JEE Main chapters · Class 11 · Physics

Laws of Motion: JEE Main Physics formulas and notes

Laws of Motion is the heart of mechanics: Work and Energy, Gravitation, Rotation and Oscillations all use it. JEE Mains usually gives 2 to 3 questions from it. The favourites are friction (incline and block on block), connected bodies and pulleys, circular motion with banking, and impulse. Aim for 90 seconds per question. About 11 hours in total is enough.

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Quick revision

ΣF = ma: draw the free-body diagram first
Lift: N = m(g + a) going up, m(g − a) going down
Impulse = Δp = area under the F–t graph
Atwood: a = ((m₁ − m₂)g)/(m₁ + m₂), T = (2m₁m₂g)/(m₁ + m₂)
Two blocks pushed: contact force = (m₂F)/(m₁ + m₂)
Friction: f(s) ≤ μ(s)N; at rest f = the applied force
tanθ(repose) = μ
Smooth incline a = g sinθ; rough a = g(sinθ − μ cosθ)
Flat turn v = √(μrg); smooth banking tanθ = (v²)/(rg)
Vertical circle: bottom √(5gr), top √(gr)
Pseudo force = −ma in an accelerating frame
Gun: F = n m v; stream: F = ρAv²
Spring cut: spring force stays, tension changes

Formulas with solved examples

Newton’s laws

Newton’s three laws

F⃗ = (dp⃗)/(dt) = ma⃗ (constant mass)
first law: no net force ⇒ constant velocity
third law: F(AB) = −F(BA), on different bodies, at the same time

How to apply. Draw a free-body diagram for each body and mark only the forces that act on that body. Choose axes, then write ΣF = ma along each axis. Action and reaction act on different bodies, so they never cancel in one diagram.

Example. A 5 kg block is pushed with 20 N and friction of 5 N opposes it: a = (20 − 5) ÷ 5 = 3 m/s².

Equilibrium and Lami’s theorem

at rest or at constant velocity: ΣF = 0
three forces in equilibrium: (F₁)/(sinα) = (F₂)/(sinβ) = (F₃)/(sinγ) (each angle is opposite its force)
the tension in a light string is the same all along it

How to apply. For a point held by strings, resolve every force along two axes and set each sum to zero. Lami’s theorem is a shortcut for exactly three concurrent forces.

Example. A lamp of weight W hangs from two strings, each at 45° to the vertical: 2T cos45° = W, so T = W ÷ √2.

Weight, normal force and the lift

weight W = mg
the normal force N is perpendicular to the surface and takes whatever value is needed
lift going up with acceleration a: N = m(g + a)
acceleration down: N = m(g − a)

How to apply. The reading of a weighing machine is the normal force N. In a lift, the apparent weight is larger when the acceleration is upward and smaller when it is downward.

Example. A 60 kg man in a lift (g = 10 m/s²): accelerating up at 2 m/s² gives 720 N; accelerating down at 2 m/s² gives 480 N; in free fall 0.

Springs and string cut

F = kx
series: (1)/(k) = (1)/(k₁) + (1)/(k₂)
parallel: k = k₁ + k₂
a spring cut into n equal parts: k becomes nk
when a string is cut, the spring force does not change at that instant

How to apply. A spring force depends on the stretch, and the stretch cannot change in an instant. The tension of a string can change instantly. So right after a cut, keep the spring force and drop the tension.

Example. A 2 kg block hangs from a spring of k = 200 N/m: extension = 20 ÷ 200 = 0.1 m. Two springs of 200 N/m give 100 N/m in series and 400 N/m in parallel.

Momentum and impulse

p = mv
impulse J = FΔt = Δp = ∫F dt = area under the F–t graph
average force = (Δp)/(Δt)

How to apply. Give directions a sign. For a ball that rebounds, the change in momentum is m(v₂ − v₁) with v₂ opposite to v₁, so the sizes add.

Example. A 0.2 kg ball hits a wall at 10 m/s and comes back at 8 m/s: Δp = 0.2 × (8 + 10) = 3.6 N s. If the contact lasts 0.01 s, the average force is 360 N.

Conservation of momentum

no external force: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
explosion from rest: m₁v₁ = −m₂v₂
recoil of a gun: v(gun) = (m)/(M) v(bullet)
the lighter body gets the larger kinetic energy

How to apply. Use it when the bodies push on each other and nothing from outside acts along that line. Choose the positive direction first. Kinetic energy is not conserved in a collision that sticks.

