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Newton's Laws of Motion & Friction — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Newton's Laws of Motion & Friction MCQs with step-by-step solutions covering First & Second Laws, Force & Equilibrium, Connected Bodies, Strings & Pulley Systems, Constrained Motion (Kinematic Linkages), Non-Inertial Frames & Pseudo Forces, Friction, Inclined Planes & Block-on-Block, Variable Mass Systems & Advanced Mechanics. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — First & Second Laws, Force & Equilibrium · hard · theory
Which statement about Newton's first and second laws is correct?
A. The first law is a special case of the second: if F_net = 0 then a = 0 ✓ Correct
B. The second law is a special case of the first law
C. The first law contradicts the second law
D. The two laws are completely independent
Solution: 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.
Q2 — First & Second Laws, Force & Equilibrium · hard · theory
An inertial frame of reference is one in which:
A. Newton's laws never hold
B. The frame is necessarily at absolute rest
C. Pseudo forces must always be added
D. A body with no net force on it has zero acceleration (Newton's laws hold) ✓ Correct
Solution: 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.
Q3 — First & Second Laws, Force & Equilibrium · hard · theory
Lami's theorem, used for three concurrent forces in equilibrium, states that each force is:
A. Proportional to the cosine of the angle between the other two
B. Equal to the sum of the other two
C. Proportional to the sine of the angle between the other two forces ✓ Correct
D. Inversely proportional to the angle between the other two
Solution: 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.
Q4 — First & Second Laws, Force & Equilibrium · hard · numerical
A time-varying force F = 6t (N) acts on a 3 kg body at rest. Its speed at t = 2 s is:
A. 2 m/s
B. 12 m/s
C. 8 m/s
D. 4 m/s ✓ Correct
Solution: 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.
Q5 — First & Second Laws, Force & Equilibrium · hard · numerical
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:
A. 100 N
B. 1 N
C. 0.1 N
D. 10 N ✓ Correct
Solution: 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).
Q6 — First & Second Laws, Force & Equilibrium · hard · numerical
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:
A. 10 N·s
B. 2.5 N·s
C. 7.5 N·s ✓ Correct
D. 5 N·s
Solution: 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.
Q7 — First & Second Laws, Force & Equilibrium · hard · numerical
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:
A. 60 N
B. 240 N
C. 120 N ✓ Correct
D. 20 N
Solution: a = v²/(2s) = 6²/(2 × 0.3) = 36/0.6 = 60 m/s²; F = ma = 2 × 60 = 120 N.
Q8 — Connected Bodies, Strings & Pulley Systems · hard · theory
In a string that has appreciable mass (a massful rope), the tension:
A. Is the same everywhere
B. Acts perpendicular to the string
C. Varies from point to point along the string ✓ Correct
D. Is zero at the point of support
Solution: For a heavy rope, tension at a section must support the weight of the rope below/behind it, so it varies along the length.
Q9 — Connected Bodies, Strings & Pulley Systems · hard · theory
When force F pushes block m₁ which is in contact with block m₂ (smooth floor), the contact (normal) force between them equals:
A. Zero
B. The weight of the second block
C. The full applied force F
D. The force needed to accelerate the second block: m₂·a ✓ Correct
Solution: The contact force is what accelerates the block that is NOT pushed directly: N = m₂·a = m₂F/(m₁+m₂). Pitfall: the contact force depends on which block is pushed.
Q10 — Connected Bodies, Strings & Pulley Systems · hard · theory
The tension in the string of an ideal Atwood machine (masses m₁ and m₂) is:
A. $\dfrac{(m_1 - m_2)g}{2}$
B. $\dfrac{m_1 m_2 g}{m_1 + m_2}$
C. $\dfrac{2 m_1 m_2 g}{m_1 + m_2}$ ✓ Correct
D. $(m_1 + m_2)g$
Solution: Substituting a into either mass's equation gives T = 2m₁m₂g/(m₁ + m₂) (the harmonic-mean-like result).
Q11 — Connected Bodies, Strings & Pulley Systems · hard · theory
A uniform rope hangs from a ceiling. The tension in the rope:
A. Is greatest at the top and least at the bottom ✓ Correct
B. Is zero at the top
C. Is uniform throughout
D. Is greatest at the bottom
Solution: At any section the tension supports the weight of the rope below it, so it increases from bottom (≈0) to top (= full weight).
Q12 — Connected Bodies, Strings & Pulley Systems · hard · theory
A block on a smooth incline is connected over a pulley to a hanging block. The system accelerates in the direction of the hanging block only if:
A. The hanging block is lighter
B. The two masses are equal
C. The weight of the hanging block exceeds the incline component m₁g sinθ ✓ Correct
D. The incline is frictionless only
Solution: The net driving force is m₂g − m₁g sinθ; the hanging mass descends when m₂g > m₁g sinθ.
