ScienceCBSE Class 9 · How Forces Affect Motion

Newton's Second and Third Laws: F = ma, Action-Reaction and Numericals (Class 9)

Updated 17 October 2026 · Fact-checked

Newton's second law says that a net force makes an object accelerate in the direction of the force, with acceleration directly proportional to the net force and inversely proportional to the mass: F = ma. One newton is the force that gives a 1 kg mass an acceleration of 1 m s⁻². Newton's third law says that when one object exerts a force on a second, the second exerts an equal and opposite force on the first at the same time; the two forces act on different objects and never cancel.

Net force gives acceleration along the force (F = ma); in the third law the equal, opposite pair acts on different objects, so it never cancels.

Illustration drawn for QuestionBuddy.

Understand Newton's Second and Third Laws with Numericals

Second law. With a cart pulled by a hanging cup, the same mass with a bigger force gives a bigger acceleration (a ∝ F), and the same force on a bigger mass gives a smaller acceleration (a ∝ 1/m). Together: a = F ÷ m, or F = ma, where F is the net force and the acceleration is along the net force. 1 N = 1 kg m s⁻². In the cart activity the cart starts from rest and covers the same distance s, so ½a₁T₁² = ½a₂T₂² and a₂ ÷ a₁ = T₁² ÷ T₂². Weight W = mg, with g = 9.8 m s⁻² (about 10 for quick estimates); g does not depend on mass. Momentum = mv. In the fuller form, the rate of change of momentum is proportional to the net force.

Stopping forces. From F = ma and a = (v − u) ÷ t, F = m(v − u) ÷ t. For the same change in velocity, a longer stopping time means a smaller force. A fielder draws the hands back while catching a cricket ball to lengthen the time; airbags and seat belts lengthen the time over which the head and chest stop; bubble wrap protects glass; sand beds cushion a high jumper; bending the knees on landing reduces the force. A coconut dropped on a hard floor stops in a very short time, so the ground exerts a very large force and the shell breaks.

Using the second law with kinematics. Find a from the net force (a = F ÷ m), then use v = u + at, s = ut + ½at² or v² = u² + 2as for speeds, times and distances.

Third law. Forces arise from the interaction of two objects. Whenever one object exerts a force on a second, the second exerts a force of equal magnitude in the opposite direction on the first at the same time. The two forces act on different objects, so they never cancel (balanced forces act on the same object). Equal forces do not mean equal accelerations, because the masses may differ: a falling fruit and the Earth pull each other equally, but the Earth's huge mass gives it a negligible acceleration. The law holds for contact and non-contact forces. Examples: walking (feet push the ground back, friction pushes you forward), rowing, climbing a coconut tree, a balloon or rocket (gas pushed out backwards pushes the rocket forwards), the recoil of a gun, a hose pipe pushing back on a fire fighter.

Connected objects. Two boxes of masses m₁ and m₂ on a smooth floor are joined by a string and a force F pulls box 1. Treat both boxes and the string as one system: internal forces (the tension) cancel for the whole system, and only the external force F acts horizontally, so a = F ÷ (m₁ + m₂). The tension is then found from the box that is only pulled by the string: T = m₂a for the rear box. The tension is the action-reaction pair acting through the string.

Newton's Second and Third Laws with Numericals: formulas and key facts

Second law: $F = ma, a = F \div m, w i t h F t h e n e t f or ce$
One newton: $1 N = 1 k g \times 1 m s^{- 2}$
Force from change in velocity: $F = m (v - u) \div t$
Weight: $W = m g, g = 9.8 m s^{- 2} (ab o u t 10 f or es t ima t es)$
Connected objects: $a = F \div (m_{1} + m_{2}) ∣ t e n s i o n o n t h er e a r b o x T = m_{2} a$
Third law: $F o n B b y A = - F o n A b y B (e q u a l s i z e, o pp os i t e d i r ec t i o n, o n d i f f er e n t o bj ec t s)$

How to solve Newton's Second and Third Laws with Numericals questions

Use this method for force and acceleration numericals.

1Draw the object and mark every force; find the net force in the direction of motion.
2Write m, u, v, t, s with units; convert g to 9.8 m s⁻² unless 10 is stated.
3Use a = F ÷ m to get the acceleration, or F = m(v − u) ÷ t to get the force from the change in velocity.
4Use an equation of motion to go from a to speed, distance or time.
5For connected objects, find a for the whole system first, then T from one box.
6For the third law, name the two objects in each pair and state that the forces are equal and opposite and act on different bodies.

Newton's Second and Third Laws with Numericals 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.

Why bending the knees helps

Suppose a 50 kg boy jumps down from a low wall and lands at 4 m s⁻¹. Landing with stiff legs, he stops in 0.05 s; landing with bent knees, he stops in 0.4 s.

