NDA Physics · Laws of Motion and Forces

Conservation of Momentum and Collisions

In the absence of external forces, total linear momentum is conserved; this governs recoil, collisions, and any system where mass is being added or ejected.

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Why this matters

Roughly 8 PYQs across 2019–2024 — the second-biggest computation pocket in the chapter. Every problem reduces to one rule: total momentum before = total momentum after. Recoil of a gun, a boy jumping onto a cart, sand on a conveyor belt, equal-mass elastic collisions — all are the same conservation statement with the algebra rearranged.

Concept 1 of 3: Conservation of linear momentum

If no outside force pushes on a system, its total momentum cannot change — internal forces (like a chemical explosion or two balls colliding) only shuffle momentum between the parts. So before-and-after, the total stays the same. This single rule solves recoil, explosions, and 'stick-together' collisions.

Definition

Law of conservation of linear momentum: for a system with no net external force, the total momentum is constant: $\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}$ . Internal forces (explosions, collisions, chemical reactions) cannot change the total momentum or the velocity of the centre of mass. It follows directly from Newton's third law: internal action-reaction pairs cancel.

Conservation of momentum (two bodies)

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

m₁, m₂masses of the two bodies
u₁, u₂velocities before
v₁, v₂velocities after

Worked example

A 10 g bullet is fired at 300 m/s from a 1 kg pistol initially at rest. Find the recoil velocity of the pistol.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2022 · Q149Moderate

Example 1 · Laws of Motion and Forces · Conservation of Momentum and Collisions

A bullet of mass 10 g is horizontally fired with velocity 300 m s

^{-1}

from a pistol of mass 1 kg. What is the recoil velocity of the pistol?

Internal forces can't move the centre of mass

NDA 2019: an object moving at velocity v has a chemical reaction inside it. The reaction is internal, so it cannot change the velocity of the centre of mass (statement 1 correct). It CAN, however, redistribute kinetic energy among the particles, so the energy-conservation statement is false. Internal forces redistribute, never reset the total.

Drill 4 more on conservation of linear momentum

Concept 2 of 3: Force when mass is added or ejected (variable mass)

When mass piles onto a moving object — sand falling onto a belt, rain filling a wagon — momentum must still be accounted for. Falling material that has no horizontal motion adds horizontal mass but no horizontal momentum, so the speed drops to keep momentum constant. To keep the belt moving steadily you must supply a force equal to the rate at which momentum is being added.

Definition

When mass is added or ejected, force is the rate of change of momentum: $F = \dfrac{dp}{dt} = v\,\dfrac{dm}{dt}$ when the added mass arrives with no momentum along the motion. If material falls vertically onto a horizontally moving body, horizontal momentum is conserved: $M_1 v_1 = M_2 v_2$ , so the body slows as it gains mass.

Force to maintain speed while loading mass at rate dm/dt

F = v\,\frac{dm}{dt}

Fforce needed to keep speed constant
v(constant) speed of the body
dm/dtrate at which mass is added

Worked example

Sand falls vertically onto a conveyor belt at 0.1 kg/s. The belt must keep moving at a steady 2 m/s. What horizontal force is needed on the belt?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2023 · Q122Moderate

Example 2 · Laws of Motion and Forces · Conservation of Momentum and Collisions

Sand falls vertically on a conveyor belt at a rate of 0·1 kg/s. In order to keep the belt moving at a uniform speed of 2 m/s, the force required to be applied on the belt is :

Vertically-falling mass adds no horizontal momentum

NDA 2023 (rain into a moving wagon): the rain has zero horizontal speed, so horizontal momentum is conserved as M₁v₁ = M₂v₂ — the wagon SLOWS. Don't assume the speed stays the same; the added mass must be dragged up to speed, and with no external horizontal force the wagon pays for it by slowing.

Drill 1 more on force when mass is added or ejected (variable mass)

Concept 3 of 3: Collisions — elastic and the equal-mass result

In a collision, momentum is always conserved. In an ELASTIC collision, kinetic energy is conserved too. The most-tested special case: when two equal masses collide head-on elastically and one was at rest, they simply swap velocities — the moving one stops dead and the struck one moves off with the full original speed.

Definition

In all collisions momentum is conserved. In an elastic collision kinetic energy is also conserved; in an inelastic collision some KE is lost (to heat/deformation), and in a perfectly inelastic collision the bodies stick and move together. Special case — equal masses, one at rest, elastic, head-on: the velocities are exchanged. The moving body stops and the target moves off with the incoming speed.

Equal-mass elastic head-on collision (m₂ initially at rest)

v_1' = 0, \qquad v_2' = u_1

u₁speed of the incoming body (mass m)
v₁'speed of body 1 after — it stops
v₂'speed of body 2 after — it takes the full speed

Worked example

A 0.2 kg ball moving at 6 m/s collides head-on with a stationary 0.3 kg ball. After the collision the 0.2 kg ball stops dead. Find the speed of the 0.3 kg ball.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q133Moderate

Example 3 · Laws of Motion and Forces · Conservation of Momentum and Collisions

A 100 g sphere is moving at a speed of 20 m/s and collides with another sphere of mass 50 g. If the second sphere was at rest prior to the collision and the first sphere comes at rest immediately after the collision, considering the collision to be elastic, the speed of the second sphere would be

Equal-mass elastic collision: velocities are EXCHANGED

NDA 2024 (bob X hits identical bob Y at rest): X does not bounce back or rise on the other side — it STOPS at the collision point and Y carries off all the speed. This swap only happens for equal masses in an elastic head-on hit; unequal masses share the velocity differently.

Use momentum (not KE) to find the unknown speed

When a problem tells you the post-collision state (e.g. the first sphere stops), use conservation of MOMENTUM to find the other speed. Plugging into kinetic-energy conservation is unnecessary and a common time-sink; momentum alone gives the answer directly.

Drill 1 more on collisions — elastic and the equal-mass result

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Conservation of linear momentum
Conservation of momentum (two bodies)
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
Force when mass is added or ejected (variable mass)
Force to maintain speed while loading mass at rate dm/dt
$F = v\,\frac{dm}{dt}$
Collisions — elastic and the equal-mass result
Equal-mass elastic head-on collision (m₂ initially at rest)
$v_1' = 0, \qquad v_2' = u_1$

Watch out for (4)

Internal forces can't move the centre of mass
→ Conservation of linear momentum
Vertically-falling mass adds no horizontal momentum
→ Force when mass is added or ejected (variable mass)
Equal-mass elastic collision: velocities are EXCHANGED
→ Collisions — elastic and the equal-mass result
Use momentum (not KE) to find the unknown speed
→ Collisions — elastic and the equal-mass result

Test yourself on Laws of Motion and Forces

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