Laws of Motion And Units And Measurements For NEET by PW

Solving Laws of Motion and Units and Measurements questions for NEET can be challenging when you are unsure which forces to include, how to resolve tension and friction, or which equations to apply. Pulley constraints, circular motion, accelerating reference frames and measurement errors require you to understand the physical conditions before substituting values into formulas. 

PW supports NEET preparation with live and recorded classes, study notes, Daily Practice Problems (DPPs) with video solutions, regular tests and a 24/7 doubt engine. These resources help you revise Newton’s laws, practise numerical questions, strengthen dimensional analysis and identify concepts that need more revision. Combining concept learning with regular MCQ practice can help you approach mechanics problems more systematically.

Force From Position-Dependent Velocity

When velocity depends on position coordinates, acceleration must be calculated by differentiating velocity with respect to time. Consider a particle of mass m moving in the xy-plane with velocity:

v = alpha y i + 2 alpha x j

 Acceleration is always calculated with respect to time: a = dv/dt

Since the velocity depends on x and y:

dx/dt = vx

dy/dt = vy

 From the given velocity:

vx = alpha y

vy = 2 alpha x

 Therefore:

a = alpha vy i + 2 alpha vx j

Substituting the velocity components: a = 2 alpha² x i + 2 alpha² y j

Using Newton’s second law: F = ma

Hence: F = 2m alpha² (x i + y j)

The relation a = v dv/dx cannot be directly used because the displacement depends on both x and y.

Tension In A String Lifted At Its Midpoint

Two equal masses are connected by a string of length 2a on a frictionless horizontal surface. The midpoint is pulled vertically upward by force f. When the separation between the masses is 2x, each half of the string forms a right triangle with:

• Hypotenuse: a

• Base: x

• Perpendicular side: square root of (a² - x²)

 For each mass, the horizontal component of tension provides acceleration:

T cos theta = m acceleration

 At the massless midpoint, vertical components balance the applied force: f = 2T sin theta

Therefore:

T = f / (2 sin theta)

Combining the equations: acceleration = (f / 2m) cot theta

 From geometry: cot theta = x / square root of (a² - x²)

Thus, the acceleration of each mass is: acceleration = fx / [2m square root of (a² - x²)]

It is directed toward the other mass.

Dimensional analysis alone cannot identify the correct expression when several options have the same dimensions. Force analysis and geometry are necessary.

Inertial And Non-Inertial Reference Frames

A frame is inertial when it has no acceleration. A frame is non-inertial when it accelerates or rotates.

The Earth is treated as approximately inertial in many elementary problems. Strictly, it is non-inertial because:

1. It revolves around the Sun.

2. It rotates about its own axis.

Newton’s laws are fundamentally valid in an inertial frame. A body is in equilibrium only when:

Net force = 0

and therefore: Acceleration = 0

In the midpoint problem, the massless midpoint is assumed to be in equilibrium, while the two masses accelerate and are not in equilibrium.

Conical Pendulum And Maximum Angular Velocity

In a conical pendulum, a ball moves in a horizontal circle while the string makes a constant angle with the vertical.

The tension components are:

• Vertical component: T cos theta

• Horizontal component: T sin theta

The horizontal component provides centripetal force:

T sin theta = m omega² r

Using the geometry: r = l sin theta

Therefore: T sin theta = m omega² l sin theta

After cancellation: T = m omega² l

Hence: omega maximum = square root of [T maximum / (m l)]

For the given values, the reported result is: omega maximum = 36 rad/s

Friction And Its Direction

Friction opposes relative motion or the tendency of relative motion between contacting surfaces. Its direction must be determined from the free-body diagram, not simply from the direction in which the object moves.

For an insect slowly climbing a hemispherical surface, the tendency to slip is downward. Hence, friction acts upward along the tangent.

At the limiting condition: f = mu N

The radial balance gives: N = mg cos alpha

The tangential balance gives: mg sin alpha = mu mg cos alpha

Therefore: tan alpha = mu

For mu = 1/3:

tan alpha = 1/3

This problem mainly tests the correct free-body diagram and the direction of limiting friction.

Friction On A Banked Road

On a banked road, friction changes direction depending on whether the vehicle tends to slide up or down the slope. This direction must be identified before resolving the forces.

The vertical force balance is:

N cos theta = mg + f sin theta

At limiting friction:

f = mu N

Thus:

N = mg / (cos theta - mu sin theta)

For minimum speed, friction acts up the slope because the vehicle tends to slide downward.

Negative Lift On A Racing Car

Negative lift is a downward aerodynamic force that presses a racing car against the track. It increases the normal reaction and therefore the maximum friction available to provide centripetal force.

For a car moving on a circular track, friction provides centripetal force: f = mv²/r

A downward aerodynamic force called negative lift increases the normal reaction: N = mg + FL

At limiting friction: mu N = mv²/r

Therefore: FL = mv²/(mu r) - mg

Negative lift presses the car against the track, increasing the maximum available friction.

Pulley Constraints And Different Accelerations

In a pulley system, the accelerations of connected blocks are not always equal. The string-length constraint must be written before applying Newton’s laws.

For a system containing a movable pulley, if one block has acceleration a, another block may have acceleration a/2.

For a 2 kg and 8 kg system, the equations are:

T - 20 = 2a

80 - 2T = 4a

Solving: a = 5 m/s²

Therefore, the acceleration of the 8 kg block is: a8 = 2.5 m/s²

If it falls through 0.2 m from rest:

s = 1/2 at²

t = square root of (2s/a)

Thus:

t = 0.4 s

Circular Motion And Radius Of Curvature

At a curved portion of a track, the centripetal force depends on the radius of curvature:

centripetal force = mv²/r

A smaller radius of curvature requires a greater centripetal force. Therefore, the normal reaction is generally larger for a sharper curve.

For a block moving against a smooth circular groove: N = mv²/r

Hence: N is proportional to v²

The graph of N against v is a parabola passing through the origin, not a straight line.

Dimensions Of The Poynting Vector

The Poynting vector represents electromagnetic energy flow per unit area per unit time: S = energy / (time × area)

Dimensions of energy are: [ML²T⁻²]

Therefore: [S] = [MT⁻³]

The same dimensions are obtained for power per unit area and force per unit length per unit time.

Error Propagation In Current Measurement

If current is a function of voltage, a small voltage error produces a current error according to:

dI = (dI/dV) dV

For: I = exponential of (1000V/T)

the derivative is: dI/dV = exponential of (1000V/T) × 1000/T

Given: exponential of (1000V/T) = 6

T = 300 K

dV = 0.01 V

Then: dI = 6 × (1000/300) × 0.01

Therefore: dI = 0.2 mA

Laws of Motion and Units and Measurements for NEET require a clear understanding of force, friction, tension, pulley constraints, circular motion, dimensional analysis and error propagation. PW supports your preparation through live and recorded classes, study notes, DPPs with video solutions, regular tests and a 24/7 doubt engine to help you revise concepts and identify areas that need more practice.

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