A 2.0\ kg trolley moves at 3.0\ m s^-1. Find its momentum. [2 marks]

SQA SQA AH style-style practice: A 0.40\ kg ball travelling at 12\ m s^-1 strikes a wall and rebounds at 8.0\ m s^-1 in the opposite direction. The contact lasts 0.020\ s. Calculate the impulse on the ball and the average force from the wall.

Impulse equals the change in momentum. Take the initial direction as positive, so the rebound velocity is negative. Relationship: impulse = p = m(v - u). Substitution: p = 0.40(-8.0 - 12) = 0.40 × (-20) = -8.0\ kg m s^-1. The impulse is 8.0\ N s directed away from the wall. Average force from impulse = Ft, so F = pt = -8.00.020 = -400\ N, that is 400\ N away from the wall. Markers reward treating the rebound velocity as negative, the impulse of 8.0\ N s, and the average force of 400\ N using F = p / t.

ScotlandEngineering ScienceSyllabus dot point

How do forces change motion, and how are momentum, impulse, work and energy used to analyse a system?

Dynamics with Newton's laws of motion, momentum and impulse, the conservation of momentum, and work, energy and power applied to moving bodies.

An SQA Advanced Higher Engineering Science answer on dynamics, covering Newton's three laws, momentum and impulse, the conservation of momentum in collisions, and the work-energy and power relationships for moving bodies.

Generated by Claude Opus 4.816 min answerUpdated 2026-06-16

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

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Quick answer

Newton's first law: a body stays at rest or at constant velocity unless acted on by a resultant force. Second law: the resultant force equals the rate of change of momentum, which for constant mass gives $F = ma$ . Third law: every force has an equal and opposite reaction on the other body. Momentum is $p = mv$ (a vector, in $\text{kg m s}^{-1}$ ). Impulse is the change in momentum, $\text{impulse} = Ft = \Delta(mv)$ , in $\text{N s}$ , so a longer contact time means a smaller force for the same change. Conservation of momentum: with no external force, total momentum before a collision equals total momentum after. Work is $W = Fd$ , kinetic energy is $E_k = \tfrac{1}{2}mv^2$ , and power is the rate of doing work, $P = W/t = Fv$ .

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What this key area is asking
Newton's three laws
Momentum and impulse
Conservation of momentum
Work, energy and power
Examples in context
Try this

What this key area is asking

The SQA wants you to apply Newton's three laws of motion, to use momentum and impulse and the conservation of momentum in collisions, and to apply work, energy and power to moving bodies. This is the dynamics that explains how forces accelerate the components of a mechanism and how energy moves through a system.

Newton's three laws

The word resultant is again decisive: in the second law $F$ is the net force after combining all forces, so friction and drag must be subtracted from the driving force first. The third law (equal and opposite reaction) is what makes a rocket thrust, a gun recoil, and a foot push the ground backwards to walk forwards.

Momentum and impulse

Impulse links a force acting for a time to the change in momentum it produces. Because $Ft = \Delta(mv)$ , the same change in momentum can be achieved with a large force for a short time or a small force for a long time. This is why crumple zones, airbags and crash mats work: they lengthen the contact time, reducing the peak force on the body for the same momentum change.

Conservation of momentum

This principle solves collision problems even when the forces during contact are unknown, because only the momenta before and after matter. For coupling collisions (the bodies stick together) the right-hand side becomes a single combined mass moving at one velocity. Momentum is a vector, so opposite directions take opposite signs.

Work, energy and power

Work is energy transferred by a force moving its point of application; the work done on a body increases its kinetic energy. Power is how fast that work is done. These tie the dynamics back to efficiency: friction does negative work, removing kinetic energy as heat, which is the energy that never reaches the output of a real machine.

