A 0.058\ kg tennis ball is struck, changing its velocity from -30\ m s^-1 to +40\ m s^-1 in 5.0\ ms. Find the average force on the ball. [3 marks]

SEAB Original-style practice: A ball of mass 0.15\ kg travelling at 20\ m s^-1 strikes a wall horizontally and rebounds at 15\ m s^-1. The contact lasts 0.040\ s. (a) Calculate the change in momentum. (b) Calculate the average force exerted by the wall on the ball.

Take the initial direction as positive, so the rebound velocity is negative. (a) p = m v - m u = 0.15(-15) - 0.15(20) = -2.25 - 3.00 = -5.25\ kg m s^-1. The magnitude of the momentum change is 5.25\ kg m s^-1, directed away from the wall. (b) By Newton's second law, F = p t = -5.250.040 = -131\ N. The wall exerts an average force of 131\ N on the ball, directed away from the wall. Markers reward a consistent sign convention treating the rebound as negative, the correct momentum change including the sign, and the force from the rate of change of momentum.

SingaporePhysicsSyllabus dot point

How do Newton's three laws relate force, mass and motion, and how is the second law expressed through momentum?

State and apply Newton's three laws of motion, expressing the second law as the rate of change of momentum, and identify Newton's third-law force pairs

A focused answer to the H2 Physics learning outcome on Newton's laws. The three laws, the second law as rate of change of momentum, the impulse-momentum link, and identifying genuine third-law force pairs.

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
The answer
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Try this

What this dot point is asking

SEAB wants you to state Newton's three laws, apply the second law in its momentum form $F = \dfrac{\Delta p}{\Delta t}$ , connect this to impulse, and correctly identify third-law force pairs. The momentum form of the second law is the version the syllabus emphasises, because it handles changing mass and impulsive collisions naturally.

The answer

Newton's first law

A body remains at rest or moves with constant velocity unless acted on by a resultant external force. This defines the idea of inertia and identifies that a force is needed to change motion, not to maintain it.

Newton's second law

The rate of change of momentum of a body is proportional to the resultant force acting on it, and occurs in the direction of that force:

F = \frac{\Delta p}{\Delta t} = \frac{\Delta(mv)}{\Delta t}

For constant mass this reduces to the familiar:

F = ma

The momentum form is more general: it also describes situations where mass changes, such as a rocket ejecting fuel.

Impulse and the impulse-momentum theorem

Rearranging the second law over a time interval gives impulse:

F\,\Delta t = \Delta p

The impulse (force multiplied by time) equals the change in momentum. On a force-time graph, the area under the curve is the impulse. This explains why crumple zones and airbags reduce force: extending $\Delta t$ for a fixed $\Delta p$ reduces $F$ .

Newton's third law

When body A exerts a force on body B, body B exerts an equal and opposite force on body A. A genuine third-law pair:

acts on two different bodies,
is the same type of force (both gravitational, both contact, and so on),
is equal in magnitude and opposite in direction.

Two forces acting on the same body (such as weight and normal force on a resting book) are never a third-law pair, even when they balance.

Worked example

A $1200\ \text{kg}$ car travelling at $25\ \text{m s}^{-1}$ is brought to rest in a collision lasting $0.15\ \text{s}$ . Find the average force on the car, and explain how a crumple zone reduces this force.

Step 1: Find the change in momentum

\Delta p = m(v - u) = 1200(0 - 25) = -3.0 \times 10^4\ \text{kg m s}^{-1}

Step 2: Apply the second law in momentum form

F = \frac{\Delta p}{\Delta t} = \frac{-3.0 \times 10^4}{0.15} = -2.0 \times 10^5\ \text{N}

The average force is $2.0 \times 10^5\ \text{N}$ , directed opposite to the motion.

Step 3: Explain the crumple zone

The change in momentum is fixed by the car's mass and speed. A crumple zone extends the collision time $\Delta t$ . Since $F = \dfrac{\Delta p}{\Delta t}$ , a larger $\Delta t$ for the same $\Delta p$ gives a smaller average force, reducing the force on the occupants.

Examples in context

Example 1. Rocket propulsion. A rocket ejects exhaust gases backward; by the third law the gases push the rocket forward. Because the rocket's mass falls as fuel burns, the momentum form $F = \dfrac{\Delta p}{\Delta t}$ is needed, with the thrust equal to the rate of change of momentum of the ejected gas.

Example 2. Catching a cricket ball. A fielder draws their hands back while catching, lengthening the time over which the ball's momentum changes. Since the change in momentum is fixed, increasing $\Delta t$ reduces the average force $F = \dfrac{\Delta p}{\Delta t}$ on the hands, preventing injury, the same principle as a car crumple zone.

Try this

Q1. State Newton's second law in terms of momentum, and show how it reduces to $F = ma$ for constant mass. [3 marks]

Cue. $F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta(mv)}{\Delta t}$ ; for constant $m$ , $F = m\dfrac{\Delta v}{\Delta t} = ma$ .

Q2. A $0.058\ \text{kg}$ tennis ball is struck, changing its velocity from $-30\ \text{m s}^{-1}$ to $+40\ \text{m s}^{-1}$ in $5.0\ \text{ms}$ . Find the average force on the ball. [3 marks]

Cue. $\Delta p = 0.058(40 - (-30)) = 4.06\ \text{kg m s}^{-1}$ ; $F = \dfrac{4.06}{5.0 \times 10^{-3}} = 812\ \text{N}$ .

Q3. Identify the third-law pair of the upward push of the ground on a person standing still. [2 marks]

Cue. The downward push of the person's feet on the ground, equal in magnitude, opposite in direction, same (contact) type, acting on the ground.

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 ball of mass

0.15\ \text{kg}

travelling at

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

strikes a wall horizontally and rebounds at

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

. The contact lasts

0.040\ \text{s}

. (a) Calculate the change in momentum. (b) Calculate the average force exerted by the wall on the ball.

Show worked answer →

Take the initial direction as positive, so the rebound velocity is negative.

(a) $\Delta p = m v - m u = 0.15(-15) - 0.15(20) = -2.25 - 3.00 = -5.25\ \text{kg m s}^{-1}$ .

The magnitude of the momentum change is $5.25\ \text{kg m s}^{-1}$ , directed away from the wall.

(b) By Newton's second law, $F = \dfrac{\Delta p}{\Delta t} = \dfrac{-5.25}{0.040} = -131\ \text{N}$ .

The wall exerts an average force of $131\ \text{N}$ on the ball, directed away from the wall.

Markers reward a consistent sign convention treating the rebound as negative, the correct momentum change including the sign, and the force from the rate of change of momentum.

Original3 marksA book rests on a table. State the weight of the book and identify the Newton's third-law reaction to that weight. Explain why the normal contact force from the table is not the third-law pair of the weight.

Show worked answer →

The weight of the book is the gravitational pull of the Earth on the book.

Its third-law pair is the gravitational pull of the book on the Earth, equal in magnitude and opposite in direction, acting on the Earth.

The normal contact force from the table is not the pair of the weight because a third-law pair must act on two different bodies and be of the same type. The weight (gravitational, Earth on book) and the normal force (contact, table on book) both act on the book and are different types, so they are not a pair; they happen to balance only because the book is in equilibrium.

Markers reward identifying the weight's pair as the book's pull on the Earth, and the explanation that a true pair acts on different bodies, is the same type of force, and need not relate to equilibrium.