Physics · Work, Energy And Power · NEET
Take masses m1, m2 with initial velocities u1, u2 and final velocities v1, v2 along one line. Write two conservation laws. Momentum: m1u1 + m2u2 = m1v1 + m2v2, which rearranges to m1(u1 - v1) = m2(v2 - u2). Kinetic energy: (1/2)m1u1^2 + (1/2)m2u2^2 = (1/2)m1v1^2 + (1/2)m2v2^2, which rearranges to m1(u1^2 - v1^2) = m2(v2^2 - u2^2). Divide the KE equation by the momentum equation. Using a^2 - b^2 = (a-b)(a+b), this gives u1 + v1 = u2 + v2, so u1 - u2 = v2 - v1. Substitute this back into the momentum equation and solve for v1 and v2 to get v1 = [(m1 - m2)/(m1 + m2)]u1 + [2m2/(m1 + m2)]u2 and v2 = [(m2 - m1)/(m1 + m2)]u2 + [2m1/(m1 + m2)]u1.
A collision is called elastic by definition when no kinetic energy is lost to heat, sound, or permanent deformation. During the impact the bodies briefly deform and store energy like a compressed spring, then push apart and return all of that energy as kinetic energy. In inelastic collisions some of this stored energy stays as heat or deformation, so kinetic energy is not conserved there. Momentum is always conserved in both types because no external force acts along the line of motion.
This is a very useful result: the relative velocity of approach before collision equals the relative velocity of separation after collision. In words, the speed at which the two bodies come toward each other before hitting equals the speed at which they move apart after hitting. This holds only for elastic collisions (coefficient of restitution e = 1). It is often faster to use this equation with momentum conservation than to plug into the long formulas.
No. Just keep signs consistent. Pick one direction as positive; any body moving the opposite way gets a negative velocity in u1 or u2. Substitute the signed values into the same formulas. The answer's sign then tells you the final direction: positive means it moves in your chosen positive direction, negative means it moves the other way.
Body A of mass 4m moving with speed u collides with another body B of mass 2m at rest. The collision is head-on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Momentum is conserved because there is no external force along the line of collision. Kinetic energy is conserved because, by definition, an elastic collision loses no energy to heat or deformation.
It is exactly 1. This is why the relative velocity of separation equals the relative velocity of approach: e = (v2 - v1)/(u1 - u2) = 1.
Yes. For elastic collisions, u1 - u2 = v2 - v1 combined with momentum conservation m1u1 + m2u2 = m1v1 + m2v2 gives two simple equations that are usually faster to solve than the full formulas.
NEET regularly asks for final velocities, fraction of energy transferred, or the height a struck bob rises after an elastic hit. All of these come from this one derivation, so knowing it saves time on multiple question types.
The same formulas apply. Just insert the correct signed values of u1 and u2. No body needs to be at rest for the derivation to hold.
