Laws of Motion

1. Newton's First Law — Law of Inertia

Newton's First Law states: Every object continues in its state of rest or uniform motion in a straight line unless acted upon by an external unbalanced force. This is also called the Law of Inertia.

Inertia is the natural tendency of an object to resist any change in its state of rest or motion. It is directly proportional to mass — heavier objects have greater inertia.

Three types of inertia:

Inertia of rest: tendency to remain at rest (e.g., passengers jerk backward when a bus starts suddenly)
Inertia of motion: tendency to continue moving (e.g., passengers jerk forward when a bus brakes)
Inertia of direction: tendency to continue in the same direction (e.g., mud flies off a rotating wheel tangentially)

Inertial Reference Frame: A frame of reference in which Newton's first law holds — i.e., a frame that is either at rest or moving with constant velocity. No pseudo forces exist in inertial frames.

Non-Inertial Frame: An accelerating frame of reference. In such frames, Newton's laws do not directly apply. We introduce a pseudo force (also called fictitious force) equal to $-m ec{a}_{frame}$ (opposite to the frame's acceleration) to make Newton's second law applicable.

ec{F}_{pseudo} = -m ec{a}_{frame}

where $m$ is the mass of the object and $ec{a}_{frame}$ is the acceleration of the non-inertial frame.

NEET tip: Newton's first law defines force and inertia, and also defines an inertial frame. If no net force acts, velocity (not just speed) is constant. A body moving in a circle is NOT in an inertial frame situation — it has centripetal acceleration.

2. Newton's Second Law — Force and Momentum

Newton's Second Law states: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.

ec{F} = rac{d ec{p}}{dt}

For constant mass: $ec{F} = m ec{a}$

For variable mass (rocket, falling chain): $ec{F} = m rac{d ec{v}}{dt} + ec{v} rac{dm}{dt}$

Key points about $ec{F} = m ec{a}$ :

It is a vector equation — force and acceleration are in the same direction
$ec{F}$ here is the net (resultant) force, not any individual force
1 Newton = force that gives 1 kg mass an acceleration of 1 m/s² → $1 ext{ N} = 1 ext{ kg·m·s}^{-2}$
Weight: $W = mg$ (gravitational force on object), acts downward toward Earth's centre

Impulse: When a large force acts for a very short time, we use impulse:

ec{J} = ec{F} cdot Delta t = Delta ec{p} = m ec{v}_f - m ec{v}_i

SI unit of impulse: N·s = kg·m·s⁻¹ (same as momentum)

Impulse is the area under the Force–time graph. This is extremely useful in problems involving collisions, kicks, and impacts where the exact force profile is unknown but the time interval is given.

Pro tip: Dimensional formula of force:

[MLT^{-2}]

. For the impulse-momentum theorem, always write the vector form — signs (direction) are crucial in NEET problems involving bouncing balls or collisions.

3. Newton's Third Law — Action and Reaction

Newton's Third Law states: For every action, there is an equal and opposite reaction. More precisely: if body A exerts a force on body B, then body B simultaneously exerts a force on body A that is equal in magnitude, opposite in direction, and along the same line of action.

ec{F}_{AB} = - ec{F}_{BA}

Critical understanding — action-reaction pairs act on DIFFERENT bodies:

Your foot pushes backward on the ground → ground pushes your foot forward (you walk)
Rocket exhaust pushes backward → gases push rocket forward (propulsion)
Your hand pushes water backward → water pushes hand forward (swimming)
Earth pulls you down (gravity) → you pull Earth up (same force, tiny effect on Earth)

Caution — Common Misconception: Action and reaction forces do not cancel each other because they act on different bodies. To cancel, forces must act on the same body. When you draw a Free Body Diagram for one object, only forces ON that object are shown — the reaction force is on the other body and is not drawn.

NEET tip: Action-reaction pairs are always of the same type — both gravitational, both normal, both tension, etc. They are simultaneous — neither is the "cause" of the other. If one ceases, both cease simultaneously.

4. Free Body Diagram (FBD)

A Free Body Diagram is a sketch of a single isolated object showing all external forces acting on it. Mastering FBDs is the most important skill for solving mechanics problems.

Steps to draw a correct FBD:

Identify the object (system) to be isolated
Draw the object as a point mass or simple shape
Draw Weight (W = mg) vertically downward from the centre
Draw Normal force (N) perpendicular to the contact surface
Draw Tension (T) along the string, away from the object
Draw Friction (f) along the surface, opposing relative motion (or tendency of motion)
Draw any other applied forces with correct direction
Choose convenient axes (along and perpendicular to motion, or along and perpendicular to incline)
Resolve all forces along chosen axes and apply $Sigma F = ma$

Example — Block on an inclined plane:

After drawing the FBD, resolve forces: along the incline — $mgsin heta - f = ma$ ; perpendicular to incline — $N - mgcos heta = 0$ , so $N = mgcos heta$ .

