If you can answer one physics idea confidently in the CDS General Studies paper, make it momentum. The law of conservation of momentum explains why a gun recoils, why a rocket lifts off, and what happens when two bodies collide. It flows straight out of Newton’s laws, the formulas are short, and the questions are predictable. This page builds the concept from scratch and drills the exact patterns examiners love.
Why momentum is a guaranteed scorer
Across CDS and OTA General Studies papers, mechanics supplies a steady stream of questions, and momentum is among the most repeated single ideas. It is rule-based, the arithmetic is light, and most questions test the same conservation principle in slightly different clothing — a gun and bullet, two trolleys, a rocket, or a person jumping from a boat.
The reason momentum repeats is that it links three big chapters: Newton’s laws, collisions, and rocket propulsion. Learn it once and you unlock all three. Unlike topics that demand memorising long data, momentum rewards a clear understanding of one conserved quantity and the ability to track direction (sign) carefully.
Momentum is a property of moving mass. A stationary object has zero momentum no matter how heavy it is, and a light object moving fast can carry large momentum.
What momentum actually means
Linear momentum (symbol p) is the product of an object’s mass and its velocity. It measures ‘how much motion’ a body carries and how hard it is to stop.
p = m × v
Mass m in kilograms, velocity v in metres per second, so momentum is measured in kg·m/s (also written kg m s−1).
Two features matter for the exam:
- Momentum is a vector. It has both magnitude and direction — the same direction as the velocity. This is why direction (and sign) decides almost every numerical.
- It depends equally on mass and velocity. Doubling either one doubles the momentum. A 10-tonne truck crawling and a bullet flying can have comparable momentum.
This is the reason a fast cricket ball stings the palm while a gently rolled ball of the same mass does not: the faster ball brings more momentum that your hand must absorb.
A loaded truck is harder to stop than an empty one at the same speed because it carries more momentum; equally, the same truck is harder to stop at high speed than at low speed. Both mass and velocity pull in the same direction, which is why heavy fast-moving vehicles are the most dangerous on the road. Keeping this dual dependence in mind helps you reason out qualitative comparison questions without any calculation.
Momentum and Newton's second law
Newton actually stated his second law in terms of momentum, not the popular F = ma. The true statement is: the rate of change of momentum of a body is directly proportional to the applied force and acts in the direction of the force.
F = (change in momentum) ÷ time = Δp / Δt
For constant mass this reduces to F = m(v − u)/t = m × a, the familiar form.
This wider definition matters when mass itself changes — as in a rocket burning fuel. There you cannot simply use F = ma, but F = Δp/Δt still works.
If a question mentions a changing mass (rocket, water jet, conveyor of sand), reach for F = Δp/Δt, not F = ma.
Impulse: force acting over time
When a force acts for a short time, the product of force and time is called impulse (symbol J). Impulse equals the change in momentum it produces.
Impulse J = F × t = Δp = m(v − u)
Unit of impulse: newton-second (N·s), which is the same as kg·m/s.
The impulse idea explains many everyday safety designs. To reduce the force on a body, we increase the time over which momentum changes:
- A cricketer pulls the hands back while catching, lengthening t so the force on the palms is smaller.
- Air bags and crumple zones in cars extend the stopping time, cutting the force on passengers.
- We bend our knees on landing after a jump, again stretching t.
Same change in momentum, longer time → smaller force. Shorter time → larger, more damaging force.
The law of conservation of momentum
This is the heart of the chapter. It follows directly from Newton’s third law (action and reaction are equal and opposite).
In the absence of an external force, the total momentum of a system stays constant.
Total momentum before = Total momentum after
m1u1 + m2u2 = m1v1 + m2v2
The phrase ‘no external force’ is crucial. Internal forces between the two bodies (the push of a bullet on a gun, two trolleys colliding) cancel in pairs because of Newton’s third law, so they cannot change the total momentum of the whole system.
Think of it this way. When body A pushes body B, body B pushes back on A with an equal and opposite force for exactly the same length of time. So the impulse A gives B is cancelled by the impulse B gives A. Whatever momentum one body gains, the other loses an equal amount, and the grand total stays fixed. This is why the conservation law is so dependable in the exam: as long as the only forces act between the members of the system, you can equate total momentum before and after without worrying about the messy details of the interaction.
Conservation applies to the whole system, never to a single body. A bullet’s momentum clearly changes when fired; it is the bullet-plus-gun system whose total momentum is conserved.
Recoil of a gun and walking on a boat
Before firing, the gun and bullet are at rest, so the total momentum is zero. After firing, momentum must still total zero, so the gun must move backward to balance the forward bullet.
0 = mbulletvbullet + mgunVgun
So Vgun = − (mbulletvbullet) / mgun
The minus sign shows the gun moves opposite to the bullet — that backward kick is the recoil velocity. Because the gun is much heavier than the bullet, its recoil speed is small, but its momentum exactly matches the bullet’s.
The same logic explains other situations:
- When you step forward off a small boat, the boat slides backward.
- A balloon released with its neck open zips around as air rushes out one way and the balloon moves the other.
- A jet of water from a hose pushes back on the fireman holding it.
