1. Physics
  2. Laws of Motion
  3. Impulse and Conservation of Momentum

Laws of Motion · JEE & NEET Physics

Impulse and Conservation of Momentum: notes and previous year questions

Impulse and force–time graphs, why soft catches hurt less, conservation of momentum, collisions, restitution and explosions.

Impulse and Conservation of Momentum in short

  • Impulse J = ∫F dt is the area under the force–time graph and equals the change in momentum.
  • Average force is Δp/Δt.
  • For a fixed change in momentum, a longer contact time means a smaller force: catches, airbags, bent knees.
  • With no net external force, the total momentum of a system is conserved; this follows from the third law.

1Impulse

In a hit, a large force acts for a very short time. What changes the motion is the force and the time together: the impulse.

J⃗=∫t1t2F⃗ dt\vec J = \int_{t_1}^{t_2}\vec F\,dtFor a steady force, J = F Δt. A vector along the force; unit N s = kg m/s.

From the second law, F⃗=dp⃗/dt\vec F = d\vec p/dt, so F⃗ dt=dp⃗\vec F\,dt = d\vec p. Adding up over the whole collision gives the impulse–momentum theorem:

J⃗=Δp⃗=p⃗f−p⃗i\vec J = \Delta\vec p = \vec p_f - \vec p_i

Example: 50 N for 0.2 s gives J=10J = 10 N s. And 1 N s is exactly 1 kg m/s, since 1 N = 1 kg m/s².

2Force–time graphs

The impulse is the area under the force–time graph, whatever its shape: a rectangle for a steady force, a triangle or a spike for a hit.

Favg=JΔt=ΔpΔtF_{\text{avg}} = \frac{J}{\Delta t} = \frac{\Delta p}{\Delta t}The steady force that gives the same impulse in the same time.

3Soft landings

To stop a body of mass mm moving at vv you must remove momentum mvmv, however you do it. So F=mv/ΔtF = mv/\Delta t: stretch the time and the force drops.

  • Airbags and crumple zones lengthen a crash, so the force on people is smaller.
  • Bending your knees lengthens a landing.
  • Follow-through in golf, cricket or tennis lengthens contact for more control with a lower peak force.
  • A hammer does the opposite: a large force for a very short time.

4Conservation of momentum

When A and B interact, F⃗AB=−F⃗BA\vec F_{AB} = -\vec F_{BA} (third law), so dp⃗Adt+dp⃗Bdt=0\frac{d\vec p_A}{dt} + \frac{d\vec p_B}{dt} = 0.

F⃗ext=0  ⇒  p⃗total=constant\vec F_{\text{ext}} = 0 \;\Rightarrow\; \vec p_{\text{total}} = \text{constant}

Internal forces come in cancelling pairs, so they cannot change the total. Two trolleys at rest pushed apart by a spring: 2 kg at −3 m/s (−6 kg m/s) and 3 kg at +2 m/s (+6 kg m/s); the total stays zero.

Recoil: mbulletvbullet+mgunvgun=0m_{\text{bullet}}v_{\text{bullet}} + m_{\text{gun}}v_{\text{gun}} = 0. A 5 kg gun firing 20 g at 300 m/s recoils at 1.2 m/s; a 4 kg rifle firing 10 g at 400 m/s, at 1 m/s.

5Kinds of collision

CollisionMomentumKinetic energyeAfter
Elasticconservedconserved1bounce apart
Inelasticconservedsome lost0 < e < 1bounce, slower
Perfectly inelasticconservedmost lost0stick together

Lost kinetic energy becomes heat, sound and deformation; total energy is still conserved. Examples: hard steel balls and atoms (nearly elastic), car crashes and bouncing balls (inelastic), a bullet embedding in wood or clay balls (perfectly inelastic).

v=m1u1+m2u2m1+m2v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}Perfectly inelastic: the common velocity.
ΔK=12 m1m2m1+m2 (u1−u2)2\Delta K = \frac{1}{2}\,\frac{m_1m_2}{m_1 + m_2}\,(u_1 - u_2)^2Kinetic energy lost when they stick: the largest loss possible.