NEET tip: Every NEET mechanics problem should start with a FBD. Common error: forgetting to include all forces, or drawing the reaction force on the same body. The FBD is for ONE body only.

5. Normal Force & Apparent Weight

The normal force N is the contact force exerted by a surface on an object, always perpendicular to the surface. It adjusts itself to prevent the object from passing through the surface — it is a reaction force, not always equal to mg.

A person of mass m in a lift (elevator):

Situation	Equation (↑ positive)	Apparent Weight
Lift at rest or uniform velocity	$N - mg = 0$	$N = mg$
Lift accelerating upward	$N - mg = ma$	$N = m(g+a)$ > mg
Lift accelerating downward	$mg - N = ma$	$N = m(g-a)$ < mg
Free fall ( $a = g$ downward)	$mg - N = mg$	$N = 0$ (weightlessness)

On an inclined plane (angle θ, no motion perpendicular to incline):

N = mgcos heta

As θ increases, N decreases. At θ = 90° (vertical wall), N = 0 (no contact force).

Caution: N ≠ mg in general. Always derive N from the FBD by applying Newton's second law perpendicular to the surface. In a banking curve, on a rough incline with external forces, or in a lift — N can be very different from mg.

Pro tip: "Apparent weight" is what a weighing scale reads — it measures the normal force N, not the true gravitational force mg. In an upward-accelerating rocket, astronauts feel heavier; in free fall, they feel weightless.

6. Friction — Static, Kinetic, and Rolling

Friction is the force that opposes relative motion (or tendency of relative motion) between two surfaces in contact. It acts parallel to the contact surface.

Static Friction ( $f_s$ ): Acts when there is no relative motion. It is self-adjusting — it matches the applied force up to a maximum limit:

0 leq f_s leq mu_s N quadquad f_{s, ext{max}} = mu_s N

$mu_s$ = coefficient of static friction (dimensionless, depends only on the nature of surfaces)

Kinetic (Sliding) Friction ( $f_k$ ): Acts when there is relative sliding motion between surfaces:

f_k = mu_k N quadquad (mu_k < mu_s ext{ always})

Kinetic friction is approximately constant, independent of speed and area of contact.

Property	Static Friction	Kinetic Friction
Value	0 to $mu_s N$	$mu_k N$ (fixed)
Nature	Self-adjusting	Constant
When it acts	No relative motion	Relative sliding motion

Angle of friction (λ): The angle the resultant of N and $f_{s, ext{max}}$ makes with N:

anlambda = mu_s

Angle of repose (θ): The maximum angle of an incline at which an object remains just on the verge of sliding:

an heta = mu_s quadRightarrowquad heta = lambda ext{ (angle of repose = angle of friction)}

Rolling friction is much smaller than sliding friction, which is why wheels are used. Order: $f_{rolling} ll f_{kinetic} < f_{static, ext{max}}$ .

NEET tip: Friction always opposes relative motion or tendency of motion, but it can act in the forward direction for a system (e.g., friction on the driving wheels of a car acts forward, propelling the car). Never assume friction always opposes the direction of movement of the whole system.

7. Inclined Plane Problems

Inclined plane problems are extremely common in NEET. The key is to choose axes along and perpendicular to the incline, resolve all forces, then apply $Sigma F = ma$ .

Block sliding DOWN a rough incline (angle θ, coefficient μ):

N = mgcos heta

f_k = mu mgcos heta ext{ (up the incline)}

ma = mgsin heta - mu mgcos heta

oxed{a = g(sin heta - mucos heta)} quad ext{(condition: } an heta > mu ext{)}

Block pushed UP a rough incline: Both gravity component and friction act down the incline, so:

a = g(sin heta + mucos heta) quad ext{(deceleration, opposing upward motion)}

Block remains stationary on incline (limiting condition):

ext{Block just starts to slide when } an heta = mu_s quadRightarrowquad heta = ext{angle of repose}

Time to slide a distance L from rest:

L = rac{1}{2}at^2 quadRightarrowquad t = sqrt{ rac{2L}{a}} = sqrt{ rac{2L}{g(sin heta - mucos heta)}}

Force required to push a block up an incline at constant velocity (zero acceleration):

F = mgsin heta + mu mgcos heta = mg(sin heta + mucos heta)

Pro tip: On a frictionless incline (

mu = 0

), acceleration is simply

gsin heta

regardless of mass. On a smooth incline, all objects (light or heavy) slide with the same acceleration — mass cancels out, just like in free fall.

Caution: If

mucos heta > sin heta

(i.e.,

mu > an heta

), the block does not slide even without an applied force — static friction is sufficient. Always check whether sliding actually occurs before applying the kinetic friction formula.

8. Connected Bodies & Pulley (Atwood Machine)

When two or more bodies are connected by strings (assumed massless and inextensible) over pulleys (assumed massless and frictionless), they form a connected system. Key principle: all connected bodies have the same magnitude of acceleration (if the string is inextensible).

