Perfectly Inelastic Collision
Imagine two lumps of wet clay flying toward each other. When they hit, they don't bounce apart — they squash together into one bigger lump and keep moving as a single object. That's the core picture: the colliding bodies stick together and move with a common velocity after the collision.
This is the defining feature of a perfectly inelastic collision. It's the most extreme case of inelasticity — the bodies lose their separate identities and become one.
Why "Perfectly Inelastic"?
The word "inelastic" means kinetic energy is not conserved. In any real collision, some kinetic energy transforms into heat, sound, or deformation. A perfectly inelastic collision takes this to the maximum: the bodies deform so much that they lock together, and the kinetic energy loss is as large as it can possibly be — while still obeying the law of conservation of momentum.
Do not confuse "perfectly inelastic" with "total energy loss." Momentum is always conserved in any collision (in the absence of external forces). Kinetic energy is not — and in this case, it's reduced to the minimum possible value.
The Precise Statement
For two masses m1 and m2 moving with initial velocities u1 and u2 along a straight line, after a perfectly inelastic collision they move together with a common velocity v.
Conservation of momentum gives:
m1u1+m2u2=(m1+m2)v
So the common velocity is:
v=m1+m2m1u1+m2u2
This is simply the velocity of the centre of mass of the system — which remains unchanged throughout the collision.
The Kinetic Energy Loss
Before collision, the total kinetic energy is:
Ki=21m1u12+21m2u22
After collision, with both masses moving together:
Kf=21(m1+m2)v2
The loss in kinetic energy is:
ΔK=Ki−Kf
You can show (by substituting v and simplifying) that:
ΔK=21m1+m2m1m2(u1−u2)2
This is the maximum possible kinetic energy loss for a given pair of masses and initial velocities, consistent with momentum conservation. Notice it depends only on the relative speed (u1−u2) and the reduced mass m1+m2m1m2.
If one body is initially at rest (u2=0), the loss simplifies to 21m1+m2m1m2u12. The larger the mass of the stationary body, the more energy is lost.
A Quick Check: The Coefficient of Restitution
The coefficient of restitution e is defined as: …