JEE Mains Physics · Work, Energy and Power

Impulse, Explosions and Perfectly Inelastic Collisions

Impulse is the change in momentum; in an explosion, a recoil or a collision where the bodies stick, total momentum is conserved but kinetic energy is not.

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Why this matters

Twenty-one PYQs, sixteen of them multiple choice, and two from 2026. Seven are about impulse: a ball rebounding from the ground, a bat or a wall, the area under a force–time graph, and the average force over a contact time; nine conserve momentum in explosions, recoils, blocks that split and bodies that stick together; five fire a bullet into a pendulum bob and follow the swing. Momentum is a vector in every one of them, so a rebound or a sideways fragment changes the arithmetic.

Concept 1 of 3: Impulse and change in momentum

Impulse is force multiplied by the time it acts, and it equals the change in momentum. Momentum is a vector, so a body that bounces straight back has its momentum reversed: the change is the sum of the two momenta, not their difference. The same change spread over a longer time needs a smaller force.

Definition

$\vec J = \displaystyle\int \vec F\,dt = \Delta\vec p = m(\vec v - \vec u)$ .
Straight back from u to v: $|\Delta p| = m(u + v)$ ; with the same speed, $2mu$ .
Dropped from $h_1$ , rebounding to $h_2$ : $|\Delta p| = m\left(\sqrt{2gh_1} + \sqrt{2gh_2}\right)$ .
Hitting a wall at angle θ to the normal and leaving at the same speed and angle: only the normal part reverses, $|\Delta p| = 2mv\cos\theta$ , directed along the normal.
Average force: $F_{\text{avg}} = \Delta p/\Delta t$ ; N bodies striking together give N times that.
On an F–t graph the impulse is the area under the curve.

Impulse

\vec J = \int \vec F\,dt = \Delta\vec p \qquad F_{\text{avg}} = \frac{\Delta p}{\Delta t}

Worked example

A 0.2 kg ball is dropped from 5 m. It hits the floor and rebounds to 1.25 m. The contact lasts 0.01 s. Find the impulse from the floor and the average force, ignoring gravity during the contact. (

g = 10

m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q17Moderate

Example 1 · Work, Energy and Power · Impulse, Explosions and Perfectly Inelastic Collisions

A spherical body of mass

100\text{ }g

is dropped from a height of

10\text{ }m

from the ground. After hitting the ground, the body rebounds to a height of

5\text{ }m

. The impulse of force imparted by the ground to the body is given by : (given

g = 9.8\text{ }m/s^{2}

)

A rebound adds the two speeds

Momentum reverses on a rebound, so the change is m(u + v). Subtracting the speeds treats the ball as if it carried on in the same direction, and gives far too small an impulse.

At an angle, only the normal part reverses

A wall pushes only along its normal. A ball hitting at 45° to the normal has an impulse of 2mv cos 45°, which is 1/√2 of the impulse for a head-on hit at the same speed.

Same impulse, smaller force

A body brought to rest from the same speed always has the same impulse. A longer stopping time lowers the average force, not the impulse.

Drill 6 more on impulse and change in momentum

Concept 2 of 3: Conservation of momentum in explosions, recoil and sticking collisions

With no outside force in a direction, the total momentum in that direction cannot change. A body at rest that explodes has pieces whose momenta add to zero as vectors. Two bodies that stick move off together with the total momentum, and some kinetic energy is lost as heat and sound.

Definition

$\sum m\vec u = \sum m\vec v$ , added as vectors.
From rest into two pieces: equal and opposite momenta, so $K_1 : K_2 = m_2 : m_1$ . A gun recoils at $V = mv/M$ .
Three pieces from rest: the third piece's momentum cancels the vector sum of the other two. Two equal momenta p at right angles add to $\sqrt2\,p$ .
Bodies that stick: $v = \dfrac{m_1u_1 + m_2u_2}{m_1 + m_2}$ , with directions as signs.
Kinetic energy lost: $\tfrac12\dfrac{m_1m_2}{m_1 + m_2}(u_1 - u_2)^{2}$ . As heat it can raise the temperature: $Q = (m_1 + m_2)\,s\,\Delta T$ .
A moving body that splits can GAIN kinetic energy: the extra comes from the internal energy released.

