Physics
Laws of Motion
Newton's Second Law – Force and Momentum

Laws of Motion · JEE & NEET Physics

Newton's Second Law – Force and Momentum: notes and previous year questions

How a force changes momentum, how to use F = ma with free body diagrams, and how to solve inclines, pulleys and impulse.

41 JEE Main questions (2002–2026)
3 JEE Advanced questions (2006–2009)
17 NEET questions (2000–2020)

Newton's Second Law – Force and Momentum in short

Momentum p = mv is a vector along the velocity; unit kg m/s = N s.
The net force equals the rate of change of momentum, F = dp/dt.
For a constant mass this becomes F = ma; 1 N = 1 kg m/s².
The acceleration is proportional to the net force and inversely proportional to the mass.

1Momentum

Two trolleys roll at the same speed; the heavier one is harder to stop. The "quantity of motion" of a body is its momentum:

\vec p = m\vec v

A vector, in the direction of the velocity. Unit: kg m/s, the same as N s.

Momentum depends on both mass and speed: a 1 kg trolley at 6 m/s and a 3 kg trolley at 2 m/s both have 6 kg m/s. Two bodies with the same momentum and masses in the ratio 1 : 4 have speeds in the ratio 4 : 1.

Example: a cricket ball of 150 g = 0.15 kg at 20 m/s has $p = 0.15 \times 20 = 3$ kg m/s.

2Force changes momentum

Push a 2 kg trolley with a steady 4 N and its momentum grows in a straight line, by 4 kg m/s each second: the slope of the p–t graph is the force.

If the mass is constant it comes out of the derivative: $F = \frac{d(mv)}{dt} = m\frac{dv}{dt} = ma$ .

\vec F = m\vec a

Only for a constant mass. F = dp/dt is the general form (rockets, conveyor belts).

This defines the newton: 1 N is the force that gives 1 kg an acceleration of 1 m/s², so $1\ \text{N} = 1\ \text{kg m s}^{-2}$ . A force that gives 2 kg an acceleration of 3 m/s² is $2 \times 3 = 6$ N.

3F = ma in action

a = \frac{F_{\text{net}}}{m}

Same mass: $a \propto F$ . Doubling (or tripling) the force doubles (or triples) the acceleration.
Same force: $a \propto 1/m$ . Doubling the mass halves the acceleration (12 N gives 2 kg 6 m/s², but 4 kg only 3 m/s²).
The acceleration always points along the net force.

Example: a 20 N force on a 4 kg body gives $a = 20/4 = 5$ m/s².

Signs and components. Choose a positive direction; forces along it are +, against it −. Example: 30 N right and 10 N left on a 4 kg crate give a net 20 N, so $a = 5$ m/s² to the right. In two dimensions use components: $F_x = ma_x$ and $F_y = ma_y$ , independently.

4Free body diagrams

To use F = ma you need every force on a body. A free body diagram shrinks the body to a dot and draws every force acting on it.

Isolate one body at a time.
List every force on it: weight $mg$ , normal force $N$ , tension $T$ , friction $f$ , any applied force $F$ .
Draw each as an arrow in its direction.
Choose axes (usually horizontal and vertical, or along and across a slope).
Resolve forces into components, then apply F = ma along each axis.

A book at rest on a table feels exactly two forces: its weight (from the Earth) and the normal force (from the table). A 5 kg box at rest on a floor has $N = mg = 50$ N.

5Blocks on inclines

On a slope, use one axis along the slope and one perpendicular to it. The weight splits into:

Part of the weight	Direction	Size
Along the slope	down the slope	$mg\sin\theta$
Into the slope	perpendicular	$mg\cos\theta$

Nothing moves perpendicular to the slope, so $N = mg\cos\theta$ . Along it, $mg\sin\theta - f = ma$ . On a smooth slope ( $f = 0$ ):

a = g\sin\theta

The mass cancels: heavy and light blocks slide down together.

Examples: 10 kg on a smooth 30° slope: $a = 10 \times 0.5 = 5$ m/s². For $a = 6$ m/s², $\sin\theta = 0.6$ and $\theta \approx 37°$ . On 53° ( $\sin = 0.8$ ): 8 m/s².

Sine or cosine? Check with a vertical cliff, $\theta = 90°$ : $\sin 90° = 1$ , so the whole weight acts along the "slope", as it should.

A 5 kg block on a smooth 30° slope pulled up it with 40 N: net force along the slope $40 - 25 = 15$ N, so $a = 3$ m/s² up the slope.

