A ball X of mass 1 kg traveling at 2 m/s has a head-on collision with an identical ball Y at rest. X stops and Y moves off. Calculate the velocity of Y after the collision.

Here given

Before collision:

Mass of the ball X, $m_X=1\ kg$

Speed of ball X, $u_X=2\ m/s$

Mass of ball Y, $m_Y=1\ kg$

Speed of ball Y, $u_Y=0$   [at rest]

After collision:

Velocity of ball X after the collision, $v_X=0$

Let $v_Y$ be the velocity of ball Y after the collision.

According to the law of conservation of momentum

$m_Xu_X+m_Yu_Y=m_Xv_X+m_Yv_Y$

Or $1\times 2+1\times0=1\times0+1\times v_Y$

Or $2+0=0+v_Y$

Or $v_Y=2\ m/s$

Therefore, the velocity of Y after the collision is $2\ m/s$. 

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Updated on: 10-Oct-2022

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