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NLM formulas

Master NLM through 32 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

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NLM, every formula

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Newton's first law (inertia)
$F_{n e t} = 0 \Rightarrow v = constant$
Q1MCQFirst law
A body in uniform motion continues so unless acted on by:
- A
  a net external force
- B
  its weight
- C
  friction only
- D
  gravity only
Linear momentum
$p = m v$
Newton's second law
$F = \frac{d p}{d t}$
Q1MCQSecond law
Newton's second law states $\vec F$ equals:
- A
  $\dfrac{d\vec p}{dt}$
- B
  $m\vec v$
- C
  $\dfrac{d\vec v}{dt}$
- D
  $\vec p\,t$
Second law (constant mass)
$F_{n e t} = m a$
Newton's third law
$F_{A B} = - F_{B A}$
Q1MCQThird law
Action and reaction forces:
- A
  are equal and opposite, on different bodies
- B
  act on the same body
- C
  cancel out
- D
  are unequal
Impulse
$J = \int F d t = Δ p$
Weight
$W = m g$
Contact force between two blocks
$f_{1} = \frac{m _{2} F}{m _{1} + m _{2}}$
F applied on m1
Q1MCQContact force
When $F$ pushes block $m_1$ against $m_2$ on a smooth surface, the contact force on $m_2$ is:
- A
  $\dfrac{m_2 F}{m_1+m_2}$
- B
  $\dfrac{m_1 F}{m_1+m_2}$
- C
  $F$
- D
  $\dfrac{F}{2}$
Tension in a massive rope
$T = F (1 - \frac{x}{L})$
Q1MCQMassive rope tension
For a force $F$ on a uniform rope of length $L$, the tension at distance $x$ from the pulled end is:
- A
  $F\left(1-\dfrac{x}{L}\right)$
- B
  $F\dfrac{x}{L}$
- C
  $F$
- D
  $\dfrac{F}{L}$
Atwood machine acceleration
$a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$
Atwood machine tension
$T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}$
Force on pulley (Atwood)
$R = 2 T = \frac{4 m _{1} m _{2} g}{m _{1} + m _{2}}$
Q1MCQForce on pulley
The force on the Atwood pulley's support is:
- A
  $2T$
- B
  $T$
- C
  $\dfrac{T}{2}$
- D
  $4T$
Spring force (Hooke's law)
$F = - k x$
Q1MCQHooke's law
The restoring force of a spring is:
- A
  $-kx$
- B
  $kx^{2}$
- C
  $\dfrac{k}{x}$
- D
  $k$
Spring cut in ratio
$k_{1} = \frac{( m + n ) k}{m}, k_{2} = \frac{( m + n ) k}{n}$
cut in ratio m:n
Q1MCQSpring cut
If a spring of constant $k$ is halved, each half has constant:
- A
  $2k$
- B
  $\dfrac{k}{2}$
- C
  $k$
- D
  $4k$
Springs in series
$\frac{1}{k _{e q}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$
Springs in parallel
$k_{e q} = k_{1} + k_{2} + \dots$
Apparent weight in a lift (up)
$W^{'} = m (g + a)$
Apparent weight in a lift (down)
$W^{'} = m (g - a)$
Weightlessness (free fall)
$a = g \Rightarrow W^{'} = 0$
Q1MCQWeightlessness
A person in a freely falling lift feels:
- A
  weightless
- B
  heavier
- C
  normal weight
- D
  double weight
Wedge constraint
$V_{3} = V_{1} sin θ$
velocities ⟂ contact equal
Q1MCQWedge constraint
For a block on a moving wedge (angle $\theta$) staying in contact, the velocity relation is:
- A
  $V_3=V_1\sin\theta$
- B
  $V_3=V_1\cos\theta$
- C
  $V_3=V_1$
- D
  $V_3=V_1\tan\theta$
Newton's law for a system
$F_{e x t} = \sum m_{i} a_{i}$
Q1MCQSystem law