Example. A 50 g bullet leaves a 5 kg gun at 200 m/s: recoil speed = 0.05 × 200 ÷ 5 = 2 m/s.

Force from a stream or a machine gun

F = (mass per second) × (change of velocity)
a stream that stops on a wall: F = ρAv²
a stream that rebounds with the same speed (elastic): F = 2ρAv²
machine gun: F = n m v for n bullets per second

How to apply. The force is the momentum delivered per second. Find the mass arriving each second (ρAv for a stream) and multiply by the change in velocity.

Example. A gun fires 10 g bullets at 250 m/s and is held in place by 125 N: n = 125 ÷ (0.01 × 250) = 50 bullets per second.

Variable force and the F–t graph

impulse = area under the F–t graph
Δv = (J)/(m)
average force = (J)/(Δt)

How to apply. Split the area into triangles, rectangles and trapeziums. If the force changes sign, the area below the axis is negative impulse.

Example. A force rises uniformly from 0 to 10 N in 2 s: impulse = ½ × 10 × 2 = 10 N s. A 5 kg body starting from rest reaches 2 m/s.

Connected bodies

Two blocks pushed or pulled together

a = (F)/(m₁ + m₂)
force between the blocks when m₁ pushes m₂: N = (m₂F)/(m₁ + m₂)
rope tension when m₂ is pulled and m₁ is behind: T = (m₁F)/(m₁ + m₂)

How to apply. Treat the two blocks as one body to find a. Then take one block alone: the force on it must equal its mass times a.

Example. Blocks of 2 kg and 3 kg on a smooth table, F = 10 N pushes the 2 kg block against the 3 kg block: a = 2 m/s², contact force = 3 × 2 = 6 N.

Atwood machine

a = ((m₁ − m₂)g)/(m₁ + m₂)
T = (2m₁m₂g)/(m₁ + m₂)
force on the pulley support = 2T

How to apply. The heavier mass goes down. Write ΣF = ma for each mass, with the same a and the same T for a light string and a smooth pulley. Add the two equations to find a.

Example. m₁ = 3 kg, m₂ = 2 kg (g = 10): a = 10 ÷ 5 = 2 m/s², T = 2 × 3 × 2 × 10 ÷ 5 = 24 N.

Block on a table with a hanging mass

a = (m₂g)/(m₁ + m₂)
T = (m₁m₂g)/(m₁ + m₂) (smooth table)
with friction μ on m₁: a = (m₂g − μm₁g)/(m₁ + m₂)

How to apply. The pull on the whole system is the weight of the hanging mass only. The tension is smaller than m₂g, because the hanging mass is also accelerating.

Example. m₁ = 4 kg on a smooth table and m₂ = 1 kg hanging (g = 10): a = 10 ÷ 5 = 2 m/s², T = 4 × 2 = 8 N.

Pulley constraints and movable pulleys

a string of fixed length: the total length is constant, so differentiate it to relate speeds and accelerations
a movable pulley held by one string: 2T = Mg at rest
for a movable pulley whose string has one end fixed, the acceleration of the pulley is half that of the free end

How to apply. Write the length of the string as a sum of distances, differentiate it twice, and use the result as an extra equation. A light pulley has zero net force on it.

Example. A mass M hangs from a movable pulley supported by a string over a fixed pulley: at rest 2T = Mg, so T = Mg ÷ 2.

Block on an incline with a hanging mass

a = (m₂g − m₁g sinθ − μm₁g cosθ)/(m₁ + m₂)
T = m₂(g − a)
smooth: drop the μ term

How to apply. Take m₁ on the slope and m₂ hanging over the pulley. Write ΣF = ma for each mass along its own line of motion and add.

Example. m₁ = 4 kg on a smooth 30° slope, m₂ = 3 kg hanging (g = 10): a = (30 − 20) ÷ 7 = 1.43 m/s², T = 3 × (10 − 1.43) = 25.7 N.

Friction

Static and kinetic friction

f(s) ≤ μ(s)N
f(k) = μ(k)N
μ(s) > μ(k)
friction opposes the relative motion, or the tendency to move
if the body does not move, friction equals the applied force

How to apply. First check whether the body moves: compare the applied force with μ(s)N. If it stays at rest, the friction is the applied force, not μN. Once it slides, use μ(k)N.

