Momentum and collisions for JEE Main

Collision questions look varied but rest on two equations. Momentum is conserved in every collision, and one more condition tells you how bouncy it was: the coefficient of restitution. Once you can write those two lines quickly, most of the chapter becomes algebra.

In the syllabus, impulse and momentum conservation sit under the laws of motion, and elastic and inelastic collisions in one and two dimensions sit under work, energy and power. Expect them to combine with energy, circular motion and the centre of mass.

Momentum and impulse

Linear momentum is p = mv, a vector. Newton's second law in its general form is F = dp/dt.

Impulse is the change in momentum: J = ∫F dt = Δp. On an F–t graph, impulse is the area under the curve. Average force = Δp / Δt.

This explains everyday examples. A cricketer draws the hands back while catching, and a car's crumple zone collapses. Both stretch Δt, so the same Δp needs a smaller force.

Conservation of linear momentum

If the net external force on a system is zero, its total momentum is constant. Internal forces come in action–reaction pairs and cancel.

Two useful refinements:

• If the external force is zero along one direction only, momentum is conserved along that direction.
• During a very short collision or explosion, finite external forces such as gravity deliver negligible impulse. So you may conserve momentum across the collision even though gravity acts.

Recoil: a gun of mass M fires a bullet of mass m at speed v. Initial momentum is zero, so the gun recoils at V = mv/M in the opposite direction.

Collisions in one dimension

Take masses m₁ and m₂ with velocities u₁ and u₂ before, v₁ and v₂ after, all along one line.

• Momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
• Restitution: e = (v₂ − v₁)/(u₁ − u₂), the velocity of separation divided by the velocity of approach.

Solving the two equations together gives:

• v₁ = [(m₁ − em₂)u₁ + (1 + e)m₂u₂] / (m₁ + m₂)
• v₂ = [(m₂ − em₁)u₂ + (1 + e)m₁u₁] / (m₁ + m₂)

Kinetic energy lost: ΔK = ½ × [m₁m₂/(m₁ + m₂)] × (1 − e²) × (u₁ − u₂)². It is zero when e = 1 and largest when e = 0.

Special cases of an elastic collision with m₂ at rest

With e = 1 and u₂ = 0: v₁ = (m₁ − m₂)u₁/(m₁ + m₂) and v₂ = 2m₁u₁/(m₁ + m₂).

• Equal masses exchange velocities: the moving ball stops, the other moves off with u₁.
• A very heavy body hitting a light one carries on at about u₁; the light one flies off at about 2u₁.
• A light body hitting a very heavy one rebounds at about −u₁.
• The fraction of kinetic energy transferred to the target is 4m₁m₂/(m₁ + m₂)², which is 1 only for equal masses.

Collisions in two dimensions

Resolve momentum along two perpendicular axes and conserve each component separately. Kinetic energy is conserved as well only if the collision is elastic.

A classic result: when two equal masses collide elastically and one was at rest, they move off at 90° to each other (unless the collision is head-on). Momentum gives p = p₁ + p₂ as vectors; energy gives p² = p₁² + p₂². Both hold only if p₁·p₂ = 0.

A ball bouncing on the floor

Dropped from height h onto a floor with restitution e:

• Speed after the nth bounce = eⁿ × √(2gh); height after the nth bounce = e²ⁿh.
• Total distance travelled before it stops = h(1 + e²)/(1 − e²).

Worked problems

Problem 1: general 1D collision. A 2 kg ball moving at 4 m/s hits a 1 kg ball at rest head-on, with e = 0.5. Find both final velocities and the energy lost.

v₁ = [(2 − 0.5 × 1) × 4]/3 = 6/3 = 2 m/s.
v₂ = [(1 + 0.5) × 2 × 4]/3 = 12/3 = 4 m/s.
Check: momentum 2 × 2 + 1 × 4 = 8 kg m/s, as before; e = (4 − 2)/4 = 0.5.
KE before = 16 J; after = 4 + 8 = 12 J; 4 J lost. The formula agrees: ½ × (2/3) × 0.75 × 16 = 4 J.

