PhysicsCBSE Class 11 · Laws of Motion

Momentum, Impulse and Conservation of Momentum Class 11

Updated 17 October 2026 · Fact-checked

Momentum is p = mv. Newton's second law is F = dp/dt, so impulse J = ∫F dt = Δp. When no external force acts on a system its total momentum is conserved, which explains recoil and explosions; the centre of mass of a system with no external force keeps moving uniformly. For a steady stream of mass, the force is F = v dm/dt.

With no external force, total momentum stays zero, so the lighter body recoils with greater speed in the opposite direction (m·3v = 3m·v).

Illustration drawn for QuestionBuddy.

Understand Momentum, Impulse and Conservation of Momentum

Momentum and Newton's second law. Linear momentum is $p = m v$ . Newton's second law in its general form is $F = \frac{d p}{d t}$ , which for constant mass gives $F = m a$ .

Impulse. The impulse of a force is $J = \int F d t$ and equals the change in momentum, $J = Δ p$ (impulse-momentum theorem). For a short, large force, the average force is $F_{a v} = Δ p /Δ t$ ; a longer contact time means a smaller force for the same change in momentum (why cricketers draw hands back). The area under an F-t graph is the impulse. The SI unit is N s, equal to kg m/s.

Conservation of linear momentum. If the net external force on a system is zero, $\sum p$ is constant. Internal forces occur in equal and opposite pairs (third law) and cancel in the sum. Applications:

*Recoil:* initially at rest, $m_{1} v_{1} + m_{2} v_{2} = 0$ , so the lighter body moves faster in the opposite direction.
*Explosion:* total momentum just after equals that just before; the total kinetic energy increases.
*Perfectly inelastic collision:* bodies stick together, $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ , and kinetic energy is lost.
A shell exploding in flight: the centre of mass follows the original parabola until a piece lands.

Relative velocity trap. When a person jumps from a boat with speed $u$ *relative to the boat*, the boat recoils, so the person's ground speed is $u - v_{b o a t}$ ; apply momentum conservation in the ground frame.

Continuous flow. For a stream of mass $d m$ per second hitting a surface and stopping, $F = \frac{d p}{d t} = v \frac{d m}{d t}$ . For a jet of density $ρ$ , area $A$ and speed $v$ , $d m / d t = ρ A v$ , so $F = ρ A v^{2}$ .

Momentum, Impulse and Conservation of Momentum: formulas and key facts

Momentum: $p = m v$; kg m/s.
Second law: $F = \frac{d p}{d t}$; Constant mass: $F = ma$ .
Impulse: $J = \int F d t = Δ p$; Area under F-t graph.
Conservation: $\sum p_{i} = \sum p_{f}$; No external force.
Perfectly inelastic collision: $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$; Bodies stick together.
Force from a jet: $F = v \frac{d m}{d t} = ρ A v^{2}$; Jet stopped by a wall.

How to solve Momentum, Impulse and Conservation of Momentum questions

Pick the system so that the forces between its parts are internal, then equate the total momentum before and after along each axis.

1Choose the system and check that the net external force along the axis is zero.
2Fix a positive direction and write the momentum of each body with sign.
3Equate total momentum before and after (use the ground frame).
4For impulse, find Δp (watch the sign change on a rebound) and divide by Δt for the average force.
5For energy questions, compute KE before and after separately.
6For flow problems use F = v(dm/dt).

Momentum, Impulse and Conservation of Momentum in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Landing from a jump

Suppose a 50 kg child jumps off a low wall and lands at 4 m/s. On bent knees she stops in 0.5 s; on stiff legs she stops in 0.05 s.

Compare the average force.

Try it first, then show the solutionSolution

Δp = 50 × 4 = 200 N s.
Bent knees: F = 200/0.5 = 400 N; stiff legs: F = 200/0.05 = 4000 N.

Answer: 400 N versus 4000 N, ten times larger.

Memory hook: Spread the impulse over a longer time.

A skater pushing off

Suppose two skaters of 40 kg and 60 kg stand at rest on smooth ice and push apart. The 60 kg skater moves at 2 m/s.

Find the speed of the 40 kg skater.