6Connected bodies

For bodies tied together, write F = ma for each body, add the constraint, and solve together. With a light string over a smooth, light pulley, the tension is the same all along the string.

a = \frac{(m_2 - m_1)g}{m_1 + m_2}

T = \frac{2m_1m_2g}{m_1 + m_2}

Equal masses: a = 0 and T = mg. 2 kg and 3 kg: T = 24 N. 2 kg and 4 kg: a = g/3.

String constraint: an inextensible string has a fixed length, so the accelerations of the bodies on it are linked.
Wedge constraint: a block that stays on a moving wedge must match the wedge's motion along the contact normal.
Pulley constraint: a movable pulley's acceleration is the average of the accelerations of its two strands.

7Impulse and rockets

From $F = dp/dt$ , $F\,dt = dp$ . A force acting for a time gives an impulse equal to the change in momentum:

J = \int F\,dt = \Delta p

For a constant force, J = F Δt = m(v − u). Unit N s = kg m/s.

The same change in momentum can come from a large force for a short time (hammer on a nail) or a small force for a long time (pushing a car). On a force–time graph, the impulse is the area under the line.

Catching with moving hands, airbags and bending your knees all lengthen the stopping time, so the same change in momentum needs a smaller force. The change in momentum itself is not smaller.

Rockets lose mass all the time, so use $F = dp/dt$ . The thrust is $v_{\text{rel}}\,dm/dt$ : gas at 2000 m/s relative to the rocket, 5 kg each second, gives 10 000 N. In free space the speed gained is the rocket equation below.

v - u = v_{\text{rel}}\ln\frac{m_i}{m_f}

Summary

Key ideas

Momentum p = mv is a vector along the velocity; unit kg m/s = N s.
The net force equals the rate of change of momentum, F = dp/dt.
For a constant mass this becomes F = ma; 1 N = 1 kg m/s².
The acceleration is proportional to the net force and inversely proportional to the mass.
Use signs along one line, and components (F_x = ma_x, F_y = ma_y) in two dimensions.
Mass is an amount of matter; weight mg is a force.
A free body diagram shows only the forces acting ON one body.
On a smooth incline a = g sin θ, whatever the mass; N = mg cos θ.
For connected bodies, write F = ma for each body and add the string or pulley constraint.
Atwood tension lies between the two weights.
Impulse FΔt equals the change in momentum; it is the area under the F–t graph.
Variable-mass systems such as rockets need F = dp/dt.

Every equation

Momentum: $\vec p = m\vec v$
Second law: $\vec F = d\vec p/dt$
Constant mass: $\vec F = m\vec a$
The newton: $1\ \text{N} = 1\ \text{kg m s}^{-2}$
Components: $F_x = ma_x,\ F_y = ma_y$
Weight: $W = mg$
Horizontal pull: $F - f = ma,\ N = mg$
Incline: normal force: $N = mg\cos\theta$
Incline: along the slope: $mg\sin\theta - f = ma$
Smooth incline: $a = g\sin\theta$
Atwood acceleration: $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$
Atwood tension: $T = \frac{2m_1m_2g}{m_1 + m_2}$
Wedge on a smooth floor: $a_M = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}$
Chain leaving a table: $t = \sqrt{L/g}\,\cosh^{-1}(L/b)$
Impulse: $J = F\Delta t = \Delta p$
Rocket thrust: $F = v_{\text{rel}}\,dm/dt$
Rocket equation: $v - u = v_{\text{rel}}\ln(m_i/m_f)$

Previous year questions with solutions

Real JEE and NEET questions on newton's second law – force and momentum. Try each one before you open the solution.

Q1JEE Main 2025One correct option

A massless spring gets elongated by amount $x_{1}$ under a tension of 5 N . Its elongation is $x_{2}$ under the tension of 7 N . For the elongation of $(5x_{1}-2x_{2})$ , the tension in the spring will be,

A20 N
B39 N
C11 N
D15 N

Show answer and solution

Answer: Option C

Hooke's law makes the elongation proportional to the tension, so $x_{1} = \dfrac{5}{k}$ and $x_{2} = \dfrac{7}{k}$ . Then $5x_{1} - 2x_{2} = \dfrac{25 - 14}{k} = \dfrac{11}{k}$ , and the tension that produces an elongation of $\dfrac{11}{k}$ is $k \times \dfrac{11}{k} = 11\ \mathrm{N}$ . The unknown $k$ never had to be found and never could be. Because the relation is linear, any combination $ax_{1} + bx_{2}$ of elongations belongs to the same combination of tensions — here $5(5) - 2(7) = 11$ , which is the whole question in one line.

