JEE Mains Physics · Laws of Motion

Newton's Second Law, Impulse and Variable Mass

The net force on a body is the rate at which its momentum changes: ma when the mass is fixed, an impulse when the force acts briefly, and v dm/dt when mass is gained or thrown off.

All Laws of Motion notes JEE Mains Physics strategy

Why this matters

Sixteen PYQs, all multiple choice, and none from 2026. Four apply F = ma directly: a sum of force vectors, a balloon that drops ballast, a monkey on a rope. Five are impulse questions: a catch, a bounce, a bullet that embeds, a mass dropped onto a moving body. Seven are variable-mass questions: a rocket, a machine gun, a conveyor belt, a water jet. Every one of them is F = dp/dt read a different way.

Concept 1 of 3: Second law as a vector and along a rope

Add every force on the body as a vector, then divide by its mass. For a body on a rope, pick up as positive: the rope must pull harder than the weight only when the acceleration points up. The direction of motion does not matter; the direction of the acceleration does.

Definition

$\vec F_{\text{net}} = m\vec a$ : add the forces component by component, then divide by m.
If three forces balance and one is removed, the other two add to minus the removed one, so $a = F_{\text{removed}}/m$ , pointing opposite to it.
Body on a rope, up positive: $T - mg = ma$ . Accelerating up, $T = m(g + a)$ ; accelerating down, $T = m(g - a)$ ; steady speed, $T = mg$ .
Balloon: the upthrust F does not change when ballast is dropped. $F - Mg = Ma$ , then $F - (M - x)g = (M - x)a'$ , so $x = \dfrac{M(a' - a)}{g + a'}$ .

Second law

\vec F_{\text{net}} = m\vec a, \qquad T = m(g \pm a)

Worked example

A 40 kg boy climbs a rope that breaks above 600 N. Find the tension when (a) he climbs up with acceleration

3\ \text{m/s}^{2}

and (b) he slides down with acceleration

2\ \text{m/s}^{2}

. What is the largest upward acceleration the rope allows? (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q18Moderate

Example 1 · Laws of Motion · Newton's Second Law, Impulse and Variable Mass

A monkey of mass

50\text{ }kg

climbs on a rope which can withstand the tension

(T)

350\text{ }N

. If monkey initially climbs down with an acceleration of

4\text{ }m/s^{2}

and then climbs up with an acceleration of

5\text{ }m/s^{2}

. Choose the correct option

\left( g = 10\text{ }m/s^{2} \right)

Steady climbing needs only mg

A rope pulls harder than the weight only when the acceleration points up. Climbing at a steady speed, up or down, the tension is mg. It is the upward acceleration that adds ma.

The upthrust stays the same when ballast is dropped

A balloon's upthrust depends on its volume, not on its load. Keep F fixed and change only the mass. Scaling F with the mass gives one of the wrong options.

A removed force leaves its opposite

If three forces balance and one is taken away, the net force is that force reversed. With 10 N removed from a 5 kg body, a = 2 m/s². The other two forces' sizes do not enter.

Drill 3 more on second law as a vector and along a rope

Concept 2 of 3: Impulse and change of momentum

A brief force is measured by what it does to momentum: average force × contact time = change in momentum. Stopping a ball takes away mv. Sending it back at the same speed takes away mv and then gives mv the other way, so the change is 2mv. When no outside force acts along the motion, momentum is conserved instead.

Definition

Impulse $\vec J = \vec F_{\text{avg}}\,\Delta t = \Delta \vec p$ . It is a vector.
Catch (the ball stops): $|\Delta p| = mv$ .
Rebound from speed v to speed v′ the other way: $|\Delta p| = m(v + v')$ ; same speed back gives 2mv.
Third law: the ball pushes the hand, or the wall, with the same force the hand or wall pushes the ball.
No outside horizontal force: momentum is conserved. Mass m dropped onto M moving at v: $v' = \dfrac{Mv}{M + m}$ . A bullet that embeds: $v' = \dfrac{mu}{m + M}$ .

Impulse–momentum

F_{\text{avg}}\,\Delta t = \Delta p, \qquad |\Delta p|_{\text{rebound}} = m(v + v')

Worked example

A 0.2 kg ball hits a wall at 15 m/s and comes back at 10 m/s. The contact lasts 0.04 s. Find the average force on the ball.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q93Moderate

Example 2 · Laws of Motion · Newton's Second Law, Impulse and Variable Mass

A ball of mass

0.15\text{ }kg

hits the wall with its initial speed of

12{\text{ }ms}^{- 1}

and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is

100\text{ }N

. calculate the time duration of the contact of ball with the wall.

