JEE Mains Physics · Formula sheet

Laws of Motion formulas

13 formulas, 1 reference table and 37 common traps for JEE Mains Physics Laws of Motion, grouped by subtopic.

Newton's Second Law, Impulse and Variable Mass

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}

Common traps

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.

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.

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.

Equilibrium of Forces

Learn this subtopic in the notes

Strings, chains and forces at a point

Equilibrium at a point

\sum F_x = 0,\quad \sum F_y = 0, \qquad T_0 = \frac{mg}{2}\cot\theta

Normal reactions, smooth inclines and supports

Normal reaction

N = mg \pm F\sin\theta, \qquad F_{\text{horizontal}} = mg\tan\theta

Common traps

The lower half of the rope carries only the weight

With a sideways force at the midpoint, the rope below that point still hangs straight and holds mg. Only the upper half tilts, and its angle comes from tan θ = F/mg.

The lowest-point tension is not half the weight

At the lowest point the chain is horizontal, so its tension is the horizontal part of the support tension, (mg/2) cot θ. Half the weight is the vertical part at each support.

The steeper string carries more

A string close to vertical does most of the lifting. If an answer gives the larger tension to the flatter string, the cosines have been swapped.

N is not always mg

Any force with a vertical part changes the normal reaction. A push down at an angle adds F sin θ; a pull up takes it away. Writing N = mg here gives one of the wrong options.

Constant velocity means zero net force

A body moving at steady speed is in equilibrium, exactly as if it were at rest. The applied force only cancels the other forces; it does not exceed them.

A smooth contact pushes perpendicular to the surface

A bar on a smooth shoulder or a ladder on a smooth wall feels a force at right angles to the contact surface, not straight up. Taking it as vertical changes the answer.

Connected Bodies and Pulleys

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One acceleration, then one tension at a time

Atwood machine

a = \frac{(m_2 - m_1)g}{m_1 + m_2}, \qquad T = \frac{2m_1m_2\,g}{m_1 + m_2}

Movable pulleys and string constraints

Constraint

\sum T_i\,a_i = 0, \qquad a_{\text{load}} = \frac{a_{\text{end}}}{2}

Friction inside a pulley system

Table block and hanging block

a = \frac{m_h g - \mu_k m_t g}{m_h + m_t}

Common traps

The tension is not the hanging weight

A hanging mass that accelerates down pulls with m(g − a), less than mg. Only when nothing moves is the tension equal to the weight.

Divide by the total mass

The driving force accelerates every body on the string. Dividing by the hanging mass alone gives an acceleration that is far too large.

Read which ratio is asked

Atwood questions ask for m₁/m₂ in one paper and m₂/m₁ in another, and the stem may name either mass as the heavier one. Decide which is heavier before writing the ratio: the heavier over the lighter is always above 1.

Equal accelerations do not hold with a movable pulley

Only bodies on the same straight run of string share an acceleration. The load on a movable pulley moves at half the rate of the free end; giving both the same a makes every option wrong.

The movable pulley feels 2T

Two segments of the string pull the movable pulley up, so its force equation has 2T. The body on the free end feels only T.

Check that it moves before using kinetic friction

If the driving force is smaller than the largest static friction, the answer is a = 0. Subtracting μN anyway gives a negative acceleration, which means the check was skipped.

No sliding, no friction

A block lying freely on another block that moves at constant velocity feels no friction, because nothing tries to make it slide. Adding μmg there adds a force that is not present.

Static and Kinetic Friction

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Friction on a level surface

Friction and the least starting pull

f_s \le \mu_s N,\quad f_k = \mu_k N, \qquad F_{\min} = \frac{\mu mg}{\cos\theta + \mu\sin\theta}

Holding and pushing a block on a rough incline

Push up and hold

F_{\text{up}} = mg(\sin\theta + \mu\cos\theta), \qquad F_{\text{hold}} = mg(\sin\theta - \mu\cos\theta)

Sliding down a rough incline

Sliding down

a = g(\sin\theta - \mu\cos\theta), \qquad \mu = \tan\theta\left(1 - \frac{1}{n^{2}}\right)

Common traps

N is not mg when the force is at an angle

An angled pull lifts part of the weight and an angled push adds to it. Using μmg for the friction here gives the horizontal-force answer, which is always among the options.

