Physics 9702 · Paper 2

Conservation of Momentum

When objects collide, the single velocities change. One total does not change: the total momentum.
This single rule solves almost every collision problem in the paper.

The principle

The principle of conservation of momentum: the total momentum of a is constant, in any direction.
A closed (or isolated) system means no resultant external force acts on the group of objects. The objects inside may push each other very hard: those internal pushes are third-law pairs and cancel.
For a collision this becomes: total momentum before = total momentum after.

The written answer has two marked parts: (1) total momentum is constant / before equals after, and (2) for a closed system, or when no resultant external force acts. Leaving out part (2) loses a mark (9702/23/M/J/25 Q4(a)).

Before: a moving ball approaches a stationary ball. After: the two move apart. — Individual velocities change in the collision. The total momentum does not.

The before/after method

Sketch two pictures: before and after.
Choose a positive direction and mark it with an arrow.
Write the total momentum of each picture: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ . Any velocity pointing the negative way gets a minus sign.
Solve for the unknown.

The exam sketch: one picture before, one picture after, and an arrow that fixes the positive direction. Total momentum is 2.4 kg m s⁻¹ in both pictures.

Worked example

Smallest case: trolleys that stick

A trolley of mass 0.80 kg moves at 3.0 m s⁻¹ and hits a stationary trolley of mass 1.60 kg. They stick together. Find their common velocity.

Before: $p = 0.80 \times 3.0 + 1.60 \times 0 = 2.4 \text{ kg m s}^{-1}$ .
After, the moving mass is $0.80 + 1.60 = 2.40 \text{ kg}$ : $2.4 = 2.40\,v$ .
$v = 1.0 \text{ m s}^{-1}$ in the original direction.

Answer

v = 1.0 \text{ m s}^{-1}

Worked example

One change: both moving before

Object A (4.0 kg) moves at 6.0 m s⁻¹ and catches object B (2.0 kg) moving at 3.0 m s⁻¹ in the same direction. They join and move off together. (9702/23/M/J/25 Q4)

Before: $p = 4.0(6.0) + 2.0(3.0) = 24 + 6 = 30 \text{ kg m s}^{-1}$ .
After: $30 = 6.0\,v$ , so $v = 5.0 \text{ m s}^{-1}$ .

Answer

v = 5.0 \text{ m s}^{-1}

Worked example

One change: head-on (a minus sign appears)

Two railway carriages, each of mass 5000 kg, move toward each other: one at 2.00 m s⁻¹, the other at 1.00 m s⁻¹. They join in the collision. Find their velocity afterwards. (9702/12/O/N/24 Q10)

Take the first carriage's direction as positive: $p = 5000(2.00) + 5000(-1.00) = 5000 \text{ kg m s}^{-1}$ .
After: $5000 = 10\,000\,v$ , so $v = 0.50 \text{ m s}^{-1}$ in the first carriage's direction.

Answer

v = 0.50 \text{ m s}^{-1}

Common mistake

Adding the sizes of the momenta when the objects move opposite ways. Momentum is a vector: one of the two velocities must carry a minus sign. Adding 2.00 and 1.00 here (giving 1.5 m s⁻¹ after) is the standard wrong answer.

Other versions of this question: the objects bounce apart and you find one final velocity; the collision runs backwards (find a speed before); one object is the Earth (its huge mass makes its velocity change invisibly small).

Try the collision yourself

Set the masses and starting velocities, then check that the total momentum is the same before and after:

Collision Explorer

Set each cart's mass and velocity (negative = moving left), then pick the collision type. Total momentum stays the same either way; kinetic energy only survives an elastic collision.

Mass A2 kg

Velocity A4 m/s

Mass B3 kg

Velocity B-2 m/s

momentum before

2.0kg m/s

momentum after

2.0kg m/s

KE before

22.0J

KE after

0.4J

Kinetic energy is NOT conserved: 98% lost to heat and sound.Momentum: always conserved.

PreviousMomentum NextElastic & Inelastic Collisions

Conservation of Momentum

When objects collide, the single velocities change. One total does not change: the total momentum.
This single rule solves almost every collision problem in the paper.

The principle

The principle of conservation of momentum: the total momentum of a is constant, in any direction.
A closed (or isolated) system means no resultant external force acts on the group of objects. The objects inside may push each other very hard: those internal pushes are third-law pairs and cancel.
For a collision this becomes: total momentum before = total momentum after.

The before/after method

Sketch two pictures: before and after.
Choose a positive direction and mark it with an arrow.
Write the total momentum of each picture: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ . Any velocity pointing the negative way gets a minus sign.
Solve for the unknown.

The exam sketch: one picture before, one picture after, and an arrow that fixes the positive direction. Total momentum is 2.4 kg m s⁻¹ in both pictures.

Worked example

Smallest case: trolleys that stick

A trolley of mass 0.80 kg moves at 3.0 m s⁻¹ and hits a stationary trolley of mass 1.60 kg. They stick together. Find their common velocity.

Before: $p = 0.80 \times 3.0 + 1.60 \times 0 = 2.4 \text{ kg m s}^{-1}$ .
After, the moving mass is $0.80 + 1.60 = 2.40 \text{ kg}$ : $2.4 = 2.40\,v$ .
$v = 1.0 \text{ m s}^{-1}$ in the original direction.

Answer

v = 1.0 \text{ m s}^{-1}

Worked example

One change: both moving before

Object A (4.0 kg) moves at 6.0 m s⁻¹ and catches object B (2.0 kg) moving at 3.0 m s⁻¹ in the same direction. They join and move off together. (9702/23/M/J/25 Q4)

Before: $p = 4.0(6.0) + 2.0(3.0) = 24 + 6 = 30 \text{ kg m s}^{-1}$ .
After: $30 = 6.0\,v$ , so $v = 5.0 \text{ m s}^{-1}$ .

Answer

v = 5.0 \text{ m s}^{-1}

Worked example

One change: head-on (a minus sign appears)

Two railway carriages, each of mass 5000 kg, move toward each other: one at 2.00 m s⁻¹, the other at 1.00 m s⁻¹. They join in the collision. Find their velocity afterwards. (9702/12/O/N/24 Q10)

Take the first carriage's direction as positive: $p = 5000(2.00) + 5000(-1.00) = 5000 \text{ kg m s}^{-1}$ .
After: $5000 = 10\,000\,v$ , so $v = 0.50 \text{ m s}^{-1}$ in the first carriage's direction.

Answer

v = 0.50 \text{ m s}^{-1}

Common mistake

Other versions of this question: the objects bounce apart and you find one final velocity; the collision runs backwards (find a speed before); one object is the Earth (its huge mass makes its velocity change invisibly small).

Try the collision yourself

Set the masses and starting velocities, then check that the total momentum is the same before and after:

Collision Explorer

Set each cart's mass and velocity (negative = moving left), then pick the collision type. Total momentum stays the same either way; kinetic energy only survives an elastic collision.

Mass A2 kg

Velocity A4 m/s

Mass B3 kg

Velocity B-2 m/s

momentum before

2.0kg m/s

momentum after

2.0kg m/s

KE before

22.0J

KE after

0.4J

Kinetic energy is NOT conserved: 98% lost to heat and sound.Momentum: always conserved.