Momentum conserved

m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2 \qquad \Delta K_{\text{lost}} = \frac12\frac{m_1m_2}{m_1 + m_2}\left(u_1 - u_2\right)^{2}

Worked example

A 6 kg shell at rest explodes into pieces of 1 kg, 2 kg and 3 kg. The 1 kg piece flies east at 24 m/s and the 2 kg piece flies north at 9 m/s. Find the velocity of the 3 kg piece.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q2Moderate

Example 2 · Work, Energy and Power · Impulse, Explosions and Perfectly Inelastic Collisions

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio

2:2:3

. The two pieces of equal masses fly off perpendicular to each other with a speed of

18\text{ }m/s

each. The velocity of the heavier fragment is

\_\_\_\_

m/s

Momenta add as vectors

Two fragments with momentum p each, flying at right angles, have a total momentum of √2·p, not 2p. The third fragment must carry √2·p back the other way.

Kinetic energy is not conserved when bodies stick

Only momentum carries through a sticking collision. Writing ½m₁u₁² = ½(m₁ + m₂)v² gives a wrong common speed; the lost energy goes into heat, sound and deformation.

An explosion can increase kinetic energy

When a moving block splits, the pieces can together have more kinetic energy than the block had. Momentum is conserved; kinetic energy is not, in either direction.

Drill 8 more on conservation of momentum in explosions, recoil and sticking collisions

Concept 3 of 3: Bullet and pendulum: momentum in the impact, energy in the swing

A bullet hitting a pendulum bob is two separate events. The impact is very short, so momentum is conserved but kinetic energy is not. The swing that follows is slow, the string does no work, and mechanical energy is conserved. Use one law for each stage, in order.

Definition

Bullet embeds: $mu = (M + m)V$ . Then the swing: $V^{2} = 2gh$ . Together: $u = \dfrac{M + m}{m}\sqrt{2gh}$ .
Bullet passes through and leaves at v: $mu = mv + MV$ .
Bullet bounces back at v: $mu = -mv + MV$ , so the bob gets more momentum than the bullet brought.
To take the bob round a full vertical circle on a string, it needs $V = \sqrt{5gL}$ at the bottom.
Kinetic energy lost in an embedding impact: $\tfrac12 mu^{2}\cdot\dfrac{M}{M + m}$ , nearly all of it when $M \gg m$ .

Ballistic pendulum

mu = (M + m)V \qquad V = \sqrt{2gh} \qquad V_{\text{full circle}} = \sqrt{5gL}

Worked example

A 20 g bullet moving at 450 m/s embeds itself in a 2.98 kg block hanging on a long string. How high does the block rise? (

g = 10

m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q16Moderate

Example 3 · Work, Energy and Power · Impulse, Explosions and Perfectly Inelastic Collisions

A simple pendulum of length

1\text{ }m

has a wooden bob of mass

1\text{ }kg

. It is struck by a bullet of mass

10^{- 2}\text{ }kg

moving with a speed of

2 \times10^{2}{\text{ }ms}^{- 1}

. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use

g = 10\text{ }m/s^{2}

)

Do not conserve energy through the impact

Setting ½mu² = (M + m)gh skips the impact and claims the bullet's kinetic energy all goes into the swing. Most of it becomes heat. Use momentum for the impact, then energy for the swing.

A bullet that bounces back gives the bob extra momentum

If the bullet recoils at v, the bob's momentum is MV = m(u + v), not m(u − v). The minus sign of the recoiling velocity turns into a plus.

Drill 4 more on bullet and pendulum: momentum in the impact, energy in the swing

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)

Impulse and change in momentum
Impulse
$\vec J = \int \vec F\,dt = \Delta\vec p \qquad F_{\text{avg}} = \frac{\Delta p}{\Delta t}$
Conservation of momentum in explosions, recoil and sticking collisions
Momentum conserved
$m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2 \qquad \Delta K_{\text{lost}} = \frac12\frac{m_1m_2}{m_1 + m_2}\left(u_1 - u_2\right)^{2}$
Bullet and pendulum: momentum in the impact, energy in the swing
Ballistic pendulum
$mu = (M + m)V \qquad V = \sqrt{2gh} \qquad V_{\text{full circle}} = \sqrt{5gL}$

Watch out for (8)

A rebound adds the two speeds
→ Impulse and change in momentum
At an angle, only the normal part reverses
→ Impulse and change in momentum
Same impulse, smaller force
→ Impulse and change in momentum
Momenta add as vectors
→ Conservation of momentum in explosions, recoil and sticking collisions
Kinetic energy is not conserved when bodies stick
→ Conservation of momentum in explosions, recoil and sticking collisions
An explosion can increase kinetic energy
→ Conservation of momentum in explosions, recoil and sticking collisions
Do not conserve energy through the impact
→ Bullet and pendulum: momentum in the impact, energy in the swing
A bullet that bounces back gives the bob extra momentum
→ Bullet and pendulum: momentum in the impact, energy in the swing

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