Collisions
What is conserved in a collision?
Carrom strikers, billiard balls, shunting railway wagons: collisions are everywhere. Let us see what stays the same when two bodies collide.
During the collision, each body pushes the other with equal and opposite forces. So the total momentum before equals the total momentum after. Always.
Total energy is also always conserved. But kinetic energy is a different story.
In a slow-motion replay, watch two rubber balls squash and then spring back. During the contact, some kinetic energy is stored as the balls deform.
If they spring back fully, all the kinetic energy comes back. This is an elastic collision: momentum and kinetic energy are both conserved.
Now two lumps of clay hit and stay squashed. Some kinetic energy turns into heat, sound and lasting deformation. This is an inelastic collision.
When the bodies stick together and move off as one, as railway wagons do when they couple, it is a completely inelastic collision. The most kinetic energy is lost.
Collisions in one dimension
Cart one, mass m₁, moves at u₁ towards cart two, mass m₂, which is at rest. They hit head-on and move along the same line.
If they stick together, momentum alone is enough: m₁u₁ equals m₁ plus m₂ times v. So v equals m₁u₁ over m₁ plus m₂.
For an elastic collision we have two equations: one for momentum and one for kinetic energy. Solving them gives the final velocities.
v₁ equals m₁ minus m₂ over m₁ plus m₂, times u₁. And v₂ equals 2 m₁ over m₁ plus m₂, times u₁.
Case one: equal masses. Then v₁ is zero and v₂ equals u₁. The first cart stops dead and the second moves off with its speed. They swap velocities.
Case two: a heavy cart hits a light one. The heavy cart barely slows down, and the light one shoots off at nearly twice u₁.
Case three: a light ball hits a very heavy one, like a ball thrown at a wall. The heavy one hardly moves, and the light ball bounces back with almost the same speed.
Collisions in two dimensions
On a carrom board, the striker hits a coin off-centre. The two fly apart at angles. This is a collision in two dimensions.
Momentum is a vector, so it is conserved separately along x and along y.
Along x: m₁u₁ equals m₁v₁ cos θ₁ plus m₂v₂ cos θ₂.
Along y, there was no momentum to start with, so: zero equals m₁v₁ sin θ₁ minus m₂v₂ sin θ₂.
That is two equations but four unknowns: two speeds and two angles. If we measure one angle, and the collision is elastic, we can find the rest.
Here is a neat result. When two equal masses collide elastically and one was at rest, they always fly apart at right angles: θ₁ plus θ₂ equals 90 degrees.
So a billiard player who sends the target ball off at 37 degrees knows the cue ball will leave at 53 degrees.
Try it: colliding carts
- Choose Elastic or Sticky, set the masses and speeds, then press Collide.
- Total momentum is the same before and after in every collision. Check the readout.
- Kinetic energy is lost only in the sticky collision. How much?
Equal masses
A carrom striker slides at 3 m/s and hits an identical striker at rest head-on in an elastic collision. What happens just after?
- Both move forward at 1.5 m/s
- The first stops; the second moves at 3 m/s
- The first bounces back at 3 m/s; the second stays at rest
- Both move forward at 3 m/s
Answer
B. With m₁ = m₂, the elastic formulas give v₁ = (m₁ − m₂)u₁/(m₁ + m₂) = 0 and v₂ = 2m₁u₁/(m₁ + m₂) = u₁. The bodies swap velocities. “Both at 1.5 m/s” conserves momentum but loses half the kinetic energy, so it cannot be elastic.
Sticky collision
A 2 kg trolley moving at 3 m/s hits a 1 kg trolley at rest. They lock together. How fast do they move, and how much kinetic energy is lost?
- 1.5 m/s, no energy lost
- 2 m/s, 3 J lost
- 3 m/s, 0 J lost
- 2 m/s, 6 J lost
Answer
B. Momentum: 2 × 3 = (2 + 1)v, so v = 2 m/s. KE before = ½ × 2 × 3² = 9 J. KE after = ½ × 3 × 2² = 6 J. So 3 J becomes heat, sound and deformation. Momentum is conserved; kinetic energy is not.
Summary
- Momentum (every collision):
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ - Completely inelastic, m₂ at rest:
v = m₁u₁ / (m₁ + m₂) - Elastic 1D, m₂ at rest: first body:
v₁ = (m₁ − m₂)u₁ / (m₁ + m₂) - Elastic 1D, m₂ at rest: second body:
v₂ = 2m₁u₁ / (m₁ + m₂)
- Momentum and total energy are conserved in every collision. Kinetic energy is conserved only in an elastic collision.
- In a completely inelastic collision the bodies stick together and the most kinetic energy is lost.
- Equal masses in a 1D elastic collision swap velocities.
- In 2D, conserve momentum along x and y separately. Equal masses, one at rest, elastic: they fly apart at 90°.