Paper 1

Physics9702

Paper 1

Physics9702

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

Definition

2 marks

What is meant by the principle of conservation of momentum?

Model answer: The total momentum of a system remains constant, provided no resultant external force acts on the system.

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 (9702/21/M/J/24 Q3(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.

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.

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).

Your turn9702-style

A 0.30 kg trolley moves at 2.0 m s⁻¹ and sticks to a stationary 0.20 kg trolley. Find their common velocity.

Show worked answer

Before: $p = 0.30(2.0) = 0.60\text{ kg m s}^{-1}$ .
After: $0.60 = (0.30 + 0.20)v$ .

Answer

v = 1.2\text{ m s}^{-1}

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 in this closed-system model; 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 is conserved in this closed-system model.

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

Definition

2 marks

What is meant by the principle of conservation of momentum?

Model answer: The total momentum of a system remains constant, provided no resultant external force acts on the system.

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.

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.

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).

Your turn9702-style

A 0.30 kg trolley moves at 2.0 m s⁻¹ and sticks to a stationary 0.20 kg trolley. Find their common velocity.

Show worked answer

Before: $p = 0.30(2.0) = 0.60\text{ kg m s}^{-1}$ .
After: $0.60 = (0.30 + 0.20)v$ .

Answer

v = 1.2\text{ m s}^{-1}

Try the collision yourself

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

Collision Explorer

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 is conserved in this closed-system model.

PreviousMomentum NextElastic & Inelastic Collisions