Example. 10 kg block, μ(s) = 0.5, μ(k) = 0.4 (g = 10). A 40 N push: the limit is 50 N, so the block stays and f = 40 N. A 60 N push: it slides, f = 40 N, a = 20 ÷ 10 = 2 m/s².

Angle of friction and angle of repose

tan θ(repose) = μ
tan λ = μ (λ is the angle of friction)
a block on an incline starts sliding when tanθ > μ(s)

How to apply. The angle of repose is the steepest incline on which a block stays at rest. The resultant of N and friction makes the angle of friction with the normal.

Example. μ = 1/√3 gives θ = 30°. A block on a 45° incline with μ = 0.5 slides, because tan45° = 1 > 0.5.

Block on a block

the top block moves with the bottom one if the friction needed is ≤ μ(s)m g
largest common acceleration: a(max) = μ(s)g
force on the lower block with a smooth floor: F(max) = μ(s)(m₁ + m₂)g

How to apply. Find the friction that the top block needs to get the same acceleration, and compare it with its limit. If the limit is crossed, the blocks slip and each has its own acceleration.

Example. 2 kg block on a 4 kg block, μ = 0.4 between them, smooth floor: a(max) = 4 m/s², so F(max) = 6 × 4 = 24 N.

Least force to move a block

pulling at an angle θ above the horizontal: N = mg − F sinθ and F cosθ = μN
F(min) = (μmg)/(√(1 + μ²)) at tanθ = μ
pushing at an angle increases N, so it needs more force

How to apply. Write the vertical and horizontal balance with the friction μN, and solve for F. The least value comes when the pull is along the angle of friction.

Example. m = 10 kg, μ = 0.5 (g = 10): a horizontal pull needs 50 N; the least force is 50 ÷ √1.25 = 44.7 N at tanθ = 0.5, about 27°.

Inclined plane

Block on a smooth incline

a = g sinθ
N = mg cosθ
time to slide a length L from rest: t = √(2L)/(g sinθ)
speed after falling a height h: v = √(2gh)

How to apply. Resolve the weight along the slope (mg sinθ) and perpendicular to it (mg cosθ). The speed at the bottom depends only on the height, but the time is longer on the gentler slope.

Example. A 30° smooth incline: a = 5 m/s². From rest over 10 m: t = √(2 × 10 ÷ 5) = 2 s and v = 10 m/s.

Block on a rough incline

sliding down: a = g(sinθ − μ cosθ)
thrown up the slope: a = −g(sinθ + μ cosθ) (it slows down)
it just starts to slide when tanθ = μ(s)
pushed up at constant speed by a force F along the slope: F = mg(sinθ + μ cosθ)

How to apply. Friction opposes the motion, so it changes direction when the block reverses. Going up, friction and gravity both act down the slope; coming down, they oppose each other.

Example. θ = 37° (sin 0.6, cos 0.8), μ = 0.25: down the slope a = 10(0.6 − 0.2) = 4 m/s²; up the slope a = −10(0.6 + 0.2) = −8 m/s².

Wedge and the sliding time

a smooth wedge moved with a = g tanθ keeps a block at rest on it
N = m√(g² + a²) in that case
same height, smooth slopes: t = (1)/(sinθ)√(2h)/(g) and the same final speed

How to apply. In the frame of the wedge, add a pseudo force ma opposite to the wedge’s acceleration. Then the block is in equilibrium under mg, N and ma.

Example. Smooth 30° wedge: a = 10 × tan30° = 5.77 m/s² keeps the block at rest. For h = 5 m the sliding time is 2 s at 30° and 1 s straight down.

Circular motion

Centripetal force

F(c) = (mv²)/(r) = mω²r
it is the net force towards the centre, not an extra force to be added
stone on a string in a horizontal circle: T = (mv²)/(r)

How to apply. Find which real forces point towards the centre (tension, friction, normal force, a part of the weight) and set their sum equal to mv²/r. Never draw the centripetal force as a separate arrow.

Example. A 0.5 kg stone on a 1 m string whirled at 4 m/s in a horizontal circle: T = 0.5 × 16 ÷ 1 = 8 N.