Find the average force on his legs in each case and compare.

Try it first, then show the solutionSolution

Stiff legs: F = m(v − u) ÷ t = 50 × 4 ÷ 0.05 = 4000 N.
Bent knees: F = 50 × 4 ÷ 0.4 = 500 N.
The stopping time is 8 times longer, so the force is 8 times smaller.

Answer: 4000 N and 500 N; 8 times smaller with bent knees.

Memory hook: Stretch the time and shrink the force.

Weighing with a spring balance

Suppose a spring balance reads 24.5 N when a pumpkin hangs from it in a school lab.

Find the mass of the pumpkin (g = 9.8 m s⁻²).

Try it first, then show the solutionSolution

A spring balance measures the force, here the weight W = 24.5 N.
m = W ÷ g = 24.5 ÷ 9.8 = 2.5 kg.

Answer: 2.5 kg.

Memory hook: Weight in newtons, divide by 9.8 for kilograms.

Tricks to remember Newton's Second and Third Laws with Numericals

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

Force is mass times acceleration
Cover F in 'F = ma' to see m × a; cover m to see F ÷ a; cover a to see F ÷ m.
Use it when: Rearranging the second law.
Stretch the stop
Same change in velocity, longer time, smaller force. Airbags, bubble wrap and bending knees.
Use it when: Stopping-force reasoning.
Different bodies, no cancelling
Action-reaction pair: two forces, two objects. Balanced forces: two forces, one object.
Use it when: Third-law traps.
System first, tension later
To find T in a pulled train of boxes, take the whole train for a and then one box for T.
Use it when: Connected-objects problems.

Newton's Second and Third Laws with Numericals: solved examples

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

Example 1: Direct use of F = ma · 3 marks · Short answerBoard level

(a) A net force of 3000 N acts on a 1200 kg car. Find its acceleration and its speed 6 s after starting from rest. (1)
(b) What force gives a 0.4 kg ball an acceleration of 12 m s⁻²? (1)
(c) What mass accelerates at 4 m s⁻² under a 20 N force? (1)

Try it first, then show the solutionSolution

(a) a = F ÷ m = 3000 ÷ 1200 = 2.5 m s⁻². v = u + at = 0 + 2.5 × 6 = 15 m s⁻¹. (1)
(b) F = ma = 0.4 × 12 = 4.8 N. (1)
(c) m = F ÷ a = 20 ÷ 4 = 5 kg. (1)

Answer: (a) 2.5 m s⁻² and 15 m s⁻¹; (b) 4.8 N; (c) 5 kg.

Example 2: Mass and weight · 3 marks · Short answerBoard level

Take g = 9.8 m s⁻². (a) Find the weight of a 60 kg student. (1)
(b) An object has a weight of 147 N. Find its mass. (1)
(c) A 5 kg object has a mass of 5 kg on the Moon too. Why does its weight change there? (1)

Try it first, then show the solutionSolution

(a) W = mg = 60 × 9.8 = 588 N. (1)
(b) m = W ÷ g = 147 ÷ 9.8 = 15 kg. (1)
(c) Mass is the amount of matter and stays the same; weight = mg depends on g, which is different on the Moon, so the weight changes. (1)

Answer: (a) 588 N; (b) 15 kg; (c) g differs, mass does not.

Example 3: Stopping force: gymnast and airbag · 5 marks · Long answerHarder board level

(a) A 60 kg gymnast lands at 6 m s⁻¹ on a thick mat and is brought to rest in 0.3 s. Find the average force on her. (1)
(b) She lands at the same speed on a hard floor and stops in 0.03 s. Find the force. (1)
(c) Find the distance she moves while stopping on the mat. (1)
(d) A 70 kg driver moving at 18 m s⁻¹ stops in 0.2 s against an airbag. Find the average force on him. How would the force change if he stopped in 0.02 s against the steering wheel? (2)

Try it first, then show the solutionSolution

(a) F = m(v − u) ÷ t = 60 × (0 − 6) ÷ 0.3 = −1200 N; the magnitude is 1200 N, opposite to her motion. (1)
(b) F = 60 × 6 ÷ 0.03 = 12 000 N, ten times as large. (1)
(c) s = ½(u + v)t = ½ × 6 × 0.3 = 0.9 m. (1)
(d) With the airbag: F = 70 × 18 ÷ 0.2 = 6300 N. Against the steering wheel: F = 70 × 18 ÷ 0.02 = 63 000 N, ten times as large. (2)

Answer: (a) 1200 N; (b) 12 000 N; (c) 0.9 m; (d) 6300 N with the airbag and 63 000 N without, ten times greater.

