On a smooth horizontal surface, a disc (of 1kg) sliding with kinetic energy $9J$ along x-axis strikes another disc (of 2kg) at rest. The lighter disc is seen to be moving with kinetic energy $3J$ along y-axis, after it strikes the heavier disc. Kinetic energy of the system lost during the collision is

If speed of the lighter disc before collision is u, speed of the lighter the collision is $\frac{u}{\sqrt{3}},$ (as kinetic energy becomes one-third). Let velocity of heavier disc after the collision is v making angle $\alpha$ with x-axis. By conservation of momentum

$1\times u=2\times v\cos \alpha -----\left(i\right)$

(b) Along y-axis

$\begin{array}{l}0=1\times \frac{u}{\sqrt{3}}-2\times v\sin \alpha \\ \Rightarrow 1\times \frac{u}{\sqrt{3}}=2v\sin \alpha \mathrm{....}\left(2\right)\end{array}$

From (1) & (2), by squaring and adding

$\begin{array}{l}{u}^2+{\left(\frac{u}{\sqrt{3}}\right)}^2=4v^2\\ \Rightarrow v=\frac{u}{\sqrt{3}}\end{array}$

Kinetic energy of heavier disc$=\frac{1}{2}\times 2\times {\left(\frac{u}{\sqrt{3}}\right)}^2=\frac{2}{3}\times \frac{1}{2}{u}^2$

$\frac{2}{3}\times 9=6J$

Total kinetic energy of the system after the collision$=3J+6J=9J$

Kinetic energy lost = zero