2 plastic balls have masses of $100 g$ and $200 g$ . The $100 g$ ball is moving at $3 m s^{- 1}$ and hits the $200 g$ stationary ball. They collide with each other and after the collision, the $100 g$ ball moves with a velocity of $1.5 m s^{- 1}$ in the same direction. Calculate the velocity of the other ball after collision.

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Solution

Let $m_{1} = 100 g = 0.1 k g$ be the mass of first ball, $m_{2} = 200 g = 0.2 k g$ be the mass of second ball.
$u_{1}$ initial velocity of first ball before collision $= 3 m s^{- 1}$
$u_{2}$ initial velocity of second ball before collision $= 0 m s^{- 1}$
$v_{1}$ final velocity of first ball after collision $= 1.5 m s^{- 1}$
$v_{2}$ final velocity of second ball after collision
Applying law of conservation of momentum we can write,
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Substituting values
$0.1 \times 3 + 0.2 \times 0 = 0.1 \times 1.5 + 0.2 \times v_{2}$
$0.3 = 0.15 + 0.2 \times v_{2}$
Or $v_{2} = 0.75 m s^{- 1}$

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