Laws of Motion

1. Newton's Laws of Motion

Fundamental laws governing the motion of objects.

First Law (Inertia): Body remains at rest or uniform motion unless acted upon by external force.
Second Law (Momentum): Force = Rate of change of momentum (F = dp/dt = ma).
Third Law (Action-Reaction): To every action, there is an equal and opposite reaction (F_AB = -F_BA).

Impulse

Large force acting for short time.
Impulse (J) = F Δt = Δp (Change in Momentum).
Area under F-t graph gives Impulse.

(Asked in NEET 2018, 2020, 2021)

2. Friction

Opposing force that comes into play when there is relative motion (or tendency) between surfaces.

Static Friction (f_s): Self-adjusting. Max value is Limiting Friction (f_L = μ_sN).
Kinetic Friction (f_k): Constant during motion (f_k = μ_kN).
Angle of Repose (α): tanα = μ_s.

(Asked in NEET 2016, 2019, 2023)

3. Circular Motion Dynamics

Force required to keep body in circular path.

Centripetal Force: F_c = mv²/r = mrω². (Directed towards center).
Centrifugal Force: Pseudo force in rotating frame (Available to observer inside car).

Banking of Roads

Raising outer edge to provide centripetal force via Normal component.

Safe velocity (No Friction): v = √(rg tanθ)
Max velocity (With Friction):
v_max = √[ rg (μ + tanθ) / (1 - μtanθ) ]

(Asked in NEET 2017, 2022)

4. Conservation of Momentum

If F_ext = 0, then p = Constant.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Recoil of Gun: V_gun = - (m_bullet v_bullet) / M_gun.

Common Forces

Weight (mg): Always downwards.
Normal (N): Perpendicular to surface.
Tension (T): Along string, away from body.
Spring Force: F = -kx.

Numericals: Laws of Motion

Q1. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Find magnitude and direction of acceleration.

Solution:
F_net = √(8² + 6²) = √(64 + 36) = √100 = 10 N.
a = F/m = 10 / 5 = 2 m/s².
Direction θ = tan^-1(6/8) = tan^-1(0.75) = 37° with 8 N force.

Q2. A cricket ball of mass 150 g moving at 12 m/s is hit by a bat and turns back at 20 m/s. Force acts for 0.01 s. Find average force.

Solution:
Change in momentum Δp = m(v - (-u)) = m(v + u).
Δp = 0.150 (20 + 12) = 0.15 × 32 = 4.8 kg m/s.
F_avg = Δp / Δt = 4.8 / 0.01 = 480 N.

Q3. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If muzzle speed of shell is 80 m/s, what is recoil speed of gun?

Solution:
Conservation of Momentum: M_gV_g + m_sv_s = 0.
100 V_g = - 0.020 × 80
V_g = - 1.6 / 100 = -0.016 m/s (or 1.6 cm/s backward).

Q4. A man of mass 70 kg stands on a weighing scale in a lift accelerating upwards with 5 m/s². What is the reading? (g=10)

Solution:
Apparent Weight R = m(g + a).
R = 70 (10 + 5) = 70 × 15 = 1050 N.
Scale Reading (Mass) = 1050 / 10 = 105 kg.

Q5. Two masses 8 kg and 12 kg are connected by a string passing over a pulley. Find acceleration of the system.

Solution:
a = (m₂ - m₁)g / (m₁ + m₂)
Assuming m₂ = 12 kg (heavier).
a = (12 - 8) × 10 / (12 + 8)
a = 40 / 20 = 2 m/s².

Q6. A block of mass 4 kg rests on a horizontal plane (μ_s=0.4, μ_k=0.3). Force of 12 N is applied. Find friction force.

Solution:
Limiting Friction f_L = μ_smg = 0.4 × 4 × 10 = 16 N.
Applied Force F = 12 N.
Since F < f_L, block does not move.
Static Friction adjusts to applied force. f = F = 12 N.

Q7. A circular road of radius 30 m is banked for 15 m/s. Calculate the angle of banking.

