Physics · Work, Energy And Power · NEET
It is a number that tells you how much a collision keeps the objects moving apart. Take the speed at which the two bodies move apart after they hit (velocity of separation) and divide it by the speed at which they came together before they hit (velocity of approach). That ratio is e. A high e means the objects bounce off fast; a low e means they barely separate. It has no unit because it is speed divided by speed.
For two bodies moving along one line, e = (v2 - v1) / (u1 - u2), where u1 and u2 are velocities before collision and v1 and v2 are velocities after collision (body 1 is behind body 2). The numerator (v2 - v1) is the velocity of separation and the denominator (u1 - u2) is the velocity of approach. Always keep a fixed sign convention (say, rightward positive) for all four velocities so the ratio comes out correct.
Kinetic energy can be lost in a collision (as heat, sound, deformation) but it can never increase on its own. If the bodies separated faster than they approached, kinetic energy would go up, which is impossible without an outside energy source. So the separation speed cannot exceed the approach speed, making e at most 1. It cannot be negative because bodies do not pass through each other. Hence 0 is less than or equal to e less than or equal to 1.
The floor (Earth) does not move, so its velocity is zero before and after. If the ball hits the floor with speed u and rebounds with speed v, then e = v / u. Using v = sqrt(2 g h2) for rebound height h2 and u = sqrt(2 g h1) for drop height h1, you get e = sqrt(h2 / h1). So drop from height h1, measure bounce height h2, and take the square root of their ratio.
Velocity of approach is how fast the two bodies come toward each other just before the collision, equal to (u1 - u2). Velocity of separation is how fast they move apart just after the collision, equal to (v2 - v1). The coefficient of restitution is exactly separation divided by approach. In a head-on collision both are measured along the same line joining the centres.
A moving block of mass m collides with a stationary block of mass 4m. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is v, the coefficient of restitution (e) is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
e = 1. In a perfectly elastic collision the velocity of separation equals the velocity of approach, and kinetic energy is fully conserved.
e = 0. The two bodies move together with a common velocity after collision, so the velocity of separation is zero and the maximum possible kinetic energy is lost.
No. It is a ratio of two velocities, so the units cancel and e is a pure (dimensionless) number with no unit.
Not for ordinary collisions, because that would mean kinetic energy increased on its own. It can only exceed 1 in special cases where an internal energy source (like a small explosion) adds energy during the collision. For NEET, take 0 <= e <= 1.
It depends mainly on the materials of the two colliding bodies (how elastic they are) and their surfaces. Harder, more elastic materials like steel give e close to 1; soft, sticky materials give e close to 0. It also varies a little with collision speed.
