Work, Energy and Power
Two spheres on a frictionless track. Drag the coefficient of restitution e from 0 to 1 and watch the same collision morph from stick-together to perfect bounce — while momentum stays conserved every single time. Exact closed-form physics, live energy bars, the COM frame, and a challenge mode where you predict the final velocities first.
Drag the e slider (0→1) and hit Replay. Toggle COM frame to see why momentum is always conserved. Try Challenge mode — predict v₁, v₂ before revealing.
5 minutes · +4 right, −1 wrong (real NEET marking) · one global leaderboard.
Yes. On a frictionless track there is no external force, so total linear momentum is conserved in every collision regardless of e — even when e = 0 and the bodies stick together. What changes with e is the kinetic energy: only e = 1 conserves KE; for any e < 1 some KE is converted to heat, sound, and deformation.
They exchange velocities. If one is moving and the other is stationary, the moving one stops dead and the stationary one moves off with the original velocity. This velocity-swap is the most-tested NEET/JEE result for 1D elastic collisions.
Momentum is a vector tied to Newton's third law — the equal-and-opposite impulses cancel, so the system total never changes. Kinetic energy is a scalar that can be converted into other forms (heat, sound, permanent deformation) during the collision. The fraction lost is fixed by e: ½·(m₁m₂/M)·(1 − e²)·v_rel².
Both bodies move together at the velocity of the centre of mass, v_cm = (m₁u₁ + m₂u₂)/(m₁ + m₂). This is the case of maximum possible kinetic-energy loss for a given pair of initial momenta.
Ek ball mass m, speed u se ek STATIONARY ball (mass m) se elastic head-on collision karti hai. Collision ke baad?