Why do objects accelerate when pushed? Why does a seatbelt save lives during sudden braking? How do banking angles prevent sports cars from skidding on sharp highway turns? The answers lie in Dynamics — the branch of physics that studies the forces causing motion. Laws of Motion forms Chapter 5 of the CBSE Class 11 Physics syllabus and is one of the most vital chapters for JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers Newton's three laws of motion, linear momentum conservation, impulse-momentum theorem, apparent weight of a person in an elevator, static/kinetic/rolling friction, angle of repose, banking of circular roads, and five step-by-step solved entrance exam numericals.
- 1. Newton's Three Laws of Motion & Momentum
- 2. Impulse-Momentum Theorem & Applications
- 3. Law of Conservation of Linear Momentum (Recoil of Gun)
- 4. Apparent Weight of a Person in an Elevator (Lift Physics)
- 5. Physics of Friction: Static, Kinetic & Angle of Repose
- 6. Circular Motion Dynamics & Banking of Roads
- 7. Solved Entrance Exam Numerical Problems (JEE / NEET)
- 8. Frequently Asked Questions (FAQ)
1. Newton's Three Laws of Motion & Momentum
| Newton's Law | Mathematical Formulation | Core Principle & Key Takeaway |
|---|---|---|
| 1st Law (Law of Inertia) | If ΣF_ext = 0 ⇒ v = constant (a = 0) | An object remains at rest or moves with uniform velocity unless acted upon by an external net force. Defines Inertia (Rest, Motion, Direction). |
| 2nd Law (Real Law of Motion) | F_net = dp/dt = d(mv)/dt = m*a | The rate of change of linear momentum is directly proportional to applied net force and takes place in force's direction. Calculates magnitude of force! |
| 3rd Law (Action & Reaction) | F_AB = - F_BA | To every action, there is an equal and opposite reaction. Action and reaction forces act on TWO DIFFERENT BODIES simultaneously. |
• Third Law is contained in Second Law: For an isolated system of two interacting bodies with no external force, total Δp = 0 ⇒ Δp1 + Δp2 = 0 ⇒ F12 = -F21.
2. Impulse-Momentum Theorem & Applications
When a large force acts on a body for a very short duration of time (e.g., hitting a cricket ball with a bat, catching a baseball), it is called an Impulsive Force.
J = F_avg · Δt = ∫ F dt (SI Unit: N·s or kg·m/s)
• Impulse-Momentum Theorem:
J = Δp = p_final - p_initial = m(v - u)
• Shock Absorbers in Vehicles: Increases time of impact during bumps, reducing force on passengers.
• Crockery Packed in Foam: Prolongs impact duration during drops.
3. Law of Conservation of Linear Momentum (Recoil of Gun)
If the net external force acting on a system of particles is zero (ΣF_ext = 0), the total linear momentum of the system remains constant in magnitude and direction.
Initial momentum before firing = 0.
Final momentum after firing = M V + m v.
By Conservation of Momentum: M V + m v = 0 ⇒ V = -(m / M) v.
The negative sign indicates that the gun recoils in the direction OPPOSITE to bullet motion!
4. Apparent Weight of a Person in an Elevator (Lift Physics)
When a person of mass m stands on a weighing scale inside an elevator, the scale reads the Normal Reaction force (N) exerted by the floor, which is the Apparent Weight (W').
| Elevator Motion State | Equation of Motion | Apparent Weight (N) vs Real Weight (mg) |
|---|---|---|
| 1. At Rest or Uniform Speed (a = 0) | N - mg = 0 | N = mg (Apparent weight = True weight). |
| 2. Accelerating UPWARD (+a) | N - mg = ma ⇒ N = m(g + a) | N > mg (Person feels HEAVIER!). |
| 3. Accelerating DOWNWARD (-a) | mg - N = ma ⇒ N = m(g - a) | N < mg (Person feels LIGHTER!). |
| 4. Free Fall (Lift cable snaps, a = g) | N = m(g - g) = 0 | N = 0 (Complete Weightlessness!). |
5. Physics of Friction: Static, Kinetic & Angle of Repose
Friction is a tangential contact force that opposes relative motion (or tendency of relative motion) between two contacting surfaces.
