Physics · Work, Energy And Power · NEET
Start from the general 1D elastic result: v1 = ((m1 - m2)/(m1 + m2)) u1 + (2 m2/(m1 + m2)) u2 and v2 = ((m2 - m1)/(m1 + m2)) u2 + (2 m1/(m1 + m2)) u1. Put m1 = m2 = m. The first bracket (m1 - m2) becomes 0, and each 2m/(2m) becomes 1. So v1 = u2 and v2 = u1. The bodies simply trade their velocities. If the second body was at rest (u2 = 0), the first stops dead and the second moves off with the full speed u1. This is exactly what you see in a Newton's cradle.
Take m1 much greater than m2, and u2 = 0. In v1 = ((m1 - m2)/(m1 + m2)) u1, the ratio is nearly 1, so v1 is approximately u1: the heavy body keeps almost its original velocity. In v2 = (2 m1/(m1 + m2)) u1, the ratio is nearly 2, so v2 is approximately 2 u1: the light body shoots off at nearly twice the heavy body's speed. Picture a moving truck grazing a football: the truck barely slows, the ball flies off fast.
Take m1 much less than m2, and u2 = 0. Now (m1 - m2)/(m1 + m2) is nearly -1, so v1 is approximately -u1: the light body rebounds with almost the same speed in the opposite direction. And 2 m1/(m1 + m2) is nearly 0, so v2 is approximately 0: the heavy body stays practically still. This is a ball bouncing off a wall. The wall (huge mass) does not move; the ball comes back.
Yes, for a target at rest. Maximum energy transfer from body 1 to body 2 happens when m1 = m2, because body 1 stops completely and hands over all its kinetic energy. When the masses are very different (heavy-to-light or light-to-heavy), only a small fraction of energy transfers. NEET 2023 tested exactly this: a bullet transfers maximum energy to a block when M = m.
Derive them. Memorise only the two general formulas for v1 and v2, then substitute the mass condition. This is safer in the exam because you will not mix up signs. The three special cases are just what those formulas reduce to when m1 = m2, m1 much greater than m2, or m1 much less than m2 with the target at rest.
Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision will respectively be
A bullet of mass m hits a block of mass M elastically. The transfer of energy is maximum when
Body A of mass 4m moving with speed u collides head-on and elastically with body B of mass 2m at rest. 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.
(1) Equal masses: velocities are exchanged. (2) Heavy body hits a light body at rest: heavy body continues at nearly u, light body moves at nearly 2u. (3) Light body hits a heavy body at rest: light body rebounds at nearly -u, heavy body stays almost still.
From v2 = (2 m1/(m1 + m2)) u1, when m1 is much larger than m2 the factor 2 m1/(m1 + m2) approaches 2. So the light body's speed approaches 2u1. The heavy body meanwhile keeps almost its full speed u1.
From v1 = ((m1 - m2)/(m1 + m2)) u1, when m1 is much smaller than m2 the factor approaches -1, so v1 approaches -u1. The minus sign means it reverses direction with nearly the same speed, like a ball off a wall.
Yes. Every elastic collision conserves both momentum and kinetic energy. The special cases are just simplified results of applying both conservation laws with particular mass conditions.
For every elastic collision the coefficient of restitution e = 1, meaning the relative speed of separation equals the relative speed of approach. The special cases here use that e = 1 plus momentum conservation. The coefficient of restitution becomes the key idea when collisions are not perfectly elastic (0 < e < 1).