Try it first, then show the solutionSolution

0 = 60 × 2 − 40 v, so v = 3 m/s.

Answer: 3 m/s in the opposite direction.

Memory hook: Smaller mass moves faster.

Tricks to remember Momentum, Impulse and Conservation of Momentum

Short memory aids. Say each one aloud once, then use it on a question the same day.

Impulse is Δp
Area under F-t graph = change of momentum.
Use it when: Graph and force-time questions.
Rebound doubles
A ball rebounding at the same speed has Δp = 2mv.
Use it when: Collision with a wall.
Recoil is opposite
Zero total momentum: m₁v₁ = −m₂v₂.
Use it when: Explosion and gun problems.
KE is not conserved
Momentum is conserved in all collisions; KE only in elastic ones.
Use it when: Collision questions.

Momentum, Impulse and Conservation of Momentum: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Impulse on a ball · 2 marks · Very short answerBoard level

A 0.2 kg ball arrives at a bat at 15 m/s and rebounds along the same line at 10 m/s. The contact lasts 0.01 s. Find the impulse and the average force.

Try it first, then show the solutionSolution

Take the rebound direction as positive: Δp = 0.2(10 − (−15)) = 5 N s. (1)
F = Δp/Δt = 5/0.01 = 500 N. (1)

Answer: Impulse 5 N s; average force 500 N.

Example 2: Boy jumping from a boat · 3 marks · Short answerHarder board level

A 60 kg boy standing on a 100 kg boat at rest on still water jumps horizontally with a speed of 5 m/s relative to the boat. Find the speed of the boat and the boy's speed relative to the ground.

Try it first, then show the solutionSolution

Let the boat recoil at v; the boy's ground velocity is 5 − v in the opposite direction. (1)
60(5 − v) = 100v, so 300 = 160v and v = 1.875 m/s. (1)
Boy's speed relative to the ground = 5 − 1.875 = 3.125 m/s. (1)

Answer: Boat 1.875 m/s; boy 3.125 m/s relative to the ground.

Example 3: Sand on a conveyor belt · 3 marks · Short answerHarder board level

Sand falls vertically on a horizontal belt at 5 kg/s. The belt moves at 2 m/s. Find the extra force needed to keep the belt moving at constant speed and the power it supplies.

Try it first, then show the solutionSolution

Each second 5 kg gains 2 m/s horizontally: F = v dm/dt = 2 × 5 = 10 N. (1)
Power = Fv = 10 × 2 = 20 W. (1)
Half of this (10 W) appears as kinetic energy of the sand; the other half is lost to friction as the sand slips on the belt. (1)

Answer: 10 N; 20 W.

Example 4: Bullet embedded in a block · 4 marks · Long answerHarder board level

A 25 g bullet moving at 400 m/s embeds itself in a 4.975 kg block at rest on a smooth table. Find the common velocity and the loss of kinetic energy.

Try it first, then show the solutionSolution

Momentum: 0.025 × 400 = 10 kg m/s = 5 v, so v = 2 m/s. (1)
KE before = ½ × 0.025 × 400² = 2000 J. (1)
KE after = ½ × 5 × 2² = 10 J. (1)
Loss = 1990 J (99.5% of the original), converted to heat and deformation. (1)

Answer: v = 2 m/s; loss 1990 J.

Example 5: Explosion at the top of the path · 4 marks · Long answerHOTS / competency

A 6 kg shell is fired at 50 m/s at an angle with cos θ = 0.6 and sin θ = 0.8 (g = 10 m/s²). At its highest point it explodes into 2 kg and 4 kg pieces; the 2 kg piece falls vertically down. Find the velocity of the 4 kg piece just after the explosion and the distance between the pieces when they land.

Try it first, then show the solutionSolution

At the top the shell has velocity 50 × 0.6 = 30 m/s horizontally. (1)
Momentum: 6 × 30 = 4v, so v = 45 m/s horizontally. (1)
u sin θ = 40 m/s: H = 40²/20 = 80 m, time to the top and time to fall are each 4 s, so the top is at x = 30 × 4 = 120 m. (1)
The 2 kg piece lands at 120 m; the 4 kg lands at 120 + 45 × 4 = 300 m. Separation = 180 m (the centre of mass lands at 240 m, the unexploded range). (1)

Answer: 45 m/s; 180 m apart.

