Laws of Motion

Laws of MotionNEET Physics · Class 11 · NCERT Chapter 4

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Laws of Motion. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Newton's second law solver

Impulse-momentum theorem

1D collision (elastic + inelastic)

Static vs kinetic friction explorer

Inclined plane with friction

Atwood machine

Banking of roads

Newton's laws and momentum

The second law in F = ma form, the impulse-momentum theorem, and 1D collisions. Conservation of momentum is the unifying idea.

Newton\'s laws

Newton's second law solver

The familiar F = m·a, with one variable left blank. The widget computes the missing one and shows the working.

Force F (N)

Mass m (kg)

Acceleration a (m/s²)

Answer

F = 10.000 N

F = m a = 5 \times 2 = 10.000 N

Try this

Leave exactly one field blank — the solver fills it in.
Type F = 100, m = 20 → a = 5 m/s².
Try F = 0, m = 5: a comes out 0 — no force, no acceleration. (Newton's first law as a special case.)
NEET tip: in many problems you have F and m, and you need a. This is the routine direction.

Impulse

Impulse-momentum theorem

Watch how the average force during a collision changes when you change the contact time. The change in momentum stays the same.

A ball of mass m hits a wall with velocity v and bounces back with the same speed. The contact time is Δt. See how the average force depends on the contact time.

Mass m: 0.50 kg

Speed v: 10 m/s

Contact time Δt: 50 ms (0.050 s)

Δp = J

10.00 N·s

Avg force F

200 N

Direction of Δp

Away from wall

J = Δ p = m (- v) - m v = - 2 m v = - 2 \times 0.5 \times 10 = - 10.00 N\cdots

F_{avg} = \frac{∣ J ∣}{Δ t} = \frac{10.00}{0.050} = 200 N

Try this

Increase the contact time → average force shrinks. Why airbags work.
Halving Δt doubles F_avg. Why a hard surface hurts more than a soft one.
Note Δp is always −2mv when the bounce returns the same speed — direction matters.

Collisions

1D collision (elastic + inelastic)

Two bodies on a frictionless line. Toggle elastic vs perfectly inelastic and adjust the masses and initial velocities to see what comes out.

m₁: 2 kg

u₁: 6 m/s

m₂: 3 kg

u₂: 0 m/s

v₁ (after)

-1.20 m/s

v₂ (after)

4.80 m/s

Initial momentum

12.00 kg·m/s

Final momentum

12.00 kg·m/s

Initial KE

36.00 J

Final KE

36.00 J

no loss (elastic)

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}} = - 1.20

v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}} = 4.80

Try this

Set m₁ = m₂, elastic → velocities exchange. Famous NEET fact.
Set m₁ ≫ m₂, target at rest → v₁ ≈ u₁, v₂ ≈ 2 u₁. Heavy ball barely slows; light ball gets bounced.
Set inelastic with u₂ = 0 → final speed is u₁ × m₁/(m₁+m₂). Lots of KE lost.

Friction

Static vs kinetic friction in action, plus the inclined-plane setup that NEET tests almost every year.

Friction

Static vs kinetic friction

Apply a horizontal force to a block and watch static friction self-adjust until the threshold, then kinetic friction kicks in.

Mass m: 5 kg

μₛ (static): 0.40

μₖ (kinetic): 0.30

Applied force F: 15 N

State

At rest (static friction balancing)

Friction force: 15.00 N

Normal N = m·g = 50 N
Max static friction = μₛ·N = 20.00 N
Kinetic friction = μₖ·N = 15.00 N
Angle of repose θᵣ = tan⁻¹(μₛ) ≈ 21.8°

(f_{s})_{max} = μ_{s} m g = 0.4 \times 5 \times 10 = 20.00 N

tan θ_{r} = μ_{s} \Rightarrow θ_{r} = 21.8°

Try this

Push softly (F small) → block stays at rest, static friction self-adjusts to F.
Increase F past μₛ·m·g → block starts sliding, kinetic friction takes over (smaller).
Lower μₛ slowly → angle of repose drops. Loose gravel has small μₛ; sandpaper has large μₛ.

Friction

Inclined plane with friction

A classic NEET setup. Adjust the angle and friction coefficient to see when the block stays put and when it slides.

Mass m: 2 kg

Incline angle θ: 30°

Friction μ: 0.30 (angle of repose ≈ 16.7°)

State

Sliding, a = 2.40 m/s²

m g sin θ = 10.00 N, μ m g cos θ = 5.20 N

a = g (sin θ - μ cos θ) = 2.40 m/s^{2}

The block slides whenever tan θ > μ. With your current μ that happens past θ ≈ 16.7°.

Try this

Set μ = 0 → a = g sin θ (frictionless incline). Mass cancels.
Set θ at exactly the angle of repose → block is on the verge. Just above, it slides.
Increase μ past tan θ → block stays put no matter how heavy it is.

Connected systems and banking

The Atwood machine and a banking-of-roads calculator with and without friction.

Pulleys

Atwood machine

Two masses on a frictionless, massless pulley. Drag the masses to see how the acceleration and tension scale with the mass difference.

Mass m₁ (left): 5 kg

Mass m₂ (right): 3 kg

Acceleration |a|

2.50 m/s²

String tension T

37.50 N

Heavier mass on the left accelerates downward at 2.50 m/s²; the other goes up at the same rate.

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = \frac{( 5 - 3 ) \times 10}{8} = 2.50 m/s^{2}

T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 \times 5 \times 3 \times 10}{8} = 37.50 N

Try this

Set m₁ = m₂ → a = 0 and T = m·g. The system is in equilibrium.
Set m₁ ≫ m₂ → a → g (free fall). Tension shrinks.
For the same difference in mass, lighter total mass means larger acceleration. Total mass appears in the denominator.

Banking

Banking of roads

A banked road lets a car turn faster than a flat one. With friction added, you also get a minimum safe speed below which the car slips inward.

Turn radius r: 100 m

Banking angle θ: 15°

Friction μ: 0.20

Ideal speed (no friction)

16.37 m/s

Max safe speed (with friction)

22.24 m/s

Min safe speed (with friction)

8.03 m/s

(below this, car slips inward)

v_{ideal} = r g tan θ = 100 \times 10 \times 0.268 = 16.37 m/s

v_{max} = \frac{r g ( tan θ + μ )}{1 - μ tan θ} = 22.24 m/s

Try this

Set μ = 0 → max and min speeds collapse onto v_ideal. Without friction, only one speed is safe.
Increase μ → both the max and the min speeds widen around the ideal speed.
Increase the radius → safe speed grows as √r. Doubling radius increases safe speed by √2 ≈ 1.41.

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Motion in a Plane

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