6Elastic collisions

Head-on and elastic: both momentum and kinetic energy are conserved.

m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
12m1u12+12m2u22=12m1v12+12m2v22\tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2 = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2
v1=(m1−m2)u1+2m2u2m1+m2v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}
v2=(m2−m1)u2+2m1u1m1+m2v_2 = \frac{(m_2 - m_1)u_2 + 2m_1u_1}{m_1 + m_2}
CaseResult
Equal massesvelocities swap: v1=u2v_1 = u_2, v2=u1v_2 = u_1
Equal masses, second at restv1=0v_1 = 0, v2=u1v_2 = u_1
Light hits heavy at rest (m1≪m2m_1 \ll m_2)v1≈−u1v_1 \approx -u_1, v2≈0v_2 \approx 0: bounces back
Heavy hits light at rest (m1≫m2m_1 \gg m_2)v1≈u1v_1 \approx u_1, v2≈2u1v_2 \approx 2u_1

Example: a 0.5 kg ball at 10 m/s hits an identical ball at rest: 0 and 10 m/s. A 0.2 kg ball hitting a wall at 6 m/s and bouncing back at 6 m/s changes its momentum by 0.2×12=2.40.2 \times 12 = 2.4 kg m/s: velocity is a vector, so the change is not zero.

7Restitution and bouncing

e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}Speed of separation over speed of approach: 0 ≤ e ≤ 1.

With ee given, solve momentum conservation together with v2−v1=e(u1−u2)v_2 - v_1 = e(u_1 - u_2). Example: 1 kg at 5 m/s hits 2 kg at rest with e=0.5e = 0.5: v1+2v2=5v_1 + 2v_2 = 5 and v2−v1=2.5v_2 - v_1 = 2.5, so v1=0v_1 = 0 and v2=2.5v_2 = 2.5 m/s.

On a fixed floor or wall (infinitely heavy), the rebound speed is v=euv = eu (e=1e = 1: the same speed back). Height goes as speed squared:

hn=e2nhh_n = e^{2n}hHeight after n bounces for a ball dropped from h.
Htotal=h 1+e21−e2H_{\text{total}} = h\,\frac{1 + e^2}{1 - e^2}Total distance before it stops: h + 2(e²h + e⁴h + …).

Example: dropped from 10 m with e=0.6e = 0.6. It hits at 200=102\sqrt{200} = 10\sqrt2 m/s, leaves at 626\sqrt2 m/s and rises to e2h=3.6e^2h = 3.6 m. Total distance: 10×1.36/0.64=21.2510 \times 1.36/0.64 = 21.25 m. From 8 m with e=0.5e = 0.5 it rises to 2 m; to bounce to 6.4 m from 10 m needs e=0.8e = 0.8.

8Explosions in two dimensions

Momentum is a vector, so each component is conserved separately:

m1u1x+m2u2x=m1v1x+m2v2xm_1u_{1x} + m_2u_{2x} = m_1v_{1x} + m_2v_{2x}And the same for y (and z).
v⃗cm=m1v⃗1+m2v⃗2m1+m2\vec v_{\text{cm}} = \frac{m_1\vec v_1 + m_2\vec v_2}{m_1 + m_2}p_total = (m₁ + m₂) v_cm, so with no external force the centre of mass keeps a constant velocity, whatever happens inside.

Summary

Key ideas

  • Impulse J = ∫F dt is the area under the force–time graph and equals the change in momentum.
  • Average force is Δp/Δt.
  • For a fixed change in momentum, a longer contact time means a smaller force: catches, airbags, bent knees.
  • With no net external force, the total momentum of a system is conserved; this follows from the third law.
  • Momentum is conserved in every collision; kinetic energy only in elastic ones.
  • Bodies that stick together lose the most kinetic energy.
  • In a head-on elastic collision, equal masses swap velocities.
  • The coefficient of restitution e is separation speed over approach speed.
  • A ball dropped on the floor rises to e²ⁿh after n bounces.
  • In two dimensions each component of momentum is conserved, and the centre of mass moves at constant velocity.

