A ball moving with a speed of $9 m / s$ strikes an identical stationary ball such that after the collision the direction of each ball makes an angle of $30^{o}$ with the original line of motion. Find the speeds of the two balls after the collision. Is the kinetic energy conserved in the collision process?

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Solution

By momentum conservation in Y-direction.
$m v_{1} sin 30 - m v_{2} 2 sin 30 = 0$
$v_{1} = v_{2}$
In X direction,
$m u_{1} = m v_{1} cos 30 + m v_{2} cos 30$
$m \times 3 = 2 m v_{1} \times \frac{\sqrt{3}}{2}$
$v_{1} = v_{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$

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A ball moving with a speed of 9m/s strikes an identical stationary ball such that after the collision the direction of each ball makes an angle of 30o with the original line of motion. Find the speeds of the two balls after the collision. Is the kinetic energy conserved in the collision process?

By momentum conservation in Y-direction.mv1sin30−mv22sin30=0 v1=v2In X direction, mu1=mv1cos30+mv2cos30 m×3=2mv1×√32 v1=v2=3√3=√3

By momentum conservation in Y-direction.
$m v_{1} sin 30 - m v_{2} 2 sin 30 = 0$
$v_{1} = v_{2}$
In X direction,
$m u_{1} = m v_{1} cos 30 + m v_{2} cos 30$
$m \times 3 = 2 m v_{1} \times \frac{\sqrt{3}}{2}$
$v_{1} = v_{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$