A 3.0\ kg object at 4.0\ m s^-1 collides with a 1.0\ kg object at -2.0\ m s^-1 and they stick together. Find their common velocity. [2 marks]

SEAB Original-style practice: A 2.0\ kg trolley moving at 3.0\ m s^-1 collides with and sticks to a stationary 4.0\ kg trolley. (a) Find their common velocity after the collision. (b) Determine whether the collision is elastic, supporting your answer with a calculation.

(a) Conserve momentum: m_1 u_1 = (m_1 + m_2)v. 2.0 × 3.0 = (2.0 + 4.0)v, so 6.0 = 6.0v, giving v = 1.0\ m s^-1. (b) Compare kinetic energy before and after. Before: E_k = 12(2.0)(3.0)^2 = 9.0\ J. After: E_k = 12(6.0)(1.0)^2 = 3.0\ J. Kinetic energy falls from 9.0\ J to 3.0\ J, so the collision is inelastic; 6.0\ J is transferred to other forms (heat, sound, deformation). Markers reward conserving momentum to get the common velocity, comparing kinetic energy before and after, and the explicit conclusion that the collision is inelastic because kinetic energy is not conserved.

SEAB Original-style practice: A stationary 0.50\ kg object explodes into two pieces. One piece of mass 0.20\ kg moves off at 6.0\ m s^-1 to the right. Find the velocity of the other piece.

Total momentum before the explosion is zero (the object is stationary). By conservation of momentum, the total momentum after must also be zero. 0 = (0.20)(6.0) + (0.30)v, so 0 = 1.2 + 0.30v, giving v = -4.0\ m s^-1. The 0.30\ kg piece moves at 4.0\ m s^-1 to the left. Markers reward setting total momentum to zero, correct masses (the second piece is 0.30\ kg), and the negative sign indicating the opposite direction.

SingaporePhysicsSyllabus dot point

Why is the total momentum of an isolated system conserved, and how does this distinguish elastic from inelastic collisions?

Apply the principle of conservation of linear momentum to collisions and explosions in one dimension, and distinguish elastic from inelastic collisions using kinetic energy

A focused answer to the H2 Physics learning outcome on momentum conservation. The principle, its basis in Newton's third law, one-dimensional collisions and explosions, and the elastic versus inelastic distinction using kinetic energy.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to apply conservation of linear momentum to one-dimensional collisions and explosions, to justify why momentum is conserved for an isolated system, and to distinguish elastic from inelastic collisions by testing whether kinetic energy is conserved. This is one of the most reliably examined topics in the mechanics section.

The answer

The principle and why it holds

The total linear momentum of a system is constant provided no resultant external force acts on it:

\sum p_{\text{before}} = \sum p_{\text{after}}

This follows directly from Newton's third law. In a collision, the forces the two bodies exert on each other are equal and opposite and act for the same time, so they produce equal and opposite changes in momentum. The internal momentum changes cancel, leaving the total unchanged.

Applying it to collisions

For a one-dimensional collision between masses $m_1$ and $m_2$ :

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

Choose a positive direction and assign signs to every velocity. The equation is a vector statement reduced to one dimension by signs.

Explosions

An explosion is the reverse of a collision: a body at rest (zero total momentum) breaks into pieces whose momenta must sum to zero. The fragments move in opposite directions with equal and opposite momenta. This is the same principle that propels a recoiling gun and a launching rocket.

Elastic and inelastic collisions

Momentum is conserved in every collision (with no external force). Kinetic energy is the discriminator:

Elastic collision: kinetic energy is conserved as well as momentum. Examples are idealised gas molecules and near-elastic ball collisions.
Inelastic collision: kinetic energy is not conserved; some is transferred to heat, sound or deformation. A perfectly inelastic collision is one where the bodies stick together.

To classify a collision, calculate the total kinetic energy before and after and compare.

The elastic test for relative speed

For a one-dimensional elastic collision, a useful result is that the relative speed of approach equals the relative speed of separation:

u_1 - u_2 = -(v_1 - v_2)

This often pairs with momentum conservation to solve both final velocities.

Worked example

A $0.40\ \text{kg}$ ball moving at $5.0\ \text{m s}^{-1}$ collides head-on and elastically with a stationary $0.40\ \text{kg}$ ball. Find the velocity of each ball after the collision.