Every equation

Impulse
J=∫F dt=FΔtJ = \int F\,dt = F\Delta t
Impulse–momentum
J=Δp=pf−piJ = \Delta p = p_f - p_i
Average force
Favg=Δp/ΔtF_{\text{avg}} = \Delta p/\Delta t
Conservation
Fext=0⇒∑p⃗=constF_{\text{ext}} = 0 \Rightarrow \sum \vec p = \text{const}
Recoil
mbvb+mgvg=0m_b v_b + m_g v_g = 0
Sticking together
v=m1u1+m2u2m1+m2v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}
KE lost
ΔK=12m1m2m1+m2(u1−u2)2\Delta K = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(u_1 - u_2)^2
Elastic: first body
v1=(m1−m2)u1+2m2u2m1+m2v_1 = \frac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}
Elastic: second body
v2=(m2−m1)u2+2m1u1m1+m2v_2 = \frac{(m_2 - m_1)u_2 + 2m_1u_1}{m_1 + m_2}
Restitution
e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}
Rebound from a wall
v=euv = eu
Bounce heights
hn=e2nhh_n = e^{2n}h
Total path
H=h1+e21−e2H = h\frac{1 + e^2}{1 - e^2}
Components
∑mux=∑mvx\sum m u_x = \sum m v_x
Centre of mass velocity
v⃗cm=m1v⃗1+m2v⃗2m1+m2\vec v_{\text{cm}} = \frac{m_1\vec v_1 + m_2\vec v_2}{m_1 + m_2}

Previous year questions with solutions

Real JEE and NEET questions on impulse and conservation of momentum. Try each one before you open the solution.

Q1NEET 2025One correct option

A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a height of 10 m . The impulse imparted to the ball during its collision with the ground is (Take g=9.8m/s2g=9.8 m/s^{2} )

  1. A0
  2. B84 NS
  3. C21 NS
  4. D7 NS
Show answer and solution

Answer: Option C

Arriving: 2×9.8×40=784=28 m/s\sqrt{2 \times 9.8 \times 40} = \sqrt{784} = 28\ \mathrm{m/s}, down. Leaving: 2×9.8×10=196=14 m/s\sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14\ \mathrm{m/s}, up. With up positive, J=0.5×(14−(−28))=0.5×42=21 N sJ = 0.5 \times \left(14 - (-28)\right) = 0.5 \times 42 = 21\ \mathrm{N\,s}, upward. D, 7 N s7\ \mathrm{N\,s}, is the trap of subtracting the speeds, 0.5×(28−14)0.5 \times (28 - 14), as though the ball kept going down more slowly. A treats the bounce as nothing happening because the ball ends up moving again. Check: the answer must lie between 0.5×28=140.5 \times 28 = 14 (no bounce) and 0.5×56=280.5 \times 56 = 28 (a perfect bounce), and 2121 does.

Q2JEE Main 2024One correct option

An artillery piece of mass M1M_{1} fires a shell of mass M2M_{2} horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:

  1. AM1/(M1+M2)M_{1}/(M_{1}+M_{2})
  2. BM2M1\frac{M_{2}}{M_{1}}
  3. CM1M2\frac{M_{1}}{M_{2}}
  4. DM2/(M1+M2)M_{2}/(M_{1}+M_{2})
Show answer and solution

Answer: Option B

Gun and shell start at rest, so just after the firing they carry equal and opposite momenta, of the same size pp. Then KgunKshell=p2/2M1p2/2M2=M2M1\dfrac{K_{\text{gun}}}{K_{\text{shell}}} = \dfrac{p^{2}/2M_{1}}{p^{2}/2M_{2}} = \dfrac{M_{2}}{M_{1}} — small, since the shell is the lighter. The trap is C, the same ratio upside down, which is what "heavier means more energy" produces; that rule holds only at equal speeds. A and D, with M1+M2M_{1} + M_{2} underneath, are fractions of the whole energy released — D is in fact the artillery's share of it — not the ratio of one energy to the other. Check with the firework: its 3 kg3\ \mathrm{kg} piece plays the artillery and its 1 kg1\ \mathrm{kg} piece the shell, and the heavy piece had 13\tfrac{1}{3} of the light one's energy, which is M2M1=13\dfrac{M_{2}}{M_{1}} = \dfrac{1}{3}.

Q3JEE Advanced 2018Numerical answer

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4m=0.4 kgkg is at rest on this surface. An impulse of 1.01.0 NsNs is applied to the block at time t=0t=0 so that it starts moving along the xx-axis with a velocity v(t)=v0e−t/τ,v(t)=v_{0}e^{-t/\tau }, where v0v_{0} is a constant and τ=4s.\tau =4s. The displacement of the block, in metres, at t=τt=\tau is ______________ Take e−1=0.37.e^{-1}=0.37.