Problem 2 (numerical answer): ballistic pendulum. A 10 g bullet at 400 m/s embeds in a 1.99 kg block hanging from a string. How high, in cm, does the block rise (g = 10 m/s²)?

Momentum during the collision: 0.01 × 400 = 2 × V, so V = 2 m/s.
Energy after the collision: h = V²/2g = 4/20 = 0.2 m = 20 cm.
The bullet's 800 J of kinetic energy drops to 4 J in the collision, so 99.5% is lost as heat and deformation. This is why you cannot use energy conservation across the impact.

Problem 3: two dimensions. A 2 kg ball moving east at 3 m/s collides and sticks with a 1 kg ball moving north at 6 m/s. Find the final velocity and the energy lost.

Momentum east = 6 kg m/s; north = 6 kg m/s. Total = 6√2 kg m/s, carried by 3 kg.
v = 2√2 ≈ 2.83 m/s at 45° north of east.
KE before = 9 + 18 = 27 J; after = ½ × 3 × 8 = 12 J; 15 J lost.

Problem 4: impulse from a graph. A force on a 0.5 kg ball at rest rises uniformly from 0 to 100 N in 0.02 s, then falls uniformly to 0 in the next 0.02 s. Find the ball's final speed.

Impulse = area of the triangle = ½ × 0.04 × 100 = 2 N s. v = 2/0.5 = 4 m/s.

Practice set

1. A ball dropped from 10 m has e = 0.5 with the floor. Find the heights after the first and second bounces.
2. A 10 g bullet at 500 m/s embeds in a 990 g block at rest on a smooth floor. Find the common velocity and the fraction of kinetic energy lost.
3. A 0.2 kg ball hits a wall at 10 m/s and rebounds at 8 m/s along the same line. Contact lasts 0.01 s. Find the average force.
4. A 5 kg gun fires a 20 g bullet at 500 m/s. Find the recoil speed.
5. A 1 kg ball at 6 m/s collides elastically head-on with a 2 kg ball at rest. Find both final velocities and the fraction of kinetic energy transferred.
6. A 3 kg shell at rest explodes into 1 kg and 2 kg pieces. The 1 kg piece moves at 20 m/s. Find the other piece's velocity and the energy released.
7. A ball dropped from 5 m has e = 0.5. Find the total distance it travels before coming to rest.
8. A ball collides elastically with an identical ball at rest and is deflected 30° from its original line. At what angle to that line does the second ball move?

Answers

1. e²h = 0.25 × 10 = 2.5 m; then 0.25 × 2.5 = 0.625 m.
2. v = 5/1 = 5 m/s. KE after/before = m/(m + M) = 0.01, so 99% is lost.
3. Δp = 0.2 × 18 = 3.6 N s; F = 3.6/0.01 = 360 N.
4. V = 0.02 × 500/5 = 2 m/s, opposite to the bullet.
5. v₁ = (1 − 2)/3 × 6 = −2 m/s (rebounds); v₂ = 2 × 1 × 6/3 = 4 m/s. Transferred fraction = 4 × 1 × 2/9 = 8/9 (16 J of 18 J).
6. 1 × 20 = 2 × v, so 10 m/s in the opposite direction. Energy = 200 + 100 = 300 J.
7. h(1 + e²)/(1 − e²) = 5 × 1.25/0.75 = 25/3 ≈ 8.33 m.
8. The two move at 90° to each other, so the second ball moves at 60° on the other side of the line.

What to do next

• Derive v₁ and v₂ from the momentum and restitution equations once.
• Solve 20 problems, writing "momentum conserved?" and "energy conserved?" before each.
• Practise two-dimensional collisions by resolving along two axes every time.
• Revise work, energy and power, then read centre of mass, which explains why momentum conservation works for whole systems.