Step 1: Write momentum conservation

0.40(5.0) + 0.40(0) = 0.40 v_1 + 0.40 v_2 \implies 5.0 = v_1 + v_2

Step 2: Use the elastic relative-speed relation

Approach speed $= 5.0 - 0 = 5.0\ \text{m s}^{-1}$ . Separation speed $= v_2 - v_1$ . For elastic: $v_2 - v_1 = 5.0$ .

Step 3: Solve the simultaneous equations

Adding $v_1 + v_2 = 5.0$ and $v_2 - v_1 = 5.0$ gives $2v_2 = 10$ , so $v_2 = 5.0\ \text{m s}^{-1}$ and $v_1 = 0$ .

Step 4: Interpret

The balls exchange velocities: the incoming ball stops and the struck ball moves off at $5.0\ \text{m s}^{-1}$ . This velocity-swap is the signature of an equal-mass elastic collision.

Examples in context

Example 1. Recoil of a rifle. A rifle and bullet start at rest with zero total momentum. When fired, the bullet's forward momentum is balanced by the rifle's backward recoil momentum. Because the rifle is far more massive, its recoil speed is small, but the two momenta are equal and opposite, illustrating the explosion case.

Example 2. Ballistic pendulum. A bullet embedding in a hanging block is perfectly inelastic: momentum conservation gives the combined speed just after impact, then energy conservation of the swinging block gives the height risen. Splitting the problem at the impact (momentum) and the swing (energy) is essential, because kinetic energy is lost in the embedding but conserved during the swing.

Try this

Q1. State the principle of conservation of linear momentum and the condition under which it applies. [2 marks]

Cue. Total momentum of a system is constant provided no resultant external force acts.

Q2. A $3.0\ \text{kg}$ object at $4.0\ \text{m s}^{-1}$ collides with a $1.0\ \text{kg}$ object at $-2.0\ \text{m s}^{-1}$ and they stick together. Find their common velocity. [2 marks]

Cue. $3.0(4.0) + 1.0(-2.0) = (4.0)v \Rightarrow 10 = 4.0v \Rightarrow v = 2.5\ \text{m s}^{-1}$ .

Q3. Explain how you would determine whether a given collision is elastic or inelastic. [3 marks]

Cue. Calculate total kinetic energy before and after; if equal the collision is elastic, if reduced it is inelastic (energy transferred to heat, sound, deformation).

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA

2.0\ \text{kg}

trolley moving at

3.0\ \text{m s}^{-1}

collides with and sticks to a stationary

4.0\ \text{kg}

trolley. (a) Find their common velocity after the collision. (b) Determine whether the collision is elastic, supporting your answer with a calculation.

Show worked answer →

(a) Conserve momentum: $m_1 u_1 = (m_1 + m_2)v$ .

$2.0 \times 3.0 = (2.0 + 4.0)v$ , so $6.0 = 6.0v$ , giving $v = 1.0\ \text{m s}^{-1}$ .

(b) Compare kinetic energy before and after.

Before: $E_k = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$ .

After: $E_k = \tfrac{1}{2}(6.0)(1.0)^2 = 3.0\ \text{J}$ .

Kinetic energy falls from $9.0\ \text{J}$ to $3.0\ \text{J}$ , so the collision is inelastic; $6.0\ \text{J}$ is transferred to other forms (heat, sound, deformation).

Markers reward conserving momentum to get the common velocity, comparing kinetic energy before and after, and the explicit conclusion that the collision is inelastic because kinetic energy is not conserved.

Original4 marksA stationary

0.50\ \text{kg}

object explodes into two pieces. One piece of mass

0.20\ \text{kg}

moves off at

6.0\ \text{m s}^{-1}

to the right. Find the velocity of the other piece.

Show worked answer →

Total momentum before the explosion is zero (the object is stationary).

By conservation of momentum, the total momentum after must also be zero.

$0 = (0.20)(6.0) + (0.30)v$ , so $0 = 1.2 + 0.30v$ , giving $v = -4.0\ \text{m s}^{-1}$ .

The $0.30\ \text{kg}$ piece moves at $4.0\ \text{m s}^{-1}$ to the left.

Markers reward setting total momentum to zero, correct masses (the second piece is $0.30\ \text{kg}$ ), and the negative sign indicating the opposite direction.