PhysicsICSE Class 9 · Newton's Laws of Motion

Newton's Second Law and Momentum: F = Δp/Δt = ma (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Momentum is p = mv (kg m s⁻¹). Newton's second law says the rate of change of momentum is proportional to the applied force and acts along it; in SI units F = Δp/Δt = m(v − u)/t = ma. The same change of momentum over a longer time needs a smaller force.

The ball's change of momentum is the same in both, so F = Δp/t: a longer stopping time means a smaller force on the hands.

Illustration drawn for QuestionBuddy.

Understand Newton's Second Law, Momentum and F = ma

Momentum. $p = m v$ is the linear momentum: the product of mass and velocity. It is a vector in the direction of the velocity; SI unit kg m s⁻¹. A heavy slow body and a light fast body can have the same momentum.

Change of momentum. For constant mass, $Δ p = m v - m u = m (v - u)$ . Take one direction as positive. If a ball bounces back, the final velocity has the opposite sign, so the change is larger than for stopping.

Second law. The rate of change of momentum of a body is directly proportional to the force applied, and takes place in the direction of the force. In SI units the constant of proportionality is 1:

$F = \frac{Δ p}{Δ t} = \frac{m ( v - u )}{t} = ma .$

Zero net force gives zero acceleration (the first law is a special case).
The same force gives a smaller acceleration to a larger mass: $a \propto \frac{1}{m}$ .
$F$ in $F = ma$ is the net (unbalanced) force: forces such as friction must be subtracted first.

Units. 1 N gives 1 kg an acceleration of 1 m s⁻²; 1 dyne gives 1 g an acceleration of 1 cm s⁻²; $1 N = 1 0^{5}$ dyne. Weight-based units: $1 kgf = 9.8$ N. Keep a calculation in either SI or cgs; do not mix.

Time of action. From $F Δ t = Δ p$ , for a fixed change of momentum a longer time means a smaller force. That is the working principle of a cricketer drawing the hands back, a high jumper's landing mat and the crumple zone of a car.

Combining with motion. The second law gives $a$ (from $F$ and $m$ ) or $F$ (from $a$ and $m$ ). The equations of motion link $a$ with $u, v, S, t$ . A stopping-distance problem therefore has two steps: motion equations for $a$ , then $F = ma$ .

Common pitfalls. Forgetting to convert grams to kilograms; forgetting the negative sign for a ball that reverses; treating weight (a force) as mass; using $F = ma$ with the applied force when friction is also acting.

Newton's Second Law, Momentum and F = ma: formulas and key facts

Momentum: $p = m v$; kg m s⁻¹; vector.
Change of momentum: $Δ p = m (v - u)$; Mass constant; signs matter.
Second law: $F = \frac{Δ p}{Δ t} = ma$; Net force, SI units.
Unit relations: $1 N = 1 0^{5} dyne; 1 kgf = 9.8 N$; 1 gf = 980 dyne.

How to solve Newton's Second Law, Momentum and F = ma questions

Decide first whether the data fits Δp/Δt or ma.

1Convert every quantity to SI (kg, m, s, N).
2Fix a positive direction and give u and v signs.
3For impact problems use $F = m (v - u) / t$ ; for pull problems find the net force then $a = F / m$ .
4If a stopping distance is given, find $a$ from $v^{2} = u^{2} + 2 a S$ first.
5State the direction of the force, and the units.

Newton's Second Law, Momentum and F = ma in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Catching a fast ball

Suppose a fielder catches a 0.16 kg ball travelling at 30 m s⁻¹.

Compare the average force if the ball is stopped in 0.01 s (stiff hands) with 0.06 s (hands drawn back).

Try it first, then show the solutionSolution

$Δ p = 0.16 \times 30 = 4.8$ kg m s⁻¹.
Stiff: $F = \frac{4.8}{0.01} = 480$ N. Giving way: $F = \frac{4.8}{0.06} = 80$ N.

Answer: 480 N against 80 N: six times less.

Memory hook: Longer time, smaller force.