Show answer and solution

Answer: 6.3 m

The impulse is delivered before the block has gone anywhere, so all of it becomes momentum: J=mv0J = mv_{0} gives v0=1.00.4=2.5 m/sv_{0} = \dfrac{1.0}{0.4} = 2.5\ \mathrm{m/s}, and that is the v0v_{0} standing in v=v0e−t/τv = v_{0}e^{-t/\tau}. After that the oil takes over and the rest is integration: x=∫0τv0e−t/τ dt=v0τ(1−e−1)=2.5×4×0.63=6.3 mx = \int_{0}^{\tau} v_{0}e^{-t/\tau}\,dt = v_{0}\tau\left(1 - e^{-1}\right) = 2.5 \times 4 \times 0.63 = 6.3\ \mathrm{m}. Do the impulse first and on its own — it is the one step that turns a force acting over a time into a number the kinematics can start from, and nothing about the impact itself ever has to be known.

Practice questions, easy to hard

Three questions from the impulse and conservation of momentum practice ladder: one easy, one medium, one hard.

Q4One correct option

For one body the mass stays fixed, so K=p22mK = \dfrac{p^{2}}{2m} says K∝p2K \propto p^{2}: whatever factor multiplies the momentum, its square multiplies the kinetic energy. A 20%20\% rise in pp multiplies pp by 1.21.2 and KK by 1.441.44 — a 44%44\% rise, not 40%40\%. Run it backwards with a square root: KK four times as large means pp twice as large.

The kinetic energy of a body rises by 69%69\%. By what percentage does its momentum rise?

  1. A30%30\%
  2. B69%69\%
  3. C13%13\%
  4. D34.5%34.5\%
Show answer and solution

Answer: Option A

A 69%69\% rise multiplies KK by 1.691.69, and p∝Kp \propto \sqrt{K}, so pp is multiplied by 1.69=1.3\sqrt{1.69} = 1.3 — a 30%30\% rise. Always turn the percentage into a factor first; the square root belongs to the factor, not to the percentage. That is the trap in C, 169=13\sqrt{169} = 13, the root of the wrong number. D halves the percentage, which is only roughly right for tiny changes, and B forgets that KK and pp grow at different rates at all. Check forwards: 1.32=1.691.3^{2} = 1.69.

Q5One correct option

If the stream strikes at an angle θ\theta to the normal and rebounds at the same speed and angle, each impact reverses only the normal part of the velocity, so it hands over 2mvcos⁡θ2mv\cos\theta, and the average force is F=n×2mvcos⁡θF = n \times 2mv\cos\theta, along the normal. Spread over an area AA, that force is a pressure, P=FAP = \dfrac{F}{A}, in N/m2\mathrm{N/m^{2}}.

A hundred balls a second, each of mass 20 g20\ \mathrm{g}, strike a plate of area 0.5 m20.5\ \mathrm{m^{2}} at 10 m/s10\ \mathrm{m/s}, at 60∘60^{\circ} to the normal, and rebound at the same speed and angle. What average pressure do they exert on the plate?

  1. A80 N/m280\ \mathrm{N/m^{2}}
  2. B69.3 N/m269.3\ \mathrm{N/m^{2}}
  3. C20 N/m220\ \mathrm{N/m^{2}}
  4. D40 N/m240\ \mathrm{N/m^{2}}
Show answer and solution

Answer: Option D

Per impact: 2×0.02×10×cos⁡60∘=0.2 kg m/s2 \times 0.02 \times 10 \times \cos 60^{\circ} = 0.2\ \mathrm{kg\,m/s}. A hundred a second: F=20 NF = 20\ \mathrm{N}. Over 0.5 m20.5\ \mathrm{m^{2}}: P=200.5=40 N/m2P = \dfrac{20}{0.5} = 40\ \mathrm{N/m^{2}}. A leaves out the cos⁡60∘\cos 60^{\circ}, as if every ball came in head-on. B uses sin⁡60∘\sin 60^{\circ} — the component along the plate, which never changes. C is what comes of counting mvcos⁡θmv\cos\theta per ball instead of 2mvcos⁡θ2mv\cos\theta, forgetting that the rebound is paid for as well. Check the limit: at θ=90∘\theta = 90^{\circ} the balls only graze the plate and the pressure is zero.