MHT-CET Physics · Laws of Motion

Impulse, Momentum and Collisions

Impulse — force times time, or the area under a force–time graph — equals the change in momentum; with no external force the total momentum of a system is conserved, and in a collision the coefficient of restitution e compares the speed of separation to the speed of approach.

All Laws of Motion notes MHT-CET Physics strategy

Why this matters

17 PYQs, one HARD. Eight are impulse and momentum rates — a force–time graph, a conveyor belt, a machine gun, balls rebounding from a surface, momentum against kinetic energy. Nine are collisions: the speeds after a head-on collision, bodies that stick, the height of a rebound, and the mass ratio for a given slowing. Two cards.

Concept 1 of 2: Impulse and Rates of Momentum

Impulse FΔt, the area under a force–time graph, equals the change in momentum; starting from rest, it gives the speed. A steady stream of mass needs a force equal to the rate of change of momentum: sand dropped at M kg/s onto a belt moving at V needs MV newtons; n bullets a second of mass m at speed v push back with nmv; balls bouncing back from a wall change momentum by 2mv each. Momentum and kinetic energy are tied by KE = p²/2m, so 20% more momentum is 44% more energy, and at equal energies the heavier body has more momentum. Gravity acting on projectiles for time 2t changes their total momentum by (m₁ + m₂)g·2t.

Definition

$J = F\Delta t = \Delta p$ = area under F–t (8 N × 0.5 s + 4 N × 0.5 s on 3 kg ⇒ 2 m/s).
Conveyor: $F = V\dfrac{dm}{dt}$ . Gun: $F = n\,mv$ . Rebound: $\Delta p = 2mv$ per ball.
$KE = \dfrac{p^2}{2m}$ : p × 1.2 ⇒ KE × 1.44; equal KE ⇒ heavier has more p.
Bullets to stop a body: $n = \dfrac{Mv}{mu}$ (60 kg at 10 m/s with 50 g at 150 m/s ⇒ 80).

Impulse

\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}, \qquad F = \frac{dp}{dt}

Worked example

A 0.5 kg ball hits a wall at 12 m/s and rebounds at 8 m/s. Impulse, and the average force if contact lasts 0.02 s?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q21Moderate

Example 1 · Laws of Motion · Impulse, Momentum, and Collisions

1000 small balls, each weighing 1 gram, strike one square cm of area per second with a velocity 50 m/s in a normal direction and rebound with the same velocity. The value of pressure on the surface will be

Forgetting the rebound doubles the change

A ball that bounces straight back at the same speed changes momentum by 2mv, not mv. The pressure of 1000 balls a second at 50 m/s on 1 cm² is 10⁶ Pa with the factor 2.

Adding the areas of a force graph without their signs

Impulse is the SIGNED area under F–t: a force pointing backward subtracts. Only then does it equal the change in momentum.

Drill 7 more on impulse and rates of momentum

Concept 2 of 2: Collisions and the Coefficient of Restitution

Momentum is conserved in every collision; kinetic energy only in an elastic one. The coefficient of restitution e = (speed of separation)/(speed of approach) is 1 when elastic and 0 when the bodies stick. Two equations — momentum and restitution — give both final velocities. A body that stops after hitting a stationary one of mass M gives e = m/M. Bodies that stick share one velocity, found component by component when they move at right angles. A ball dropped onto a surface rebounds to e²h.

Definition

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ ; $e = \dfrac{v_2 - v_1}{u_1 - u_2}$ .
Elastic, second at rest: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u$ (slowed to 2u/3 ⇒ m₁ : m₂ = 5 : 1).
Equal masses, one at rest: $\dfrac{v_2}{v_1} = \dfrac{1 + e}{1 - e}$ .
First stops: $e = \dfrac{m}{M}$ . Rebound height: $e^2h$ (0.6 from 1 m ⇒ 0.36 m).
Sticking at right angles: each component shares the total mass (m east, m north at v ⇒ $\dfrac{v}{\sqrt{2}}$ ).

Restitution

e = \frac{v_2 - v_1}{u_1 - u_2}

Worked example

A 2 kg ball at 6 m/s hits a 4 kg ball at rest, e = 0.5. Velocities after the collision?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q45Moderate

Example 2 · Laws of Motion · Impulse, Momentum, and Collisions

A sphere of mass '

m

', moving with velocity '

3u

' collides head-on with another identical sphere at rest. If '

e

' is coefficient of restitution then what will be the ratio of velocity of the second sphere to that of first sphere after collision?

Rebound height e·h

The rebound SPEED is e times the impact speed; height goes as speed squared, so the height is e²h.

Conserving kinetic energy in every collision

Only momentum is always conserved. Kinetic energy is conserved only when e = 1; bodies that stick lose the most.

Drill 8 more on collisions and the coefficient of restitution

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 (2)

Impulse and Rates of Momentum
Impulse
$\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}, \qquad F = \frac{dp}{dt}$
Collisions and the Coefficient of Restitution
Restitution
$e = \frac{v_2 - v_1}{u_1 - u_2}$

Watch out for (4)

Forgetting the rebound doubles the change
→ Impulse and Rates of Momentum
Adding the areas of a force graph without their signs
→ Impulse and Rates of Momentum
Rebound height e·h
→ Collisions and the Coefficient of Restitution
Conserving kinetic energy in every collision
→ Collisions and the Coefficient of Restitution

Test yourself on Laws of Motion

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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