Tricks to remember Newton's Second Law, Momentum and F = ma

Short memory aids. Say each one aloud once, then use it on a question the same day.

Force is the pace of momentum
F = Δp/Δt: how fast momentum changes.
Use it when: Recalling the law.
Net force only
Add forces with signs first; then F = ma on the result.
Use it when: Friction questions.
N to dyne: five zeros
1 N = 100 000 dyne; 1 kgf = 9.8 N.
Use it when: Unit conversions.

Newton's Second Law, Momentum and F = ma: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Newton to dyne · 1 mark · MCQBoard level

A force of 0.4 N is equal to:
(a) $4 \times 1 0^{3}$ dyne
(b) $4 \times 1 0^{4}$ dyne
(c) $4 \times 1 0^{5}$ dyne
(d) $4 \times 1 0^{6}$ dyne

Try it first, then show the solutionSolution

$0.4 \times 1 0^{5} = 4 \times 1 0^{4}$ dyne.

Answer: (b)

Example 2: Momentum of a car · 2 marks · Short answerBoard level

Find the momentum of a 1500 kg car moving at 12 m s⁻¹. State the unit and the direction.

Try it first, then show the solutionSolution

$p = m v = 1500 \times 12$ (1)
$= 18000$ kg m s⁻¹, in the direction of motion. (1)

Answer: 18 000 kg m s⁻¹

Example 3: Bat and ball · 3 marks · Short answerHarder board level

A 0.25 kg ball arriving at 12 m s⁻¹ is hit straight back at 18 m s⁻¹. The bat is in contact for 0.005 s. Find the change in momentum and the average force.

Try it first, then show the solutionSolution

Take the incoming direction as positive: $u = + 12$ , $v = - 18$ . (1)
$Δ p = 0.25 (- 18 - 12) = - 7.5$ kg m s⁻¹. (1)
$F = \frac{- 7.5}{0.005} = - 1500$ N: 1500 N opposite to the incoming direction. (1)

Answer: 7.5 kg m s⁻¹; 1500 N

Example 4: Force from distance covered · 3 marks · Short answerHarder board level

A body of mass 3 kg starts from rest and covers 24 m in 4 s under a constant force. Find the force and the speed at the end of 4 s.

Try it first, then show the solutionSolution

$S = \frac{1}{2} a t^{2}$ : $24 = 8 a$ , so $a = 3$ m s⁻². (1)
$F = ma = 3 \times 3 = 9$ N. (1)
$v = a t = 12$ m s⁻¹. (1)

Answer: 9 N; 12 m s⁻¹

Example 5: A block pushed then released · 4 marks · Case-basedHOTS / competency

A smooth 8 kg block rests on a frictionless table. A 20 N force acts on it for 5 s and is then removed.
(a) Find the acceleration while the force acts.
(b) Find the speed at the end of 5 s.
(c) How far does it move in those 5 s, and how far in the next 6 s?
(d) What constant opposing force would stop it in a further 5 s?

Try it first, then show the solutionSolution

(a) $a = \frac{20}{8} = 2.5$ m s⁻². (1)
(b) $v = 2.5 \times 5 = 12.5$ m s⁻¹. (½)
(c) $S_{1} = \frac{1}{2} (2.5) (25) = 31.25$ m; with no force the velocity stays 12.5, so $S_{2} = 12.5 \times 6 = 75$ m. (1½)
(d) $a = \frac{12.5}{5} = 2.5$ m s⁻² retardation; $F = 8 \times 2.5 = 20$ N. (1)

Answer: (a) 2.5 m s⁻² (b) 12.5 m s⁻¹ (c) 31.25 m; 75 m (d) 20 N

Where students lose marks in Newton's Second Law, Momentum and F = ma

The slips examiners see most often, and what to write instead.

