Laws of Motion MCQs

A body at rest or moving uniformly in a straight line continues in that state unless acted on by a net external force.

This statement describes the resistance of any object to a change in its state of motion.

That resistance is defined as inertia, so Newton’s first law is called the law of inertia. Inertia.

Normal reaction = weight = 50 N (horizontal floor).

Kinetic friction = μ × N = 0.2 × 50 N = 10 N.

Thus the frictional force opposing the motion is 10 N.

Change in momentum Δp = final − initial = 15 kg·m/s − 5 kg·m/s = 10 kg·m/s.

Average force F = Δp / Δt = 10 kg·m/s ÷ 2 s = 5 N.

Thus the average force is 5 N.

Force = mass × acceleration (F = ma).

Mass has unit kg and acceleration has unit m s⁻², so multiply: kg·(m s⁻²) = kg m s⁻².

Thus 1 newton = 1 kg m s⁻², which matches the option kg m/s².

Net force = 10 N acts on the system of mass 2 kg + 3 kg = 5 kg, so acceleration a = 10/5 = 2 m s⁻².

The tension T pulls the 3 kg block, so for that block: T = m·a = 3 kg × 2 m s⁻² = 6 N.

Thus the tension in the string is 6 N.

Maximum static friction opposes the onset of motion and is proportional to the normal reaction.

The proportionality constant is the coefficient of static friction μ_s.

Thus f_s(max)=μ_s N, where N equals the weight component perpendicular to the surface.

Hence the correct answer is μ_s N.

Downward acceleration a = g makes the net acceleration of the body equal to g downward.

In the non‑inertial frame of the lift, the pseudo force upward is m a = mg, which cancels the real weight mg.

Hence the normal reaction (apparent weight) N = mg – m a = mg – mg = 0.

Momentum = mass × velocity.

Mass is a scalar but velocity is a vector, so their product retains the direction of velocity.

Hence momentum has both magnitude and direction, making it a vector quantity. Vector quantity

Net force on the system = 10 N; total mass = 2 kg + 3 kg = 5 kg.

Acceleration a = F/m = 10 N / 5 kg = 2 m s⁻².

Thus the two blocks accelerate together at 2 m/s².

2 m/s².

Impulse = change in momentum = mass × velocity.

Mass has dimension M and velocity has dimension L T⁻¹, so momentum (and impulse) has dimension M L T⁻¹.

Thus the dimensional formula of impulse is [M L T⁻¹].

Newton’s second law states that the vector sum of external forces on a body equals the time derivative of its linear momentum.

Momentum \(p\) is defined as \(p = mv\); differentiating gives \(\frac{dp}{dt}=m\frac{dv}{dt}\) for constant mass.

Thus the net force is the rate of change of momentum, not of acceleration, displacement or the product \(mv\).

Centripetal acceleration a = v²/r points radially inward for uniform circular motion.

Net force F = m a therefore has the same direction as the acceleration.

Hence the required net (centripetal) force is directed towards the centre. Towards the centre

When a ball is caught, the change in its momentum (impulse) must equal the impulse delivered by the hand.

Impulse = Force × contact time, so for a given impulse a larger contact time allows a smaller average force.

By pulling the hands back, the cricketer extends the duration over which the ball’s momentum is brought to zero.

Thus the time of contact increases and the force on the hand is reduced, matching the correct answer.

Static friction adjusts to balance any applied horizontal force up to its maximum value (limiting friction).

Since the block remains at rest, the frictional force must exactly cancel the applied force.

Thus the static friction equals the applied force, not the weight or zero, and is less than the limiting friction. The applied force.

Component of weight perpendicular to the plane is W⊥ = mg cosθ.

Normal reaction balances this perpendicular component because there is no acceleration through the surface.

Thus N = mg cosθ, which matches the given answer.

Using Newton’s second law, F = m a.

Substitute the given values: 20 N = 4 kg × a.

Solve for a: a = 20 N / 4 kg = 5 m/s².

Hence the acceleration is 5 m/s².

For a banked curve the normal reaction N provides the centripetal force: horizontal component N sinθ = mv²/r.

The vertical component balances weight: N cosθ = mg.

Dividing the two equations eliminates N: (N sinθ)/(N cosθ) = (mv²/r)/(mg) → tanθ = v²/(r g).

Thus the correct relation is tan θ = v²/(r g).

Initial momentum = (3 kg)(4 m/s) + (1 kg)(0) = 12 kg·m/s.

After sticking together, total mass = 3 kg + 1 kg = 4 kg, and momentum is conserved.

Common velocity v = total momentum / total mass = 12 / 4 = 3 m/s.

Hence the correct answer is 3 m/s.

Weight component down the plane = mg sinθ.

Maximum static friction = μs N = μs mg cosθ.

For impending motion mg sinθ = μs mg cosθ → μs = tanθ.

Hence the coefficient of static friction equals tan θ.

Resolve the weight mg into components parallel and perpendicular to the incline.

The component parallel to the surface is mg sin θ because sin θ = opposite/hypotenuse = (parallel component)/mg.

Thus the force driving the mass down the frictionless plane is mg sin θ.

Answer: mg sin theta