Q.A ball with a velocity of $5\ \mathrm{m\,s^{-1}}$ impinges at an angle of $60^\circ$ with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.

Two-Dimensional Collision: The Intuition

Imagine two pool balls on a table. You hit one, it slides across the felt, and then it strikes the other ball at an angle — not dead centre, but off to the side. The balls don't just move along the same straight line after the hit; they scatter in different directions. That's a two-dimensional collision.

In one dimension, everything happens along a line — think of two trains bumping on a track. But in two dimensions, the collision is oblique: the objects approach each other at some angle, and after the collision they fly off in directions that are not along the original line of motion. The key insight is that momentum is a vector, and it is conserved component by component.

The Core Idea: Momentum Conservation in Two Axes

Momentum is a vector — it has both magnitude and direction. When two objects collide, the total momentum vector before the collision equals the total momentum vector after the collision. That single vector equation breaks into two independent scalar equations, one for each perpendicular direction.

We choose two perpendicular axes — typically the $x$-axis and $y$-axis — that lie in the plane of motion. Then:

• The total $x$-component of momentum before the collision equals the total $x$-component after.
• The total $y$-component of momentum before the collision equals the total $y$-component after.

These two equations are completely independent. Nothing that happens along $x$ affects the $y$-momentum conservation, and vice versa. This is the entire mathematical engine of two-dimensional collision analysis.

Here $\vec{u}$ are velocities before collision, $\vec{v}$ are velocities after, and $m$ are masses.

What About Energy?

Momentum is always conserved in any collision (provided no external forces act). Energy is a separate story.

• Elastic collision: Kinetic energy is also conserved. This gives you a third equation (scalar, not vector) that relates the speeds.
• Inelastic collision: Kinetic energy is not conserved — some is lost to heat, sound, or deformation. You only have the two momentum equations.

The Typical Problem Setup

Here is how a standard two-dimensional collision problem looks:

1. Object 1 (mass $m_1$) moves with known velocity $\vec{u}_1$ along the $x$-axis.
2. Object 2 (mass $m_2$) is initially at rest ($\vec{u}_2 = 0$).
3. They collide obliquely. After collision, object 1 moves at an angle $\theta_1$ to the $x$-axis with speed $v_1$, and object 2 moves at an angle $\theta_2$ to the $x$-axis with speed $v_2$.

You are typically asked to find some of these unknowns. The momentum equations give:

The minus sign in the $y$-equation appears because the two objects usually scatter to opposite sides of the $x$-axis — one with positive $y$, the other with negative $y$.

Why Two Axes Are Enough

You might wonder: why only $x$ and $y$? Because the collision happens in a plane — two dimensions. Any vector in that plane can be fully described by its components along two perpendicular axes. There is no third independent direction. So two equations from momentum conservation, plus possibly one from energy conservation, give you up to three equations to solve for unknowns.

A Concrete Example

A 2 kg ball moving at 3 m/s east strikes a stationary 1 kg ball. After collision, the 2 kg ball moves at 2 m/s at $30^\circ$ north of east. Find the velocity of the 1 kg ball.

Take east as $+x$, north as $+y$.

Before: $p_{1x} = 2 \times 3 = 6$ kg m/s, $p_{1y} = 0$, $p_{2x} = 0$, $p_{2y} = 0$.

After: $v_{1x} = 2\cos30^\circ = \sqrt{3}$ m/s, $v_{1y} = 2\sin30^\circ = 1$ m/s.

$x$-conservation: $6 = 2\sqrt{3} + 1 \times v_{2x} \implies v_{2x} = 6 - 2\sqrt{3} \approx 2.536$ m/s.

$y$-conservation: $0 = 2 \times 1 + 1 \times v_{2y} \implies v_{2y} = -2$ m/s (southward).

So the 1 kg ball moves with velocity $\vec{v}_2 = (2.536 \hat{i} - 2 \hat{j})$ m/s, at an angle $\theta = \tan^{-1}(-2/2.536) \approx -38.3^\circ$ (south of east).

The Big Picture

Two-dimensional collision analysis is just vector momentum conservation applied to a plane. The physics is no different from the one-dimensional case — you are simply writing the same law twice, once for each independent direction. The only new skill is resolving velocities into components and handling angles. Master that, and you master oblique collisions.

Two-dimensional collision problems are a recurring theme in Class 11 Physics numericals and JEE Main mechanics questions, and students searching for 'two-dimensional collision formula and examples' or 'oblique collision problems class 11 physics' will find this vector-momentum approach directly useful for board and competitive exam practice. This topic builds on the NCERT Class 11 Physics chapter on Work, Energy and Power, extending the one-dimensional collision idea covered there to motion in a plane.