Question 1

Newton's first law of motion is essentially a statement about:

Accepted Answer

Inertia and the existence of inertial frames of reference. The first law (law of inertia) asserts that a body keeps its state of rest or uniform motion unless a net external force acts; it also defines inertial frames — those in which this holds.

Question 2

The most general form of Newton's second law, valid even when the mass changes, is:

Accepted Answer

\vec{F} = \dfrac{d\vec{p}}{dt}. F = dp/dt is the general form. F = ma is a special case valid only when mass is constant; for variable-mass systems the full derivative of momentum must be used.

Question 3

The impulse–momentum theorem states that:

Accepted Answer

The impulse of a force equals the change in momentum it produces. Impulse J = ∫F dt = Δp. Its SI unit N·s is identical to kg·m/s. Pitfall: impulse relates to momentum change, not energy change.

Question 4

Which statement about Newton's first and second laws is correct?

Accepted Answer

The first law is a special case of the second: if F_net = 0 then a = 0. Setting F_net = 0 in F = ma gives a = 0, i.e. constant velocity — exactly the first law. So the first law is contained within the second.

Question 5

For a set of concurrent coplanar forces to keep a particle in equilibrium, it is necessary that:

Accepted Answer

Their vector sum is zero. Static equilibrium of a particle requires ΣF = 0 (vector). For three forces this is equivalent to them forming a closed triangle (triangle law).

Question 6

An inertial frame of reference is one in which:

Accepted Answer

A body with no net force on it has zero acceleration (Newton's laws hold). In an inertial frame, a free particle moves with constant velocity. Any frame moving at constant velocity relative to an inertial frame is also inertial; accelerating frames are non-inertial.

Question 7

While drawing the free-body diagram (FBD) of a chosen body, one includes:

Accepted Answer

Only the external forces acting on that body. An FBD shows only the external forces on the isolated body; internal forces occur in action–reaction pairs and cancel for the system.

Question 8

Lami's theorem, used for three concurrent forces in equilibrium, states that each force is:

Accepted Answer

Proportional to the sine of the angle between the other two forces. For three concurrent forces P, Q, R in equilibrium: P/sinα = Q/sinβ = R/sinγ, where α, β, γ are the angles opposite to each force.

Question 9

A net force of 20 N acts on a body of mass 4 kg. The acceleration produced is:

Accepted Answer

5 m/s². a = F/m = 20/4 = 5 m/s².

Question 10

A ball of mass 0.2 kg strikes a wall normally at 10 m/s and rebounds with the same speed. The magnitude of the impulse imparted to the ball is:

Accepted Answer

4 N·s. Δp = m(v_f − v_i) = 0.2(−10 − 10) = −4 kg·m/s ⇒ |impulse| = 4 N·s. Pitfall: the speed reverses direction, so you add the magnitudes, not subtract.

Question 11

A constant force of 5 N acts on a 2 kg body initially at rest for 4 s. Its final speed is:

Accepted Answer

10 m/s. a = F/m = 2.5 m/s²; v = at = 2.5 × 4 = 10 m/s. (Equivalently, impulse = 5 × 4 = 20 N·s = mv ⇒ v = 10 m/s.)

Question 12

A time-varying force F = 6t (N) acts on a 3 kg body at rest. Its speed at t = 2 s is:

Accepted Answer

4 m/s. Impulse = ∫₀² 6t dt = [3t²]₀² = 12 N·s = mv ⇒ v = 12/3 = 4 m/s. Pitfall: use the area under F–t (integral), not F×t with a single F.

Question 13

Two perpendicular forces of 3 N and 4 N act on a 1 kg body. Its acceleration is:

Accepted Answer

5 m/s². Resultant force = √(3² + 4²) = 5 N; a = 5/1 = 5 m/s².

Question 14

A particle is in equilibrium under three forces. Two of them are F₁ = (3î + 4ĵ) N and F₂ = (−î + 2ĵ) N. The magnitude of the third force is:

Accepted Answer

≈ 6.32 N. For equilibrium F₃ = −(F₁ + F₂) = −(2î + 6ĵ) N ⇒ |F₃| = √(2² + 6²) = √40 ≈ 6.32 N.

Question 15

A 1000 kg car moving at 20 m/s is brought to rest in 5 s. The average braking force is:

Accepted Answer

4000 N. F = mΔv/t = 1000 × 20/5 = 4000 N.

Question 16

A force F = (6î + 8ĵ) N acts on a body of mass 2 kg. The magnitude of its acceleration is:

Accepted Answer

5 m/s². |F| = √(6² + 8²) = 10 N; a = |F|/m = 10/2 = 5 m/s².

Question 17

A horizontal water jet of cross-sectional area 10⁻⁴ m² strikes a wall normally at 10 m/s and stops (ρ_water = 1000 kg/m³). The force on the wall is:

Accepted Answer

10 N. Force = (dm/dt)·v = (ρAv)·v = ρAv² = 1000 × 10⁻⁴ × 10² = 10 N. Pitfall: the momentum flux gives v², so the force ∝ v² (not v).

Question 18

A gun of mass 4 kg fires a bullet of mass 20 g with a muzzle speed of 400 m/s. The recoil speed of the gun is:

Accepted Answer

2 m/s. By momentum conservation: m_g v_g = m_b v_b ⇒ v_g = (0.02 × 400)/4 = 2 m/s.

Question 19

A ball of mass 0.5 kg is dropped from a height of 5 m and rebounds to a height of 1.25 m (g = 10 m/s²). The impulse imparted to the ball by the floor is:

Accepted Answer

7.5 N·s. v_down = √(2×10×5) = 10 m/s; v_up = √(2×10×1.25) = 5 m/s. Taking up positive: impulse = m(v_up − (−v_down)) = 0.5(5 + 10) = 7.5 N·s.

Question 20

A 2 kg body moving at 6 m/s is brought to rest over a distance of 0.3 m by a constant force. The magnitude of that force is:

Accepted Answer

120 N. a = v²/(2s) = 6²/(2 × 0.3) = 36/0.6 = 60 m/s²; F = ma = 2 × 60 = 120 N.

First & Second Laws, Force & Equilibrium — JEE Main Physics MCQs with Solutions

Questions with solutions