A ball of mass $m$ moving with velocity $v$ collides head-on with the second ball of mass $m$ at rest. If the coefficient of restitution is $e$ and velocity of first ball after collision is $v_{1}$ and velocity of second ball after collision is $v_{2}$ then

v_{1} = \frac{(1 - e) v}{2}, v_{2} = \frac{(1 + e) v}{2}

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v_{1} = \frac{(1 + e) v}{2}, v_{2} = \frac{(1 - e) v}{2}

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v_{1} = \frac{v}{2}, v_{2} = - \frac{v}{2}

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v_{1} = u, v_{2} = - u

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Solution

The correct option is A $v_{1} = \frac{(1 - e) v}{2}, v_{2} = \frac{(1 + e) v}{2}$
$m v + 0 = m v_{1} + m v_{2}$
$\Rightarrow v = v_{1} + v_{2} . . . . . (1)$
and $e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$e = \frac{v_{2} - v_{1}}{v - 0} . . . . (2)$
and $v_{2} - v_{1} = e v$
$v_{2} + v_{1} = v$
$2 v_{2} = v (1 + e) \Rightarrow v_{2} = \frac{v (1 + e)}{2}$
and $v_{1} = v \frac{(1 - e)}{2}$

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A ball of mass m moving with velocity v collides head-on with the second ball of mass m at rest. If the coefficient of restitution is e and velocity of first ball after collision is v1 and velocity of second ball after collision is v2 then

The correct option is A v1=(1−e)v2,v2=(1+e)v2mv+0=mv1+mv2⇒v=v1+v2.....(1)and e=v2−v1u1−u2e=v2−v1v−0....(2)and v2−v1=evv2+v1=v2v2=v(1+e)⇒v2=v(1+e)2and v1=v(1−e)2

A ball of mass $m$ moving with velocity $v$ collides head-on with the second ball of mass $m$ at rest. If the coefficient of restitution is $e$ and velocity of first ball after collision is $v_{1}$ and velocity of second ball after collision is $v_{2}$ then