Q2NEET 2011One correct option

A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m/s². If g = 10 ms⁻², the tension in the supporting cable is

A8600 N
B9680 N
C11000 N
D1200 N

Show answer and solution

Answer: Option C

The cable holds the lift and everybody in it, so the body to draw is the whole $940 + 60 = 1000\ \mathrm{kg}$ . Then $T - (M+m)g = (M+m)a$ gives $T = 1000 \times (10 + 1) = 11000\ \mathrm{N}$ . Notice what never appears: the passenger's own $N$ , which is $60 \times 11 = 660\ \mathrm{N}$ , acts up on the passenger and down on the floor, both inside the body we drew, and cancels out of it exactly. Two bodies, two questions — ask what the scale reads and you get $660\ \mathrm{N}$ , ask what the cable carries and you get $11000\ \mathrm{N}$ .

Q3JEE Main 2024One correct option

A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_{1}$ and $m_{2}$ . If the acceleration of the system is $\frac{g}{8}$ , then the ratio of the masses $\frac{m_{2}}{m_{1}}$ is :

A $5:3$
B $8:1$
C $9:7$
D $4:3$

Show answer and solution

Answer: Option C

Adding the two equations gives $a = \dfrac{(m_{2}-m_{1})g}{m_{1}+m_{2}}$ with $m_{2}$ the heavier, so $\dfrac{1}{8} = \dfrac{m_{2}-m_{1}}{m_{2}+m_{1}}$ , then $8m_{2} - 8m_{1} = m_{1}+m_{2}$ and $\dfrac{m_{2}}{m_{1}} = \dfrac{9}{7}$ . Which mass is named first changes nothing about the physics and everything about the option you tick: the formula is difference over sum, with the heavier one leading, and the question decides which of the two that is.

Practice questions, easy to hard

Three questions from the newton's second law – force and momentum practice ladder: one easy, one medium, one hard.

Q4One correct option

When several bodies are pressed together or tied together so that they all move with one acceleration, you are allowed to draw a box round the lot and treat them as a single body. The forces between them are third-law pairs — equal, opposite, and both inside the box — so they cancel in that view and never appear.

Two blocks of masses $m_{1}$ and $m_{2}$ are in contact on a smooth level floor, and a horizontal force $F$ pushes on the pair. What is their common acceleration?

A $\dfrac{F}{m_{1}}$
B $\dfrac{F}{m_{1}+m_{2}}$
C $\dfrac{F}{m_{2}}$
D $\dfrac{F}{m_{1}m_{2}}$

Show answer and solution

Answer: Option B

One body of mass $m_{1}+m_{2}$ , one external horizontal force $F$ , so $a = \dfrac{F}{m_{1}+m_{2}}$ . The push of each block on the other is in there too, twice, once each way, and the pair sums to zero — the only time third-law partners may share a diagram, because that diagram is of both bodies at once. What this view cannot tell you is how hard the two blocks press on each other. For that they have to be cut apart.

Q5Numerical answer

When a spring joins two blocks and an outside force touches only one of them, the way in is always the other block. That block has the spring force on it and nothing else, so its acceleration hands you the size of the spring force directly — and the same size, reversed, is what the spring puts on the block being pushed.

Two blocks of $3\ \mathrm{kg}$ and $5\ \mathrm{kg}$ lie on a smooth table joined by a spring. A horizontal force of $40\ \mathrm{N}$ is applied to the $5\ \mathrm{kg}$ block, directed away from the other one. At an instant when the $3\ \mathrm{kg}$ block has an acceleration of $4\ \mathrm{m/s^{2}}$ , what is the acceleration of the $5\ \mathrm{kg}$ block, in $\mathrm{m/s^{2}}$ ?

Show answer and solution

Answer: 5.6 m/s²

The $3\ \mathrm{kg}$ block has only the spring acting on it, so the spring force is $3 \times 4 = 12\ \mathrm{N}$ . The spring is stretched, so it pulls the $5\ \mathrm{kg}$ block backwards with that same $12\ \mathrm{N}$ , against the applied $40\ \mathrm{N}$ : the net force is $28\ \mathrm{N}$ and the acceleration is $\dfrac{28}{5} = 5.6\ \mathrm{m/s^{2}}$ . The two blocks do not share an acceleration here — the spring is still stretching, so they are separating — and the moment you assume a common $a$ the question is gone.