Example 4: Three bodies in a line · 5 marks · Long answerHarder board level

Three blocks of masses 2 kg, 3 kg and 5 kg are joined in a line by light strings on a smooth floor. A horizontal force of 40 N pulls the 2 kg block (the front one).
(a) Find the acceleration of the system. (1)
(b) Find the tension in the string joining the 3 kg and 5 kg blocks. (1)
(c) Find the tension in the string joining the 2 kg and 3 kg blocks. (1)
(d) Check by finding the net force on the 2 kg and the 3 kg blocks. (1)
(e) Find the speed and distance after 3 s from rest. (1)

Try it first, then show the solutionSolution

(a) a = F ÷ (2 + 3 + 5) = 40 ÷ 10 = 4 m s⁻². (1)
(b) The string behind the 3 kg block pulls only the 5 kg block: T₂ = 5 × 4 = 20 N. (1)
(c) The string between the 2 kg and 3 kg blocks pulls the 3 kg and 5 kg blocks together: T₁ = (3 + 5) × 4 = 32 N. (1)
(d) 2 kg block: 40 − 32 = 8 N = 2 × 4. 3 kg block: 32 − 20 = 12 N = 3 × 4. Both agree. (1)
(e) v = at = 4 × 3 = 12 m s⁻¹; s = ½ × 4 × 3² = 18 m. (1)

Answer: (a) 4 m s⁻²; (b) 20 N; (c) 32 N; (d) consistent; (e) 12 m s⁻¹ and 18 m.

Example 5: Recoil of a gun · 4 marks · Case-basedHOTS / competency

Suppose a gun of mass 4 kg fires a bullet of mass 0.02 kg. During firing the gases push the bullet with an average force of 800 N for 0.01 s.
(a) What force does the bullet exert on the gun? (1)
(b) Find the accelerations of the bullet and the gun. (1)
(c) Find the speeds they gain in 0.01 s, starting from rest. (1)
(d) The two forces are equal and opposite. Why do they not cancel, and why does the gun move so much less than the bullet? (1)

Try it first, then show the solutionSolution

(a) By the third law, 800 N on the gun in the opposite direction. (1)
(b) Bullet: a = 800 ÷ 0.02 = 40 000 m s⁻². Gun: a = 800 ÷ 4 = 200 m s⁻². (1)
(c) Bullet: v = 40 000 × 0.01 = 400 m s⁻¹. Gun: v = 200 × 0.01 = 2 m s⁻¹ backwards. The speed ratio is 200, equal to the mass ratio 4 ÷ 0.02. (1)
(d) The forces act on different objects (the bullet and the gun), so they do not cancel; the gun has 200 times the mass, so for the same force it gets 1/200 of the acceleration. (1)

Answer: (a) 800 N; (b) 40 000 m s⁻² and 200 m s⁻²; (c) 400 m s⁻¹ and 2 m s⁻¹; (d) different objects; larger mass gives smaller acceleration.

Where students lose marks in Newton's Second and Third Laws with Numericals

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

Mistake: Using the weight or the applied force instead of the net force in F = ma.
Friction or other forces are left out.
Wrong working
A 50 N pull on a box with 10 N friction: a = 50 ÷ m
Right working
Net force = 50 − 10 = 40 N, so a = 40 ÷ m
Do this: Find the net force first.
Usually costs: 2 marks.
Mistake: Saying the action and reaction forces cancel.
They are equal and opposite, so it looks like they should.
Wrong working
The forces cancel, so nothing moves.
Right working
They act on different objects and cannot cancel.
Do this: Name the two objects for each force.
Usually costs: 1 mark.
Mistake: Not converting grams or km h⁻¹ before using F = ma.
SI units are required.
Wrong working
F = 160 g × 25 m s⁻¹ ÷ 0.04 s
Right working
160 g = 0.16 kg, so F = 0.16 × 25 ÷ 0.04 = 100 N
Do this: Convert to kg, m, s at the start.
Usually costs: 2 marks.
Mistake: Confusing mass and weight.
Both are given in everyday words as 'weight'.
Wrong working
The weight of a 60 kg student is 60 kg.
Right working
Mass is 60 kg; weight is 60 × 9.8 = 588 N.
Do this: Weight is a force, in newtons.
Usually costs: 1 mark.
Mistake: Using T = F ÷ 2 for the tension in a string between unequal boxes.
The boxes are treated as if they had equal masses.
Wrong working
T = 50 ÷ 2 = 25 N
Right working
a = 5 m s⁻², T = m₂a = 4 × 5 = 20 N
Do this: Find a first, then T = ma of the rear box.
Usually costs: 2 marks.