A rebound doubles the change

Bouncing back at the same speed, Δp = 2mv. Using mv, the value for a catch, halves the force or doubles the time, and that wrong answer is always an option.

Kinetic energy is not conserved when mass is added

A load dropped onto a moving truck, or a bullet that embeds, keeps the momentum but loses kinetic energy. Use momentum for the new speed, then energy or kinematics for what follows.

Drill 4 more on impulse and change of momentum

Concept 3 of 3: Variable mass: rockets, guns, belts and jets

When mass is thrown off or picked up, the force is the momentum carried per second: speed × mass per second. A rocket pushes gas back and the gas pushes the rocket forward. A belt must bring each grain of sand up to its own speed, so it must keep pushing even at constant velocity.

Definition

Thrust $F = v_{\text{rel}}\dfrac{dm}{dt}$ .
Rocket at lift-off: $v_{\text{rel}}\dfrac{dm}{dt} - mg = ma$ .
Machine gun firing n bullets of mass m a second at speed v: $F = nmv$ .
Jet stopped by a wall: $\dfrac{dm}{dt} = \rho Av$ , so $F = \rho Av^{2}$ .
Conveyor belt at constant speed v: $F = v\dfrac{dm}{dt}$ and $P = Fv = v^{2}\dfrac{dm}{dt}$ . Half of P becomes the sand's kinetic energy; half is lost as heat while it slips.
If $\dfrac{dm}{dt} \propto v^{n}$ , then $F \propto v^{n+1}$ and $P \propto v^{n+2}$ .

Force from a mass flow

F = v_{\text{rel}}\frac{dm}{dt}, \qquad P_{\text{belt}} = v^{2}\frac{dm}{dt}

Worked example

Sand falls at 2 kg/s onto a belt moving at 3 m/s. Find the extra force and power needed to keep the belt at 3 m/s, and the rate at which the sand gains kinetic energy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q87Moderate

Example 3 · Laws of Motion · Newton's Second Law, Impulse and Variable Mass

A sand dropper drops sand of mass

m(t)

on a conveyer belt at a rate proportional to the square root of speed (v) of the belt, i.e.

\frac{dm}{dt}\propto\sqrt{v}

. If P is the power delivered to run the belt at constant speed then which of the following relationship is true?

Belt power is v² dm/dt, not half of it

The kinetic energy the sand gains each second is ½v² dm/dt, but the belt must supply twice that: the other half is lost as heat while the sand slips. The power asked for is Fv.

A jet's force has v twice

The mass arriving per second, ρAv, itself grows with speed, so F = ρAv². Doubling the jet speed quadruples the force.

A rocket's thrust must also lift its weight

At lift-off, v_rel dm/dt − mg = ma. Setting the thrust equal to ma alone gives a burn rate that is too small.

Drill 6 more on variable mass: rockets, guns, belts and jets

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Second law as a vector and along a rope
Second law
$\vec F_{\text{net}} = m\vec a, \qquad T = m(g \pm a)$
Impulse and change of momentum
Impulse–momentum
$F_{\text{avg}}\,\Delta t = \Delta p, \qquad |\Delta p|_{\text{rebound}} = m(v + v')$
Variable mass: rockets, guns, belts and jets
Force from a mass flow
$F = v_{\text{rel}}\frac{dm}{dt}, \qquad P_{\text{belt}} = v^{2}\frac{dm}{dt}$

Watch out for (8)

Steady climbing needs only mg
→ Second law as a vector and along a rope
The upthrust stays the same when ballast is dropped
→ Second law as a vector and along a rope
A removed force leaves its opposite
→ Second law as a vector and along a rope
A rebound doubles the change
→ Impulse and change of momentum
Kinetic energy is not conserved when mass is added
→ Impulse and change of momentum
Belt power is v² dm/dt, not half of it
→ Variable mass: rockets, guns, belts and jets
A jet's force has v twice
→ Variable mass: rockets, guns, belts and jets
A rocket's thrust must also lift its weight
→ Variable mass: rockets, guns, belts and jets

Test yourself on Laws of Motion

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Laws of Motion chapter test Drill all 16 questions