Vertical circle

at the top: T + mg = (mv²)/(r)
at the bottom: T − mg = (mv²)/(r)
string: v(top) ≥ √(gr) and v(bottom) ≥ √(5gr)
T(bottom) − T(top) = 6mg
rod or tube: v(top) can be 0 and v(bottom) ≥ 2√(gr)

How to apply. Use energy conservation between the bottom and the top, then the force equation at each point. The tension is least at the top, and zero there for the minimum speed.

Example. r = 0.9 m (g = 10): minimum speed at the bottom = √(5 × 10 × 0.9) = 6.7 m/s, and at the top 3 m/s.

Banking of roads

flat road: v(max) = √(μrg)
smooth banked road: tanθ = (v²)/(rg)
with friction: v(max) = √(rg(μ + tanθ))/(1 − μ tanθ)
v(min) = √(rg(tanθ − μ))/(1 + μ tanθ)

How to apply. On a flat road friction gives the centripetal force. On a smooth banked road the horizontal part of N does. With friction, the car can go faster or slower than the design speed.

Example. Flat road, r = 40 m, μ = 0.5 (g = 10): v(max) = √200 ≈ 14 m/s. A smooth road banked at 45° is safe at v = √(rg) = 20 m/s for r = 40 m.

Conical pendulum and a leaning cyclist

tanθ = (v²)/(rg) = (ω²r)/(g)
T cosθ = mg (T = tension)
time period of the conical pendulum: t = 2π√(L cosθ)/(g)
a cyclist leans by the same angle θ

How to apply. The horizontal part of the tension (or of the force from the ground) gives the centripetal force, and the vertical part balances the weight. Divide the two equations to remove the tension.

Example. A cyclist turns on a 20 m curve at 10 m/s: tanθ = 100 ÷ (20 × 10) = 0.5, so θ ≈ 27° from the vertical.

Bridge, dip and loop

top of a convex bridge: mg − N = (mv²)/(r)
bottom of a dip: N − mg = (mv²)/(r)
the car leaves the bridge when N = 0, at v = √(rg)

How to apply. Write the force equation along the line to the centre, with the weight and the normal force acting in opposite directions. The apparent weight is less on a hump and more in a dip.

Example. A 1000 kg car at 10 m/s on a bridge of radius 50 m (g = 10): N = 1000 × (10 − 2) = 8000 N.

Rotating table and a coin

friction gives the centripetal force: f = mω²r ≤ μmg
the body slips when ω²r = μg
ω(max) = √(μg)/(r)

How to apply. The coin slips at the smallest ω where the needed friction crosses its limit. A coin nearer the axis needs less friction, so it stays on longer.

Example. A coin 0.2 m from the axis, μ = 0.5 (g = 10): ω(max) = √(5 ÷ 0.2) = 5 rad/s.

Vertical circle at any point

at angle θ from the lowest point: T − mg cosθ = (mv²)/(r)
v² = u² − 2gr(1 − cosθ)
at the top (θ = 180°) these give T + mg = mv²/r

How to apply. Energy conservation gives v at the point, then the radial equation gives T. The string goes slack where T = 0.

Example. u² = 5gr at the bottom gives T = 3mg at the horizontal position (θ = 90°): T = mv²/r = m(5gr − 2gr)/r = 3mg.

Non-uniform circular motion

radial: a(c) = (v²)/(r)
tangential: a(t) = (dv)/(dt)
total a = √(a(c)² + a(t)²)
tension and weight components give the tangential and radial parts separately

How to apply. Resolve forces along the radius and along the tangent. The radial sum equals mv²/r, the tangential sum equals m a(t).

Example. A stone on a string of 1 m has v = 3 m/s and a(t) = 4 m/s²: a(c) = 9 m/s², a = √(81 + 16) = 9.85 m/s².

Rotor (wall of death)

the wall gives N = mω²r
the body does not slide down if μN ≥ mg, that is ω ≥ √(g)/(μr)
v(min) = √(rg)/(μ)

How to apply. The normal force from the wall is the centripetal force, and friction balances the weight.

Example. r = 2 m, μ = 0.5 (g = 10): ω(min) = √(10 ÷ 1) = 3.16 rad/s.

Pseudo force

Pseudo force in an accelerating frame

in a frame with acceleration a⃗, add a pseudo force −ma⃗ on every body
then use ΣF = ma in that frame
in a lift accelerating up, the apparent g is g + a
a pendulum in an accelerating car tilts at tanθ = (a)/(g)

How to apply. Choose a frame. If it accelerates, every body gets a force of size ma opposite to the acceleration. In an inertial frame there is no pseudo force at all.

