Discuss elastic collision in two dimension.

As shown in figure, suppose a ball of mass $m_{1}$ moving in X-direction with speed $v_{1 i}$ collides elastically with a stationary ball of mass $m_{2}$.

After the collision these ball move in the direction making angle $\theta_{1}$ and $\theta_{2}$ with the X-axis with velocities $v_{1 f}$ and $v_{2 f}$.

Momentum is conserved in collision,

$\therefore$ Momentum before collision $=$ momentum after collision.

$\ldots$ (1)

$\left[\because v_{2 i}=0\right]$

Taking X-components of momentum, $m_{1} v_{1 i}=m_{1} v_{1 f} \cos \theta_{1}+m_{2} v_{2 f} \cos \theta_{2} \quad \ldots$ (2)

Taking Y-components of momentum, $0=m_{1} v_{1 f} \sin \theta_{1}-m_{2} v_{2} f^{\sin \theta_{2}}$

$\ldots(3)$

Since the collision is elastic,

$\therefore \text { Kinetic energy before collision = kinetic energy after collision }$

$\frac{1}{2} m_{1} v_{1 i}^{2}=\frac{1}{2} m_{1} v_{1 f}^{2}+\frac{1}{2} m_{2} v_{2 f}^{2} \ldots \text { (4) }\left[\because v_{2 i}=0\right]$

Here, there are three equations (2), (3) and (4) and from these unknown quantities can be determined.

Usually the value of $m_{1}, m_{2}, v_{1 i}$ and $v_{2 i}$ are known, whereas four terms $v_{1 f} v_{2 \rho} \theta_{1}$ and $\theta_{2}$ are unknown. At least one of the four unknown quantities must be known as three equations can be give the values of only three unknown quantities.

If the motion in one dimension then unknown terms $v_{1 f}$ and $v_{2 f}$ are only be obtained because in this motion $\theta_{1}=\theta_{2}=0$ and hence unknown term can be known by the equations in one dimension.