Solution:
tanθ = v² / rg
tanθ = (15)² / (30 × 10)
tanθ = 225 / 300 = 0.75.
θ = tan^-1(0.75) = 37°.

Q8. A block slides down a 30° incline with acceleration g/4. Find coefficient of kinetic friction.

Solution:
a = g(sinθ - μcosθ)
g/4 = g(sin 30° - μ cos 30°)
0.25 = 0.5 - μ(0.866)
μ(0.866) = 0.25 → μ = 0.25 / 0.866 ≈ 0.29.

Q9. Three blocks 2kg, 3kg, 5kg are connected by strings on smooth horizontal floor & pulled by 50N force. Find acceleration.

Solution:
System Mass M = 2 + 3 + 5 = 10 kg.
a = F / M = 50 / 10 = 5 m/s².
Tension varies, but common acceleration is 5 m/s².

Q10. A stone of 0.2 kg tied to end of 1 m string is whirled in horizontal circle at 5 rad/s. Find tension.

Solution:
T = mrω²
T = 0.2 × 1 × (5)²
T = 0.2 × 25 = 5 N.

Important Formulae

1. Force & Impulse

F = ma = dp/dt

Impulse J = F_avg × Δt = Δp
Conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

2. Friction

Limiting Friction:

f_L = μ_s N

Kinetic Friction:

f_k = μ_k N

Angle of Repose α: tanα = μ_s
Accel. on rough incline: a = g(sinθ - μcosθ)

3. Circular Motion

Centripetal Force:

F_c = mv²/r

Banking Angle:

tanθ = v² / rg

4. Motion of Blocks

Connected Motion (Atwood):

a = (m₂-m₁)g / (m₁+m₂)

Lift Motion:

N = m(g + a) (Up)
N = m(g - a) (Down)

20 NEET Golden Facts

1. Inertia: Measure of mass. Higher mass → Higher inertia. Independent of velocity.
2. Action-Reaction: Act on different bodies. Never cancel each other. Simultaneous in nature.
3. Pseudo Force: Applies in Non-Inertial frames (accelerating frames). Direction opposite to acceleration of frame. F_p = -ma.
4. Friction Direction: Always opposes relative motion (or tendency of relative motion) between surfaces.
5. Static Friction: Is a variable force (0 to μ_sN). It adjusts to applied force.
6. Pull vs Push: Pulling is easier than pushing because vertical component of force reduces Normal reaction (hence friction).
7. Rocket: Works on Conservation of Linear Momentum. Mass decreases as fuel burns. Thrust F = u(dm/dt).
8. Banking: Independent of mass of vehicle. Depends on radius and angle.
9. Equilibrium: Vector sum of all forces is zero. Lami's theorem useful for 3 concurrent forces.
10. Centripetal Work: Work done by centripetal force is always Zero (Force perpendicular to displacement).
11. Limiting Friction: It is the maximum value of static friction. Once motion starts, kinetic friction (constant) acts. μ_k < μ_s.
12. Spring Scale: Measures Tension, not Mass. In a lift accelerating up, reading increases.
13. Impulse Area: Area under Force-Time graph gives Impulse (Change in Momentum).
14. Horse Cart: Horse pushes ground backward, ground pushes horse forward (Friction). Motion due to external force by ground.
15. Frame of Ref: Inertial (Non-accelerating), Newton's laws valid. Non-Inertial (Accelerating), need Pseudo force.
16. Tension: Same throughout a massless string. Different if string has mass.
17. Normal Reaction: Not always equal to mg. On incline N = mgcosθ.
18. Smooth Surface: Friction coefficient μ = 0. Reaction is always Normal (Perpendicular).
19. Rolling: Static friction acts during pure rolling (prevents slipping). No energy loss to friction.
20. Conical Pendulum: T cosθ = mg, T sinθ = mv²/r. Time Period = 2π √(Lcosθ/g).

📱 Practice MCQs for this topic inside our App