Example. A pendulum bob in a car accelerating at 5 m/s² (g = 10): tanθ = 0.5, θ ≈ 27° backwards, and the tension is m√(g² + a²).

Blocks on a moving cart or wall

a block on a truck floor: friction needed f = ma, so it slips if a > μg
a block pressed against the front wall of an accelerating truck stays up if μma ≥ mg, that is a ≥ (g)/(μ)

How to apply. Draw the pseudo force, or use the real forces in the ground frame. The wall pushes the block forward with N = ma, and friction μN must support the weight.

Example. A block against the front wall of a truck, μ = 0.5 (g = 10): it stays up if a ≥ 20 m/s².

Most asked question types

How do I solve “Net force and acceleration” questions in JEE Main?

Several forces act on a body or a block. Find the acceleration, the net force or one unknown force. 1. Draw a free-body diagram and choose the axes. 2. Write ΣF = ma along each axis. 3. Solve for the unknown and check the direction. Example: A 5 kg block is pushed with 20 N against 5 N of friction: a = 15 ÷ 5 = 3 m/s².

How do I solve “Lift and weighing machine” questions in JEE Main?

A person stands on a weighing scale in a lift, or a bucket hangs from a rope. Find the reading or the tension. 1. Take upward as positive. 2. Write N − mg = ma. 3. Put a with its sign: positive for upward acceleration. Example: A 60 kg man in a lift going up at 2 m/s²: N = 60 × (10 + 2) = 720 N.

How do I solve “Impulse and change of momentum” questions in JEE Main?

A ball hits a wall or a bat, or a force acts for a short time. Find the impulse, the average force or the new speed. 1. Write the initial and final momentum with signs. 2. Impulse = final momentum − initial momentum. 3. Average force = impulse ÷ time of contact. Example: 0.2 kg ball: 10 m/s in, 8 m/s out: Δp = 0.2 × 18 = 3.6 N s.

How do I solve “Conservation of momentum and recoil” questions in JEE Main?

A gun fires a bullet, a body explodes, or two bodies stick together. Find a speed after the event. 1. Check that no external force acts along the line. 2. Write total momentum before = total momentum after. 3. Solve for the unknown speed with the correct signs. Example: A 50 g bullet at 200 m/s from a 5 kg gun: recoil = 0.05 × 200 ÷ 5 = 2 m/s.

How do I solve “Force from a stream or a machine gun” questions in JEE Main?

Water or bullets hit a surface at a steady rate. Find the force or the number per second. 1. Find the mass arriving each second. 2. Multiply by the change in velocity. 3. Use 2v for a rebound at the same speed, v for a stop. Example: 125 N is needed to hold a gun firing 10 g bullets at 250 m/s: n = 125 ÷ 2.5 = 50 per second.

How do I solve “Connected blocks” questions in JEE Main?

Two or three blocks are pushed or pulled together on a surface. Find the acceleration, a contact force or a tension. 1. Find a for the whole system. 2. Take one block alone. 3. Its mass times a gives the net force on it. Example: 2 kg and 3 kg, F = 10 N on the 2 kg block: a = 2 m/s², contact force on the 3 kg block = 6 N.

How do I solve “Atwood machine and pulleys” questions in JEE Main?

Masses hang over a pulley, or one hangs and one is on a table. Find the acceleration and the tension. 1. Write ΣF = ma for each mass with the same a and T. 2. Add the equations to find a. 3. Put a back to find T. Example: 3 kg and 2 kg: a = 2 m/s², T = 24 N.

How do I solve “Will the block move? (friction)” questions in JEE Main?

A block is pushed with a given force. Decide whether it slides and find the friction. 1. Find the limiting friction μ(s)N. 2. If the force is smaller, friction equals the force. 3. If it is larger, the block slides with friction μ(k)N. Example: 10 kg, μ(s) = 0.5: a 40 N push leaves it at rest with f = 40 N.

How do I solve “Block on a rough incline” questions in JEE Main?

A block slides down, or is thrown up, an incline with friction. Find the acceleration, the time or the angle. 1. Resolve the weight along and across the slope. 2. Write N = mg cosθ and the friction μN. 3. Use ΣF = ma along the slope with the sign of the motion. Example: 37°, μ = 0.25: down the slope a = 10(0.6 − 0.2) = 4 m/s².

