NEET Laws of Motion Previous Year Questions with Solutions

Tyre wear and tear occurs due to sideways frictional forces between the tyres and the road surface:

• • When a vehicle travels at the design speed on a banked road, no lateral friction is demanded.
• • Centripetal acceleration is supplied entirely by the normal force from the inclined road surface.

Resolve the normal force N into horizontal and vertical components:

• • N*sin(theta) acts toward the center of the circular track.
• • N*cos(theta) balances the vehicle weight.

Divide the horizontal balance by the vertical balance:

• • Matches Option (1) exactly.

On a frictionless incline of angle θ, the acceleration is simply the component of gravity along the incline. On a rough incline with coefficient of kinetic friction μ_k, kinetic friction opposes motion.

• • Normal reaction N = mg cos θ; limiting kinetic friction f_k = μ_k mg cos θ.

Starting from rest over length L, the distance formula gives L = (1/2) a t², meaning t ∝ 1/√a. Taking the ratio of times:

• • Given n = 2, so n² = 4.

Substitute the acceleration expressions into the ratio and solve for μ_k:

• • Matches Option (4) exactly.

Using kinematic equation v² = u² + 2gh for free fall under gravity:

• • After collision, the ball rises to height h₂ = 10 m:
• • v₂ = √(2 g h₂) = √(2 × 9.8 × 10) = √196 = 14 m/s (upward: +14ĵ).

Impulse imparted by the ground equals the vector change in linear momentum of the ball:

• • Direction of impulse is vertically upward, opposing downward momentum.
• • Magnitude is exactly 21 N·s.

For a mass m whirled in a horizontal plane of radius r on a frictionless surface, the tension T in the string provides the necessary centripetal force directed toward the center of the circle:

When the speed is increased to ω_new = 2ω while keeping the radius r unchanged, the new tension T_new becomes:

Expanding the squared term:

Since both blocks slide together on a frictionless surface, they share a common horizontal acceleration a. Treating both blocks as a single system of total mass M = m_A + m_B:

Block B of mass m_B = 3 kg experiences only one horizontal force: the normal push force N exerted by block A on block B acting to the right. Applying Newton’s second law:

Substitute m_B = 3 kg and a = 2 m/s²:

For block A: F - N = m_A · a => 10 - 6 = 2 kg × 2 m/s² = 4 N, which confirms perfect dynamic equilibrium.

By Newton's second law of motion, the net external force F⃗ acting on the player is directly proportional to the rate of change of linear momentum, and therefore points in the direction of the change in velocity vector Δv⃗:

Let the positive x-axis represent East (+î), positive y-axis represent North (+ĵ), and negative y-axis represent South (-ĵ). With initial speed v:

Subtract the initial velocity vector from the final velocity vector:

The vector v î + v ĵ has equal positive components along East and North, forming an angle of 45° between East and North:

Consider the reference frame of the accelerating car (acceleration a). In this accelerated frame, a pseudo-force F_pseudo = m a acts on the body in the direction opposite to the car’s acceleration. To prevent the body from slipping backward, static friction f_s acts forward:

In the vertical direction, the body is in static equilibrium under gravity and the normal reaction from the car floor:

The maximum static friction that the floor can exert on the body is given by Amontons-Coulomb’s law of friction:

Equating the pseudo force to the limiting static friction to determine a_max:

For a body falling freely from rest from height $h = 10\text{ m}$ under gravity $g = 10\text{ m/s}^2$, taking upward as positive:

The ball rebounds back to the same height $h = 10\text{ m}$. To reach height $h$, its initial upward velocity must be:

Impulse imparted by the ground to the ball is defined as the change in its linear momentum vector:

Substitute $m = 0.15\text{ kg}$ and $\sqrt{2} \approx 1.414$:

Two bodies of masses $$m_1 = 4\text{ kg}$$ and $$m_2 = 6\text{ kg}$$ are connected by a taut, inextensible, massless string passing over a frictionless fixed pulley. Both masses accelerate with equal magnitude $$a$$.

For mass $$m_2 = 6\text{ kg}$$ descending downwards: $$m_2 g - T = m_2 a$$ For mass $$m_1 = 4\text{ kg}$$ ascending upwards: $$T - m_1 g = m_1 a$$.

Add the two equations: $$(m_2 g - T) + (T - m_1 g) = (m_1 + m_2) a$$ $$(m_2 - m_1) g = (m_1 + m_2) a$$ $$a = \frac{m_2 - m_1}{m_1 + m_2} g$$.

Substitute $$m_1 = 4\text{ kg}$$ and $$m_2 = 6\text{ kg}$$: $$a = \frac{6 - 4}{6 + 4} g = \frac{2}{10} g = \frac{g}{5}$$. Matches Option 3.