• Limiting Static Friction: Maximum static friction:
f_{s,max} = μ_s N
• Kinetic Friction (fk): Opposes actual sliding motion:
f_k = μ_k N (μ_k < μ_s).
tan λ = μ_s
• Angle of Repose (θ): Maximum angle of an inclined plane at which a body just begins to slide down under gravity:
tan θ = μ_s ⇒ Angle of Repose = Angle of Friction!
6. Circular Motion Dynamics & Banking of Roads
When a vehicle takes a turn along a curved road of radius r, it requires a Centripetal Force (Fc = m v² / r).
fs = (m v²) / r ≤ μs mg ⇒ v_max = √(μs r g).
2. Banked Curved Road (Outer edge raised by angle θ):
Resolving Normal reaction N and friction f:
• Optimum Speed (zero friction wear): v0 = √(rg tanθ).
• Maximum Safe Speed with friction μs:
v_{max} = √[ r g ( (μ_s + tanθ) / (1 - μ_s tanθ) ) ]
3. Bending of a Cyclist: A cyclist leans inward by angle θ with vertical to avoid skidding: tanθ = v² / (rg).
7. Solved Entrance Exam Numerical Problems (JEE / NEET)
Acceleration a = F / m = -50 / 20 = -2.5 m/s².
Using 1st equation: v = u + at ⇒ 0 = 15 - 2.5 t ⇒ t = 15 / 2.5 = 6 s.
(a) Moving UPWARD at a = 3 m/s²:
N1 = m(g + a) = 70(10 + 3) = 70 × 13 = 910 N (or 91 kg-wt).
(b) Moving DOWNWARD at a = 2 m/s²:
N2 = m(g - a) = 70(10 - 2) = 70 × 8 = 560 N (or 56 kg-wt).
Using 3rd equation: v² - u² = 2as ⇒ 0² - (90)² = 2 a 0.6 ⇒ -8100 = 1.2 a ⇒ a = -6750 m/s².
Resistive Force F = m a = 0.04 × (-6750) = -270 N.
(a) Limiting static friction fs,max = μs N = 0.4 × 100 = 40 N.
(b) Since F_app = 60 N > 40 N, block moves! Kinetic friction applies: fk = μk N = 0.35 × 100 = 35 N.
Net Force F_net = F_app - fk = 60 - 35 = 25 N.
Acceleration a = F_net / m = 25 / 10 = 2.5 m/s².
Using Banking Formula: v_max = √[ r g ( (μs + tanθ) / (1 - μs tanθ) ) ].
(μs + tanθ) / (1 - μs tanθ) = (0.2 + 0.268) / [1 - (0.2 × 0.268)] = 0.468 / (1 - 0.0536) = 0.468 / 0.9464 ≈ 0.494.
v_max = √(300 × 9.8 × 0.494) = √(2940 × 0.494) = √(1452.36) ≈ 38.1 m/s = 137.1 km/h.
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8. Frequently Asked Questions (FAQ)
Because both Newton's First Law and Third Law can be mathematically derived from Newton's Second Law (F = dp/dt = m*a). When F = 0, a = 0 (First Law); in isolated systems, F12 = -F21 (Third Law).
The Impulse-Momentum Theorem states that the impulse of a net force acting on a body equals the change in linear momentum produced in the body: J = F_avg · Δt = Δp.
In an upward accelerating elevator (+a), the floor must push up harder to accelerate the body mass m, increasing normal reaction N = m(g + a) > mg.
The Angle of Repose (θ) equals the Angle of Friction (λ). Both satisfy the relation tanθ = tanλ = μs (where μs is the coefficient of static friction).
Roads are banked (outer edge raised) so that the horizontal component of normal reaction (N sinθ) contributes to centripetal force, reducing reliance on friction and preventing skidding at high speeds.
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