How do I solve “Minimum force and block on a block” questions in JEE Main?

Find the least force to move a block, or the largest force before one block slips on another. 1. Write the balance of forces with friction. 2. For a block on a block, find a(max) = μg for the top block. 3. Use the total mass for the force on the lower block. Example: 2 kg on 4 kg with μ = 0.4: F(max) = 6 × 4 = 24 N.

How do I solve “Horizontal circle and banking” questions in JEE Main?

A car turns on a flat or banked road, or a body moves in a horizontal circle. Find the speed, the tension or the angle. 1. Mark the real forces towards the centre. 2. Set their sum equal to mv²/r. 3. For banking, use tanθ = v²/rg and add friction if given. Example: Flat road, r = 40 m, μ = 0.5: v(max) = √200 ≈ 14 m/s.

How do I solve “Vertical circle” questions in JEE Main?

A stone on a string, a bucket of water or a loop. Find the speed or the tension at the top or the bottom. 1. Write the force equation at the point. 2. Use energy conservation between the points. 3. For a full circle: top √(gr), bottom √(5gr). Example: r = 0.9 m: minimum speed at the bottom = √45 ≈ 6.7 m/s.

How do I solve “Pseudo force and wedge” questions in JEE Main?

A pendulum in a car, a block on a moving wedge or on a cart. Find the angle, the tension or the acceleration for rest. 1. Pick the accelerating frame. 2. Add the pseudo force −ma on each body. 3. Balance the forces in that frame. Example: Pendulum in a car at 5 m/s²: tanθ = 0.5, θ ≈ 27°.

Common myths and the truth

Myth: A body in motion needs a force to keep it moving. Truth: A body keeps its velocity if the net force is zero. A force is needed only to change the velocity.
Myth: Action and reaction cancel each other. Truth: They act on different bodies, so they cannot cancel in one free-body diagram.
Myth: Friction is always μN. Truth: Only the limiting and the kinetic friction are μN. A body at rest has friction equal to the applied force.
Myth: The normal force always equals mg. Truth: It depends on the situation: on an incline it is mg cosθ, in a lift m(g + a).
Myth: Centripetal force is an extra force on a body in a circle. Truth: It is just the name for the net force towards the centre.
Myth: The tension in a rope over a pulley equals the weight of the hanging mass. Truth: Only when the system is at rest or moving at constant speed. Otherwise T = m(g − a) for the mass going down with acceleration a.
Myth: A heavier body always falls faster on an incline. Truth: The acceleration g sinθ is independent of mass on a smooth incline.
Myth: A bigger angle of the incline always makes a block slide. Truth: It slides only when tanθ is larger than μ(s).
Myth: In a vertical circle the speed is the same at every point. Truth: Gravity changes the speed. For a string, v(bottom)² = v(top)² + 4gr.
Myth: There is a real force called pseudo force. Truth: It is a device used in an accelerating frame. In the ground frame there is no such force.

Mistakes to avoid

Friction equals the applied force while the body is at rest, not μN.
Action and reaction are on different bodies and never cancel in one free-body diagram.
Do not draw the centripetal force as an extra arrow.
The tension in a string over a pulley with a moving mass is not equal to the weight of the hanging mass.
In a lift, the normal force N is the reading of the scale, not mg.
Take the direction of motion as positive and keep the sign of friction opposite to the motion.
When a string is cut, the spring force stays the same at that instant.
Use g = 10 m/s² unless told otherwise.

Simulations for Laws of Motion

Angle of repose — Tilt a rough plane until the block begins to slide. The critical angle reveals μs.
Atwood machine — Connect two hanging masses over a pulley and predict which side descends.
Banked road turn — Bank a circular road so a car can turn without friction.
Two connected blocks — Pull a pair of blocks joined by a string and find the shared acceleration.
Apparent weight in an elevator — Accelerate an elevator and see how the floor’s normal reaction changes a weighing-scale reading.
Forces and friction — Push a block and see when the applied force overcomes friction.
Recoil of a gun — Fire a bullet from a gun on a frictionless cart. Momentum conservation fixes the recoil speed.
Impulse: catching an egg — Stop a falling object over a short or long time and compare the average force needed.
Block on an incline — Resolve gravity along a slope and include friction.
A load held by two ropes — Change the symmetry of two supporting ropes and calculate their tensions.

More: Class 11 Physics simulations · all JEE Main chapters.