📐 MODULE 2 — FORMULA BANK
All Formulas — Laws of Motion
Searchable formula bank with dimensional analysis, derivation hints, and exam importance tags.
🧮 Force Calculator (F = ma)
🧮 Friction Calculator (f = μN)
Newton's 2nd Law (Basic)
F = ma
Net force = mass × acceleration. F in Newtons, m in kg, a in m/s². Apply as net force — not individual force.
CBSENEETJEE M
Newton's 2nd Law (Momentum form)
F = dp/dt
Rate of change of momentum. More fundamental. Used for variable mass problems (rockets). Reduces to F=ma for constant mass.
JEE MJEE A
Weight
W = mg
Weight is the gravitational force on a body. g = 9.8 m/s² (use 10 for JEE calculations). Weight is a vector, mass is scalar.
CBSENEET
Apparent Weight (Lift moving up)
W_app = m(g + a)
When elevator accelerates upward, apparent weight increases. Normal force N = m(g+a). Person feels heavier.
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Apparent Weight (Lift moving down)
W_app = m(g - a)
When elevator accelerates downward, apparent weight decreases. If a = g (free fall), W_app = 0 (weightlessness).
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Tension in Atwood Machine
T = 2m₁m₂g/(m₁+m₂)
For massless string over frictionless pulley. Acceleration = (m₁-m₂)g/(m₁+m₂). One of the most tested configurations.
NEETJEE M
System Acceleration (2 blocks)
a = (F_net)/(m₁+m₂)
For connected bodies, treat as a system. Total net force divided by total mass gives common acceleration. Then find internal forces.
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Rocket Thrust Force
F_thrust = v_rel × |dm/dt|
For variable mass systems. v_rel = exhaust speed relative to rocket. dm/dt = rate of mass ejection. Net force on rocket = F_thrust - Weight.
JEE A
Inclined Plane — Acceleration
a = g(sinθ - μcosθ)
For body sliding down a rough incline. If μcosθ > sinθ, body doesn't slide. At angle of repose, a = 0.
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Derivation trick: For any multi-body system, ALWAYS treat the system as a whole first to find common acceleration (a = F_net / M_total). Then isolate each body to find internal forces (tensions, normal forces).
Static Friction (Maximum)
fs_max = μs × N
Maximum static friction. Actual static friction adjusts from 0 to this maximum. Body doesn't move until applied force exceeds fs_max.
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Kinetic Friction
fk = μk × N
Friction when body is sliding. Constant value (independent of speed). Always μk < μs. Used in all sliding problems.
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Angle of Friction
tan λ = μs
λ is the angle between resultant contact force and normal force when body is on verge of sliding. λ = angle of repose on incline.
CBSEJEE MJEE A
Angle of Repose
tan θ_r = μs
Maximum angle of incline at which body remains stationary. Above this angle, body starts sliding. θ_r = angle of friction (λ).
CBSENEET
Minimum Force to Move on Horizontal
F_min = μmg/√(1+μ²)
Minimum force (at angle φ = arctan μ) to just move body on rough horizontal surface. Pushing at this angle is most efficient.
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Normal force on incline
N = mg cosθ
Component of weight perpendicular to incline surface. This determines friction force. Decreases as θ increases — so friction also decreases at higher angles.
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When a body is on a rough incline and pushed vertically downward with force F: Normal force N ≠ mg cosθ. You must resolve F also perpendicular to the incline and add it to N. This is where JEE candidates lose marks.
Centripetal Acceleration
ac = v²/r = ω²r
Always directed toward center. v = linear speed, r = radius, ω = angular speed. ac = rω² is useful when ω is given.
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Centripetal Force
Fc = mv²/r = mω²r
Net force directed toward center. NOT a separate force — it's the role played by tension, friction, gravity etc. depending on context.
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Ideal Banking (no friction)
tan θ = v²/rg
Angle of banking for no friction. At this speed, no friction needed. Above: outer friction acts. Below: inner friction acts. Standard 3-mark question.
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Max speed on banked road (with friction)
v_max = √[rg(μ+tanθ)/(1-μtanθ)]
Upper limit of speed on a banked road. If μ=0, reduces to ideal banking speed. Derived from FBD with N and f both acting.
JEE MJEE A
Tension at bottom (vertical circle)
T_bottom = m(g + v²/r)
At the bottom of a vertical circle, tension is maximum. Both T and centripetal acceleration point upward, so T - mg = mv²/r → T = m(g + v²/r).
NEETJEE MJEE A
Minimum speed at top (vertical circle)
v_min(top) = √(gr)
When T = 0 at the top. Below this speed, the string goes slack. Used to find minimum speed at bottom: v_min(bottom) = √(5gr) via energy conservation.
NEETJEE MJEE A
Linear Momentum
p⃗ = mv⃗
Momentum = mass × velocity. Vector quantity. SI unit: kgm/s = Ns. Momentum is conserved when no net external force acts.
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Impulse
J⃗ = F⃗Δt = Δp⃗
Impulse = Change in momentum = Area under F-t graph. For variable force: J = ∫F dt. Average force Favg = J/Δt.
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Conservation of Momentum
m₁u₁+m₂u₂ = m₁v₁+m₂v₂
Total momentum conserved when net external force = 0. Apply component-wise for 2D collisions. Works for all types of collisions.
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Recoil Velocity (Gun-Bullet)
v_gun = -(m_b × v_b)/m_g
Initially at rest (p=0). After firing, gun recoils. Negative sign means opposite to bullet direction. Used in NEET numerical questions frequently.
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KE in terms of Momentum
KE = p²/(2m)
Derived from KE = ½mv² and p = mv. Useful when momentum is given and KE is asked. Also: p = √(2mKE). Used in both NEET and JEE.
NEETJEE M
Force-Momentum Relationship
F_avg = Δp/Δt
Average force = change in momentum / time. Used when F is not constant (collision, ball hitting wall). Always find Δp first, then divide by Δt.
CBSENEETJEE M
| Quantity | Formula | SI Unit | Dimensional Formula | Exam Weight |
|---|---|---|---|---|
| Force | F = ma | Newton (N) | [MLT⁻²] | Very High |
| Linear Momentum | p = mv | kgm/s or Ns | [MLT⁻¹] | Very High |
| Impulse | J = FΔt | Ns or kgm/s | [MLT⁻¹] | High |
| Friction Force | f = μN | Newton (N) | [MLT⁻²] | High |
| Coefficient of Friction (μ) | μ = f/N | Dimensionless | [M⁰L⁰T⁰] | High |
| Centripetal Force | F = mv²/r | Newton (N) | [MLT⁻²] | High |
| Angular Velocity | ω = v/r | rad/s | [T⁻¹] | Medium |
| Tension | T (contact force) | Newton (N) | [MLT⁻²] | Very High |
| Normal Force | N (contact force) | Newton (N) | [MLT⁻²] | Very High |
| Work done by friction | W = f × d | Joule (J) | [ML²T⁻²] | High |
Dimensional Formula Trick: Force [MLT⁻²] and Momentum [MLT⁻¹] are related by time dimension. Impulse [MLT⁻¹] has same dimensions as momentum — this is the Impulse-Momentum theorem in dimensional form.
Coefficient of friction (μ) is dimensionless [M⁰L⁰T⁰]. If you're asked "dimensions of μ" and write [MLT⁻²] (dimensions of force), you'll lose full marks. μ = f/N is a ratio of two forces.