A $10$ kg ball and a $20$ kg ball approach each other with velocities $20$ m/s and $10$ m/s respectively. What are their velocities after collision if the collision is perfectly elastic.

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Solution

Let the velocities of $10 k g$ & $20 k g$ ball before collision are $u_{1} = 20 m / s$ (+X axis) & $u_{2} = - 20 m / s$ ( -X axis) and after collision $v_{1} & v_{2}$ respectively. Than by the principal of momentum conservation: -

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$(10 \times 20) + (20 \times (- 20)) = (10 \times v_{1}) + (20 \times v_{2})$

$v_{1} + 2 v_{2} = - 20$ ---------------------(i)

Since the collision is perfectly elastic (e = 1)

$e = (v_{2} - v_{1}) / (u_{1} - u_{2}) = 1$

$v_{2} - v_{1} = 20 - (- 20) = 40$ ------------------------(ii)

On adding (i) & (ii): -

$3 v_{2} = 20$

$44 v_{2} = 6.67 m / s$

$v_{1} = 6.67 - 40 = - 33.33 m / s$

The $10 k g$ ball will travel along $X - a x i s$ with a velocity of $33.33 m / s$ & the $20 k g$ ball will travel along $+ X a x i s$ with a velocity of $6.67 m / s .$

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