(A) Since there is no external force acting on the system,the linear momentum of the system is conserved.Let $v_1$ and $v_2$ be the velocities of the

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(A) Since there is no external force acting on the system,the linear momentum of the system is conserved.Let $v_1$ and $v_2$ be the velocities of the blocks after the collision.Initial momentum = $mv + (2m)(0) = mv$.Final momentum = $m(0) + (2m)v_2 = 2mv_2$.By conservation of linear momentum: $mv = 2mv_2$,which gives $v_2 = v/2$.The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.Velocity of approach = $v - 0 = v$.Velocity of separation = $v_2 - 0 = v_2 = v/2$.Therefore,$e = \frac{v/2}{v} = 0.5$.

Class 11 Physics Work, Energy, Power and Collision Elastic Collision

Medium

$A$ block having mass $m$ collides with another stationary block having mass $2m$. The lighter block comes to rest after the collision. If the initial velocity of the first block is $v$,then the value of the coefficient of restitution must be:

AIIMS 2015 All AIIMS

A
$0.5$
B
$0.4$
C
$0.6$
D
$0.8$

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Work, Energy, Power and Collision

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Elastic Collision

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AIIMS 2015 Paper All AIIMS years →

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$A$ block having mass $m$ collides with another stationary block having mass $2m$. The lighter block comes to rest after the collision. If the initial velocity of the first block is $v$,then the value of the coefficient of restitution must be:

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$A$ body of mass $M_1$ collides elastically with another mass $M_2$ at rest. There is maximum transfer of energy when

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$Assertion$ : If collision occurs between two elastic bodies,their kinetic energy decreases during the time of collision.
$Reason$ : During collision,intermolecular space decreases and hence elastic potential energy increases.