For a system, the net external force equals:
- A
  $\sum m_i\vec a_i$
- B
  $\sum m_i$
- C
  $m\vec a$ of one body
- D
  $0$
Pseudo force
$F_{p se u d o} = - m a_{f r am e}$
Q1MCQPseudo force
In a frame accelerating at $\vec a$, the pseudo force on mass $m$ is:
- A
  $-m\vec a$
- B
  $m\vec a$
- C
  $mg$
- D
  $0$
Non-inertial frame law
$F_{r e a l} + F_{p se u d o} = m a$
Q1MCQNon-inertial law
In a non-inertial frame, $m\vec a$ equals:
- A
  $\vec F_{real}+\vec F_{pseudo}$
- B
  $\vec F_{real}$ only
- C
  $\vec F_{pseudo}$ only
- D
  $0$
Static friction range
$0 \leq f_{s} \leq μ_{s} N$
Q1MCQStatic friction
Static friction $f_s$ satisfies:
- A
  $0\le f_s\le\mu_s N$
- B
  $f_s=\mu_s N$ always
- C
  $f_s=\mu_k N$
- D
  $f_s>\mu_s N$
Limiting friction
$f_{s, m a x} = μ_{s} N$
Kinetic friction
$f_{k} = μ_{k} N$
Angle of friction
$tan ϕ = μ_{s}$
Q1MCQAngle of friction
The angle of friction $\phi$ satisfies:
- A
  $\tan\phi=\mu_s$
- B
  $\sin\phi=\mu_s$
- C
  $\cos\phi=\mu_s$
- D
  $\phi=\mu_s$
Contact force (max)
$F_{co n t a c t} = m g 1 + μ_{s}^{2}$
Q1MCQContact force
The maximum contact force on a block of weight $mg$ is:
- A
  $mg\sqrt{1+\mu_s^{2}}$
- B
  $mg$
- C
  $\mu_s mg$
- D
  $mg(1+\mu_s)$
Angle of repose
$tan θ_{c} = μ_{s}$
Q1MCQAngle of repose
The angle of repose $\theta_c$ satisfies:
- A
  $\tan\theta_c=\mu_s$
- B
  $\sin\theta_c=\mu_s$
- C
  $\cos\theta_c=\mu_s$
- D
  $\theta_c=\mu_s$
Block on rough incline (about to slide)
$m g sin θ = μ_{s} m g cos θ$
Q1MCQIncline sliding
A block just slides down a rough incline when:
- A
  $\tan\theta=\mu_s$
- B
  $\sin\theta=\mu_s$
- C
  $\theta=90^{\circ}$
- D
  $\mu_s=0$
Acceleration down rough incline
$a = g (sin θ - μ_{k} cos θ)$
Q1MCQAccel down incline
The acceleration of a block sliding down a rough incline is:
- A
  $g(\sin\theta-\mu_k\cos\theta)$
- B
  $g\sin\theta$
- C
  $g(\sin\theta+\mu_k\cos\theta)$
- D
  $\mu_k g\cos\theta$
Minimum force to move a block
$F_{m i n} = \frac{μ m g}{1 + μ ^{2}}$
pulled at optimum angle
Q1MCQMin force to move
The minimum force to move a block (pulled at the optimum angle) is:
- A
  $\dfrac{\mu mg}{\sqrt{1+\mu^{2}}}$
- B
  $\mu mg$
- C
  $mg\sqrt{1+\mu^{2}}$
- D
  $\dfrac{mg}{\mu}$

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NLM formulas

NLM, every formula

Newton's first law (inertia)

Linear momentum

Newton's second law

Second law (constant mass)

Newton's third law

Impulse

Weight

Contact force between two blocks

Tension in a massive rope

Atwood machine acceleration

Atwood machine tension

Force on pulley (Atwood)

Spring force (Hooke's law)

Spring cut in ratio

Springs in series

Springs in parallel

Apparent weight in a lift (up)

Apparent weight in a lift (down)

Weightlessness (free fall)

Wedge constraint

Newton's law for a system

Pseudo force

Non-inertial frame law

Static friction range

Limiting friction

Kinetic friction

Angle of friction

Contact force (max)

Angle of repose

Block on rough incline (about to slide)

Acceleration down rough incline

Minimum force to move a block

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