant velocity $v$ hits another body of the same mass moving with the same velocity $v$ but in the opposite direction and sticks to it. The velocity of the compound body after collision is

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$A$ metal ball of mass $2 \,kg$ moving with a velocity of $36 \,km/h$ has a head-on collision with a stationary ball of mass $3 \,kg$. If after the collision,the two balls move together,the loss in kinetic energy due to the collision is ........ $J$.

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$A$ 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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$A$ particle of mass $m$ moving in the $x$-direction with speed $2v$ is hit by another particle of mass $2m$ moving in the $y$-direction with speed $v$. If the collision is perfectly inelastic,the percentage loss in the energy during the collision is close to .............. $\%$

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