Force from a collision using impulse

A $1200\ \text{kg}$ car travelling at $15\ \text{m s}^{-1}$ is brought to rest in a crash lasting $0.15\ \text{s}$ . Find the average force on the car, then show how a longer crumple time would reduce it.

step 1 Find the change in momentum

$\Delta p = m(v - u) = 1200(0 - 15) = -18\,000\ \text{kg m s}^{-1}$ , a magnitude of $18\,000\ \text{N s}$ .

step 2 Find the average force from the impulse

$F = \dfrac{\Delta p}{t} = \dfrac{-18\,000}{0.15} = -120\,000\ \text{N}$ , a force of $120\ \text{kN}$ opposing the motion.

step 3 Show the effect of a longer time

If a crumple zone extended the time to $0.30\ \text{s}$ , the same momentum change gives $F = \dfrac{18\,000}{0.30} = 60\,000\ \text{N}$ , half the force. Doubling the impact time halves the force.

Examples in context

A rocket rises by Newton's third law: ejecting exhaust downwards drives the rocket up. A vehicle crumple zone uses impulse, lengthening the impact time to cut the peak force. A collision of railway wagons is solved by conservation of momentum to find the coupled speed. A pile driver does work on the pile, transferring the hammer's kinetic energy. Recognising which principle, force and acceleration, impulse, conservation of momentum, or work and energy, fits the question is the core skill.

Try this

Q1. State Newton's second law in terms of momentum. [1 mark]

Cue. The resultant force equals the rate of change of momentum.

Q2. A $2.0\ \text{kg}$ trolley moves at $3.0\ \text{m s}^{-1}$ . Find its momentum. [2 marks]

Cue. $p = mv = 2.0 \times 3.0 = 6.0\ \text{kg m s}^{-1}$ .

Q3. State why an airbag reduces the force on a passenger in a crash. [2 marks]

Cue. It lengthens the time over which the momentum change occurs, so for the same change the average force ( $F = \Delta(mv)/t$ ) is smaller.

Exam-style practice questions

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

SQA AH style5 marksA

0.40\ \text{kg}

ball travelling at

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

strikes a wall and rebounds at

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

in the opposite direction. The contact lasts

0.020\ \text{s}

. Calculate the impulse on the ball and the average force from the wall.

Show worked answer →

Impulse equals the change in momentum. Take the initial direction as positive, so the rebound velocity is negative.

Relationship: $\text{impulse} = \Delta p = m(v - u)$ .

Substitution: $\Delta p = 0.40(-8.0 - 12) = 0.40 \times (-20) = -8.0\ \text{kg m s}^{-1}$ .

The impulse is $8.0\ \text{N s}$ directed away from the wall.

Average force from $\text{impulse} = Ft$ , so $F = \dfrac{\Delta p}{t} = \dfrac{-8.0}{0.020} = -400\ \text{N}$ , that is $400\ \text{N}$ away from the wall.

Markers reward treating the rebound velocity as negative, the impulse of $8.0\ \text{N s}$ , and the average force of $400\ \text{N}$ using $F = \Delta p / t$ .

SQA AH style5 marksA

1500\ \text{kg}

railway wagon moving at

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

collides with and couples to a stationary

2500\ \text{kg}

wagon. Calculate their common velocity after coupling, and state the conservation principle used.

Show worked answer →

Use conservation of momentum: total momentum before equals total momentum after.

Relationship: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ .

Substitution: $1500(3.0) + 2500(0) = (1500 + 2500)v$ .

$4500 = 4000v$ , so $v = \dfrac{4500}{4000} = 1.125\ \text{m s}^{-1} \approx 1.1\ \text{m s}^{-1}$ .

Principle: in the absence of external forces, the total momentum of the system is conserved.

Markers reward applying conservation of momentum, the common velocity of about $1.1\ \text{m s}^{-1}$ , and stating that total momentum is conserved when no external force acts.

Related dot points

Sources & how we know this

Advanced Higher Engineering Science Course Specification (C823 77) — SQA (2019)
Advanced Higher Engineering Science Data Booklet — SQA (2019)