Where students lose marks in Momentum, Impulse and Conservation of Momentum

The slips examiners see most often, and what to write instead.

Mistake: Using the same sign for incoming and rebounding velocities.
Forgetting the direction reversal.
Do this: Take one direction as positive and put the incoming speed negative.
Usually costs: The answer mark.
Mistake: Applying conservation of momentum with an external force acting.
Ignoring gravity, friction or a wall.
Do this: Check that external forces are zero (or negligible during the short event).
Usually costs: Concept mark.
Mistake: Using relative speed as ground speed in recoil problems.
Not converting frames.
Do this: Write the ground velocity of each body before equating momentum.
Usually costs: Method marks.
Mistake: Assuming kinetic energy is conserved in a collision where bodies stick together.
Mixing conservation laws.
Do this: Only momentum is conserved in inelastic collisions.
Usually costs: 1 mark.

Exam tips

Write the system and the reason momentum is conserved in one line.
Choose the positive direction in the first line and keep signs.
State that kinetic energy is not conserved when bodies stick or when an explosion adds energy.
For impulse questions, show Δp with the sign change explicitly.
For jets use F = ρAv²; note that the assumption is that the water stops.

Practice questions on Momentum, Impulse and Conservation of Momentum

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A ball of mass m strikes a wall normally with speed v and rebounds with the same speed. The magnitude of its change in momentum is
(a) 0 (b) mv (c) 2mv (d) mv/2
Hint
Reverse the velocity.
Answer
(c) 2mv.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If a bomb at rest explodes, the velocity of its centre of mass remains zero.
Reason (R): The forces during an explosion are internal and the net external force is zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Third-law pairs cancel.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
A constant force of 20 N acts on a 4 kg body at rest for 3 s. Find the impulse and the final speed.
Hint
J = Ft.
Answer
J = 20 × 3 = 60 N s; v = J/m = 15 m/s.
Q4 · 3 marks · Short answerHarder board level
Water leaves a hose of cross-section 2 cm² at 10 m/s and strikes a wall normally without rebounding. Find the force on the wall (density of water 1000 kg/m³).
Hint
dm/dt = ρAv.
Answer
dm/dt = 1000 × 2 × 10⁻⁴ × 10 = 2 kg/s; F = v dm/dt = 10 × 2 = 20 N.
Q5 · 4 marks · Long answerHOTS / competency
Two skaters of masses 50 kg and 70 kg stand at rest on ice and push each other apart. The 50 kg skater moves at 3.5 m/s. Find (a) the speed of the other skater, (b) the total kinetic energy gained, (c) their separation 4 s later, and (d) the source of the energy.
Hint
Total momentum stays zero.
Answer
(a) 50 × 3.5 = 70v, so v = 2.5 m/s. (b) ½ × 50 × 12.25 + ½ × 70 × 6.25 = 306.25 + 218.75 = 525 J. (c) Relative speed 6 m/s, so 24 m. (d) Chemical energy of their muscles (internal work), not the momentum.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Impulse equals?Show answer
Change in momentum.
2.SI unit of impulse?Show answer
N s (or kg m/s).
3.Condition for momentum conservation?Show answer
Zero net external force.
4.Recoil rule?Show answer
m₁v₁ = −m₂v₂.
5.Force by a stopped jet?Show answer
ρAv².
6.Perfectly inelastic collision?Show answer
Bodies stick; KE is lost.

Momentum, Impulse and Conservation of Momentum: frequently asked questions

Is momentum conserved when an external force acts?

Only if the external force is zero, or the event is so brief that its impulse is negligible.

Why does the lighter body recoil faster?

Because m₁v₁ = m₂v₂ in magnitude, so a smaller mass needs a larger speed.

Is energy lost in an inelastic collision?

Kinetic energy is lost (to heat, sound and deformation) but total energy is conserved.

What does the area under an F-t graph give?

The impulse.

What is the relation between impulse and force?

The impulse equals the average force times the contact time.