Mistake: Grams not converted
F = ma needs kg for newtons.
Wrong working
F = 250 × 4 = 1000 N
Right working
F = 0.25 × 4 = 1 N
Do this: Convert first and show the line.
Usually costs: The answer.
Mistake: Rebound sign ignored
When the ball reverses, v has the opposite sign, so |v − u| becomes the sum of the speeds, not the difference.
Wrong working
Δp = 0.25 × (18 − 12) = 1.5
Right working
Δp = 0.25 × (−18 − 12) = −7.5
Do this: Write u and v with signs.
Usually costs: 2 marks of 3.
Mistake: Friction left out
F = ma uses the net force.
Wrong working
a = 6/2 = 3 m s⁻² (with 2 N friction acting).
Right working
a = (6 − 2)/2 = 2 m s⁻².
Do this: Draw the force arrows.
Usually costs: Method and answer.

Exam tips

Write "net force" and the direction in your answer to Newton’s second law numericals.
Show Δp as final minus initial momentum, with signs.
For explanations, quote F = Δp/Δt and say Δp is fixed while Δt changes.
Put units (N, kg m s⁻¹) on every line of the answer.

Practice questions on Newton's Second Law, Momentum and F = ma

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The rate of change of momentum of a body is equal to:
(a) its velocity
(b) its mass
(c) the net force acting on it
(d) its kinetic energy
Hint
Second law.
Answer
(c) the net force acting on it
Q2 · 1 mark · Very short answerBoard level
Express 3 kgf in newtons ( $g = 9.8$ m s⁻²).
Hint
1 kgf = 9.8 N.
Answer
$3 \times 9.8 = 29.4$ N.
Q3 · 2 marks · Short answerHarder board level
A 0.4 kg ball moving at 15 m s⁻¹ is brought to a velocity of 3 m s⁻¹ in the same direction by a bat in 0.06 s. Find the change in momentum and the average force.
Hint
Same direction, so no sign reversal.
Answer
$Δ p = 0.4 (3 - 15) = - 4.8$ kg m s⁻¹; $F = \frac{- 4.8}{0.06} = - 80$ N (80 N opposing the motion).
Q4 · 3 marks · Short answerHarder board level
A force of $3 \times 1 0^{5}$ dyne acts on a 2 kg body. Find the acceleration in m s⁻² and in cm s⁻².
Hint
Convert the force to newtons first.
Answer
$F = 3$ N; $a = \frac{3}{2} = 1.5$ m s⁻² $= 150$ cm s⁻².
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): It is easier to set an empty cart in motion than a loaded one.
Reason (R): For the same force, a larger mass gets a larger acceleration.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
a = F/m.
Answer
(c). For the same force $a \propto \frac{1}{m}$ , so a larger mass has a smaller acceleration.
Q6 · 3 marks · Short answerHOTS / competency
A fielder catches a 0.16 kg ball moving at 30 m s⁻¹ and brings it to rest in 0.01 s, holding his hands stiff. How many times smaller would the average force be if he draws his hands back so that the ball stops in 0.06 s? Explain in terms of momentum.
Hint
Same Δp, different time.
Answer
$Δ p = 4.8$ kg m s⁻¹ in both cases; $F = 480$ N and 80 N. The force is 6 times smaller because $F = \frac{Δ p}{Δ t}$ and $Δ t$ is 6 times longer.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.SI unit of momentum?Show answer
kg m s⁻¹.
2.Statement of the second law?Show answer
The rate of change of momentum is proportional to the force and acts in its direction.
3.1 N in dyne?Show answer
10⁵ dyne.
4.1 kgf in N?Show answer
9.8 N.
5.Is momentum a scalar or vector?Show answer
Vector.

Newton's Second Law, Momentum and F = ma: frequently asked questions

What is Newton’s second law of motion?

The rate of change of momentum of a body is directly proportional to the net force acting on it and takes place in the direction of the force: F = Δp/Δt = ma.

Why does a longer contact time reduce the force?

The change of momentum is fixed by the masses and velocities, so F = Δp/Δt falls when Δt rises.

Do I use grams or kilograms in F = ma?

Use kg to get newtons, or g with cm s⁻² to get dynes; never mix.