Exam tips

Write 'F = ma' and the net force line before substituting; units on every number.
Use 9.8 m s⁻² for g unless the question says to take 10.
For stopping-force questions give the formula F = m(v − u) ÷ t and the comparison of times.
For third-law examples always name both objects and the direction of each force.
For connected bodies give a for the system, then T for one body, and check with the other body.

Practice questions on Newton's Second and Third Laws with Numericals

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

Q1 · 2 marks · Very short answerBoard level
State Newton's third law. A horse pulls a cart with a force F and the cart pulls the horse back with an equal force. Why does the cart still move?
Hint
Think about which forces act on the cart and the horse.
Answer
Whenever one object exerts a force on a second, the second exerts an equal and opposite force on the first (1 mark). The two forces act on different objects, so they do not cancel; the cart moves because the forward pull of the horse on the cart (and not the pull of the cart on the horse) is larger than the friction on the cart, giving a net forward force on it (1 mark).
Q2 · 3 marks · Short answerHarder board level
A car of mass 1500 kg increases its speed from 54 km h⁻¹ to 72 km h⁻¹ in 10 s with constant acceleration. Find (a) the acceleration, (b) the net force, and (c) the distance covered in that time.
Hint
Convert km h⁻¹ first.
Answer
u = 15 m s⁻¹, v = 20 m s⁻¹. (a) a = (20 − 15) ÷ 10 = 0.5 m s⁻² (1 mark). (b) F = 1500 × 0.5 = 750 N (1 mark). (c) s = ½(15 + 20) × 10 = 175 m (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): The Earth pulls an apple with the same force as the apple pulls the Earth, but only the apple appears to move. Reason (R): For a given force, acceleration is inversely proportional to mass. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Apply the third law, then a = F ÷ m.
Answer
(a). The two forces are equal (third law), but the Earth's mass is enormous, so its acceleration a = F ÷ m is negligible, while the apple's acceleration is large.
Q4 · 3 marks · Short answerHarder board level
A car of mass 1500 kg pulls a trailer of mass 500 kg with a driving force of 6000 N. The friction on the car is 1000 N and on the trailer 500 N. Find (a) the net force on the car and trailer together, (b) the common acceleration, and (c) the tension in the coupling, verifying it with the car.
Hint
Use the whole system for a, then the trailer for T.
Answer
(a) Net force = 6000 − 1000 − 500 = 4500 N (1 mark). (b) a = 4500 ÷ (1500 + 500) = 2.25 m s⁻² (1 mark). (c) For the trailer: T − 500 = 500 × 2.25 = 1125, so T = 1625 N; check on the car: 6000 − 1000 − 1625 = 3375 N = 1500 × 2.25 (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
Suppose a rocket of mass 500 kg (including fuel) on a launch pad has an engine thrust of 12 000 N; take g = 9.8 m s⁻². (a) Find its weight and the net upward force at lift-off. (b) Find its initial acceleration. (c) After some fuel is burnt, the mass is 400 kg and the thrust is the same. Find the new acceleration and explain why it is greater. (d) Name the third-law pair that produces the thrust.
Hint
Net force = thrust − weight.
Answer
(a) Weight = 500 × 9.8 = 4900 N; net force = 12 000 − 4900 = 7100 N (1 mark). (b) a = 7100 ÷ 500 = 14.2 m s⁻² (1 mark). (c) Weight = 400 × 9.8 = 3920 N; net force = 8080 N; a = 8080 ÷ 400 = 20.2 m s⁻². The mass is smaller and the net force larger, so the acceleration rises (1 mark). (d) The rocket pushes the exhaust gas backwards (action), and the gas pushes the rocket forwards (reaction) (1 mark).

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.State the second law in words.Show answer
Acceleration is proportional to the net force and inversely proportional to the mass, in the direction of the net force.
2.What is one newton in base units?Show answer
1 kg m s⁻².
3.Which law explains the recoil of a gun?Show answer
Newton's third law.
4.Why do airbags reduce injury?Show answer
They lengthen the stopping time, which reduces the average force.
5.What is the formula for weight?Show answer
W = mg.
6.What force pulls the rear box of two connected boxes?Show answer
The tension in the string, equal to its mass times the common acceleration.

Newton's Second and Third Laws with Numericals: frequently asked questions

Is F = ma true when several forces act?

Yes, with F as the net force, the vector sum of all forces on the object.

Does a heavier object fall faster in a vacuum?

No. Its weight is larger but so is its mass, so a = mg ÷ m = g for every object.

How can a rocket accelerate in empty space?

It pushes exhaust gas backwards; by the third law the gas pushes the rocket forwards.

Are action and reaction forces ever on the same body?

No. They always act on two different bodies, which is why they never cancel.

Why does the Earth not move when an apple falls?

It does, but its mass is so large that the acceleration, a = F ÷ m, is negligible.