In every one of these cases the total momentum stays the same as before the event — zero, if everything started at rest. The forward momentum of the bullet, the air, or your body is precisely matched by the backward momentum of the gun, the balloon, or the boat. Examiners often dress the same idea in a new costume, so if you can spot ‘something at rest splits into two parts’, you already know the two parts carry equal and opposite momentum.
Rocket propulsion and jet engines
A rocket is the grand example of conservation of momentum. It burns fuel and throws out hot gases at high speed in one direction; to keep total momentum conserved, the rocket itself is pushed in the opposite direction.
A rocket does not need air to push against. It works perfectly in the vacuum of space because the thrust comes from ejecting its own gases, a pure momentum effect.
Because the rocket continuously loses mass as fuel burns, its acceleration grows even with the same thrust — less mass to move. This is also why questions about rockets use F = Δp/Δt rather than F = ma.
A favourite one-liner: ‘A rocket works on the principle of conservation of linear momentum.’ Memorise it word for word.
Collisions: elastic vs inelastic
In every collision, momentum is always conserved (provided no external force acts). What differs is whether kinetic energy survives.
- Elastic collision: both momentum and kinetic energy are conserved. The bodies bounce apart. Ideal example — colliding billiard or carrom-like hard balls.
- Inelastic collision: momentum is conserved but kinetic energy is not — some energy becomes heat, sound or deformation.
- Perfectly inelastic collision: the bodies stick together and move as one lump afterwards (a bullet embedding in a wooden block, two coupling railway wagons).
Students assume kinetic energy is always conserved. It is conserved only in elastic collisions. Momentum, however, is conserved in all collisions where no external force acts.
For a perfectly inelastic collision where masses stick: m1u1 + m2u2 = (m1 + m2)v, so the common final velocity v is easy to find.
Worked example: bullet and block
Let us apply conservation to a classic perfectly inelastic collision.
A bullet of mass 20 g (0.02 kg) moving at 400 m/s strikes a wooden block of mass 1.98 kg resting on a frictionless surface and gets embedded in it. Find the common velocity of the bullet-block system just after impact.
The bullet and block move together at 4 m/s. Notice momentum stayed 8 kg·m/s throughout, while the kinetic energy fell sharply — lost as heat and deformation, exactly as an inelastic collision predicts.
Units, sign and traps to avoid
Momentum questions are short, so most marks are lost to silly slips, not hard physics. Guard against these:
- Convert grams to kilograms. A 20 g bullet is 0.02 kg. Forgetting this wrecks the arithmetic.
- Track direction with signs. Take one direction as positive; the opposite is negative. Recoil and rebound answers carry a minus sign.
- Do not confuse momentum (p = mv) with kinetic energy (KE = ½mv2). Momentum is linear in v; kinetic energy is quadratic.
Equal momentum does not mean equal kinetic energy. For the same momentum p, the lighter body carries more kinetic energy because KE = p2/(2m), which rises as mass falls.
Remember the link KE = p2 / 2m. It turns several tricky comparison questions into one line.
Previous-year style question
Test yourself on the most repeated pattern — recoil velocity.
Q. A gun of mass 5 kg fires a bullet of mass 25 g (0.025 kg) with a muzzle velocity of 400 m/s. What is the recoil velocity of the gun?
Answer: Initial total momentum = 0 (both at rest). By conservation, 0 = (0.025 × 400) + (5 × V). So 10 + 5V = 0, giving V = −2 m/s. The gun recoils at 2 m/s, the minus sign showing it moves opposite to the bullet.
The gun’s momentum (5 × 2 = 10 kg·m/s) exactly equals the bullet’s (0.025 × 400 = 10 kg·m/s). Always cross-check that the two momenta balance.
Quick revision
- Momentum p = m × v, a vector measured in kg·m/s.
- Newton’s second law: F = Δp/Δt; impulse J = F×t = Δp (in N·s).
- Increase contact time to reduce force — the idea behind air bags and a cricketer’s soft hands.
- No external force → total momentum is conserved: explains recoil, boats, rockets and collisions.
- Momentum is conserved in all collisions; kinetic energy only in elastic ones.
- Useful links: Vrecoil = −mbulletv/mgun and KE = p2/2m.
Drill three recoil sums and three sticking-collision sums; that covers the bulk of what CDS asks on this topic.
Frequently asked questions
Is momentum a scalar or a vector quantity?
Momentum is a vector. It has both magnitude and direction, pointing in the same direction as the velocity. That is why tracking signs carefully is essential in recoil and collision problems.
What is the SI unit of momentum and of impulse?
Momentum is measured in kilogram-metre per second (kg·m/s). Impulse is measured in newton-second (N·s), which is dimensionally the same as kg·m/s, since impulse equals change in momentum.
On what principle does a rocket work?
A rocket works on the law of conservation of linear momentum. It ejects hot gases at high speed in one direction, and the rocket is pushed in the opposite direction. It does not need air to push against, so it works in vacuum.
Is momentum conserved in an inelastic collision?
Yes. Momentum is conserved in every collision where no external force acts, including inelastic ones. Only kinetic energy is not conserved in an inelastic collision; some of it converts to heat, sound or deformation.
Why does a gun recoil when a bullet is fired?
Before firing, the total momentum of the gun and bullet is zero. To keep the total zero after firing, the gun must gain backward momentum equal and opposite to the bullet's forward momentum, which is felt as recoil.
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