Static friction is only as large as it needs to be

A 10 kg block with μs = 0.5 pushed by 20 N does not move, and the friction is 20 N, not 50 N. μs N is the limit, reached only at the point of slipping.

Area of contact does not matter

Turning a brick on its side changes the area but not the friction. Both static and kinetic friction depend on the materials (through μ) and on N only.

Friction flips direction between push and hold

Pushing up, friction points down the slope: add μmg cos θ. Holding against sliding, it points up the slope: subtract it. Using the same sign in both gives F₁ − F₂ = 0.

The contact force is the full weight

For a block at rest or at constant velocity on an incline, N and friction together balance mg. The total contact force is mg, not mg cos θ.

μ > tan θ means it will not slide by itself

Then the force to move it down is mg(μ cos θ − sin θ). A negative answer from the 'hold' formula is the sign that the block needs a push, not a brake.

Time ratios are squared

Time goes as 1/√a. Taking 50% more time means the acceleration is smaller by a factor of 1.5² = 2.25, not 1.5. That is why the answer at 45° is 1 − 1/2.25 = 5/9.

Friction on a slope is μ mg cos θ

On an incline, N = mg cos θ. Writing the friction as μmg, the level-floor value, makes it too large and gives a wrong μ.

An extra force can change N

A horizontal electric force on a charged block has a part pressing it into the slope or lifting it off. Add that part to mg cos θ before multiplying by μ.

Lifts, Pseudo Forces and Circular Motion

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Pseudo force in an accelerating frame

Pseudo force

\vec F_{\text{pseudo}} = -m\vec a_0, \qquad \tan\theta = \frac{a_0}{g}

Forces in circular motion

Banked road with friction

v_{\max}^{2} = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}

Apparent weight in a lift

Lift's motion	Acceleration	Scale reading	For 50 kg, a = 2 m/s², g = 10 m/s²
At rest	Zero	$mg$	500 N
Moving up or down at constant speed	Zero	$mg$	500 N
Starting upward, speeding up	Upward	$m(g + a)$	600 N
Moving down and slowing to a stop	Upward	$m(g + a)$	600 N Moving down, yet the reading goes UP: the acceleration points up.
Starting downward, speeding up	Downward	$m(g - a)$	400 N
Moving up and slowing to a stop	Downward	$m(g - a)$	400 N
Cable snaps (free fall)	Downward, equal to g	Zero	0 N

The reading depends only on the direction of the acceleration, never on the direction of motion.

Common traps

Velocity does not decide the reading

A lift moving down can read more than your weight, if it is slowing down. Ask which way the acceleration points, then add or subtract a.

A lift at constant velocity changes nothing

Uniform motion adds no pseudo force. An incline or a pendulum inside such a lift behaves exactly as on the ground, so time and acceleration are the ground values.

The pseudo force points against the acceleration

A bob in a car speeding up forward swings BACK. Putting −ma₀ along the acceleration instead of against it reverses the answer, often to the other sign choice among the options.

Use it in one frame only

Either work from the ground with the real acceleration, or from the accelerating frame with the pseudo force. Doing both counts ma₀ twice.

μ = v²/(rg) only on a flat road

On a banked road the normal reaction supplies part of the inward force, so friction needs to supply less. Using the flat-road formula overstates μ.

Convert km/h before squaring

54 km/h is 15 m/s (multiply by 5/18). Squaring 54 instead gives an answer about 13 times too large.

Centrifugal force belongs to the rotating frame

From the ground, a body in a circle has an unbalanced inward force and no outward one. Add mω²r outward only when working in the rotating frame.

Laws of Motion formulas

Newton's Second Law, Impulse and Variable Mass

Second law as a vector and along a rope

Impulse and change of momentum

Variable mass: rockets, guns, belts and jets

Common traps

Equilibrium of Forces

Strings, chains and forces at a point

Normal reactions, smooth inclines and supports

Common traps

Connected Bodies and Pulleys

One acceleration, then one tension at a time

Movable pulleys and string constraints

Friction inside a pulley system

Common traps

Static and Kinetic Friction

Friction on a level surface

Holding and pushing a block on a rough incline

Sliding down a rough incline

Common traps

Lifts, Pseudo Forces and Circular Motion

Pseudo force in an accelerating frame

Forces in circular motion

Apparent weight in a lift

Common traps

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