Atwood Machine — two masses $m_1$ and $m_2$ ( $m_1 > m_2$ ) connected over a frictionless pulley:

m_1

goes down,

m_2

goes up with same acceleration

a

oxed{a = rac{(m_1 - m_2)g}{m_1 + m_2}} qquad oxed{T = rac{2m_1 m_2 g}{m_1 + m_2}}

Two blocks on a surface connected by a string, force F applied on $m_1$ :

a = rac{F}{m_1 + m_2} qquad T = rac{m_2 cdot F}{m_1 + m_2}

The tension T in the connecting string pulls $m_2$ forward and retards $m_1$ backward — both accelerate at the same rate $a$ .

Three masses in a line ( $m_1$ — $m_2$ — $m_3$ ), force F on $m_1$ :

a = rac{F}{m_1+m_2+m_3}

T_1 = rac{(m_2+m_3)F}{m_1+m_2+m_3}, quad T_2 = rac{m_3 F}{m_1+m_2+m_3}

NEET tip: For the Atwood machine, if

m_1 = m_2

, then

a = 0

and

T = m_1 g = m_2 g

— just like static equilibrium. The tension in an Atwood machine is always between

m_2 g

and

m_1 g

(i.e.,

m_{min}g < T < m_{max}g

).

9. Pseudo Force & Non-Inertial Frames

When we observe motion from a non-inertial (accelerating) reference frame, Newton's second law appears to fail — objects accelerate without visible forces. We fix this by introducing a pseudo force (fictitious force) in the direction opposite to the frame's acceleration:

ec{F}_{pseudo} = -m ec{a}_{frame}

Once pseudo force is added, we can apply Newton's second law as usual within the non-inertial frame.

Pendulum in an accelerating car (car accelerates forward with acceleration $a$ ):

ext{Pseudo force} = ma ext{ (backward on pendulum bob)}

an heta = rac{a}{g} quadRightarrowquad heta = arctan!left( rac{a}{g} ight)

g_{eff} = sqrt{g^2 + a^2} quad ext{(effective gravitational acceleration)}

Block in an accelerating truck — a block on the floor of a truck accelerating at $a$ : In the truck's frame, pseudo force $ma$ acts backward on the block. For the block to remain stationary relative to the truck, static friction must provide the forward force $ma$ . Condition: $mu_s mg geq ma$ , i.e., $mu_s geq a/g$ .

Centrifugal force — in a rotating frame (non-inertial), a pseudo force directed radially outward appears. This is the "centrifugal force" felt by passengers on a merry-go-round or in a turning car:

F_{centrifugal} = momega^2 r = rac{mv^2}{r} quad ext{(outward, in rotating frame)}

In the inertial frame, only centripetal force (inward) exists — centrifugal force is purely a pseudo force of the rotating frame.

Pro tip: For NEET, prefer solving problems from the inertial (ground) frame whenever possible — it avoids confusion about pseudo forces. Use the non-inertial frame only when it simplifies the problem (e.g., finding equilibrium position of pendulum in accelerating vehicle).

10. NEET Exam Traps & Key Reminders

Laws of Motion is one of the highest-scoring chapters in NEET Physics. These are the most common traps and must-remember points:

Trap 1 — Action-Reaction forces never cancel: They act on different bodies. To cancel, forces must act on the same body. This is the most common conceptual error in NEET.

Trap 2 — Normal force ≠ mg always: N = mg only on a flat horizontal surface with no vertical acceleration and no other vertical forces. In all other cases, derive N from the FBD.

Trap 3 — Friction can act in the direction of motion: For a system of blocks where one pushes the other, friction on the pushed block acts forward (in the direction of motion of the system). Friction opposes relative motion, not necessarily overall motion.

Trap 4 — Same acceleration for connected bodies: In a system connected by an inextensible string, all parts have the same magnitude of acceleration. If the string goes over a pulley, the directions differ but the magnitudes are equal.

Trap 5 — Impulse-momentum is a vector equation: Always include sign (direction). A ball bouncing back with the same speed has impulse

= 2mv

(not zero), because momentum changes direction.

Trap 6 — Tension in a massless string is uniform throughout: Only valid for a massless string. If the string has mass, tension varies along its length.

Quick Formula Sheet:

Newton's 2nd Law	$ec{F}_{net} = m ec{a}$
Impulse	$J = FDelta t = Delta p$
Max static friction	$f_{s,max} = mu_s N$
Kinetic friction	$f_k = mu_k N$
Incline acceleration (down)	$a = g(sin heta - mucos heta)$
Atwood acceleration	$a = (m_1-m_2)g/(m_1+m_2)$
Apparent weight (lift up)	$N = m(g+a)$
Apparent weight (lift down)	$N = m(g-a)$
Angle of repose	$an heta = mu_s$

NEET strategy: Draw the FBD first — always. Identify all forces, choose smart axes, write equations, solve. For multi-body problems, write separate equations for each body, then solve simultaneously. Never skip the FBD step, even for "simple" problems.