Q13 — Connected Bodies, Strings & Pulley Systems · hard · numerical
The same 2 kg and 3 kg blocks are in contact on a smooth floor, but now the 10 N force pushes the 3 kg block. The contact force between them is:
A. 10 N
B. 6 N
C. 5 N
D. 4 N ✓ Correct
Solution: a = 2 m/s² still, but now the contact force accelerates the 2 kg block: N = 2 × 2 = 4 N. Pitfall: the contact force changed even though a did not.
Q14 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 4 kg and a 6 kg block are in contact on a smooth floor. A horizontal force of 20 N pushes the 4 kg block. The contact force between the blocks is:
A. 20 N
B. 6 N
C. 12 N ✓ Correct
D. 8 N
Solution: a = 20/10 = 2 m/s². Contact force = 6 × 2 = 12 N.
Q15 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 2 kg block on a smooth incline of 30° is connected over an ideal pulley to a 3 kg hanging block (g = 10 m/s²). The acceleration of the system is:
A. 2 m/s²
B. 4 m/s² ✓ Correct
C. 6 m/s²
D. 10 m/s²
Solution: a = (m₂g − m₁g sinθ)/(m₁ + m₂) = (30 − 2×10×0.5)/5 = (30 − 10)/5 = 4 m/s².
Q16 — Connected Bodies, Strings & Pulley Systems · hard · numerical
For the previous system (2 kg on 30° smooth incline, 3 kg hanging, a = 4 m/s²), the string tension is:
A. 18 N ✓ Correct
B. 24 N
C. 30 N
D. 12 N
Solution: For the hanging block: m₂g − T = m₂a ⇒ T = 3(10 − 4) = 18 N. (Check on incline: T − m₁g sinθ = 18 − 10 = 8 = 2×4 ✓.)
Q17 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A uniform rope of mass 2 kg lies on a smooth floor and is pulled by a horizontal force of 12 N at one end. The tension at the midpoint of the rope is:
A. 9 N
B. 12 N
C. 6 N ✓ Correct
D. 3 N
Solution: a = F/m = 12/2 = 6 m/s². The tension at the midpoint accelerates the half of the rope beyond it (1 kg): T = 1 × 6 = 6 N.
Q18 — Connected Bodies, Strings & Pulley Systems · hard · numerical
Three blocks of 1 kg, 2 kg and 3 kg hang vertically from a ceiling, joined by strings (1 kg at top). The tension in the topmost string (g = 10 m/s²) is:
A. 30 N
B. 60 N ✓ Correct
C. 20 N
D. 10 N
Solution: The top string supports all three blocks: T = (1 + 2 + 3)g = 6 × 10 = 60 N.
Q19 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A monkey of mass 10 kg climbs up a rope with an upward acceleration of 2 m/s² (g = 10 m/s²). The tension in the rope is:
A. 20 N
B. 80 N
C. 100 N
D. 120 N ✓ Correct
Solution: T − mg = ma ⇒ T = m(g + a) = 10(10 + 2) = 120 N. Pitfall: the rope must exceed the weight to accelerate the monkey upward.
Q20 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 4 kg block on a smooth table is connected over a pulley at the edge to a 6 kg hanging block (g = 10 m/s²). The acceleration of the system is:
A. 2 m/s²
B. 10 m/s²
C. 4 m/s²
D. 6 m/s² ✓ Correct
Solution: a = m₂g/(m₁ + m₂) = 60/10 = 6 m/s².
Q21 — Connected Bodies, Strings & Pulley Systems · hard · numerical
For a 3 kg block on a smooth table connected over a pulley to a 2 kg hanging block (g = 10 m/s²), the string tension is:
A. 20 N
B. 6 N
C. 15 N
D. 12 N ✓ Correct
Solution: a = 2×10/5 = 4 m/s². Tension = m(table)·a = 3 × 4 = 12 N.
Q22 — Connected Bodies, Strings & Pulley Systems · hard · numerical
A 10 kg block is joined below a 20 kg block by a string; the pair is pulled straight up by a force of 450 N applied to the top (20 kg) block (g = 10 m/s²). The tension in the connecting string is:
A. 450 N
B. 150 N ✓ Correct
C. 100 N
D. 300 N
Solution: a = (F − (m₁+m₂)g)/(m₁+m₂) = (450 − 300)/30 = 5 m/s². The connecting string carries the lower 10 kg block: T = m(g + a) = 10(10 + 5) = 150 N.
Q23 — Constrained Motion (Kinematic Linkages) · hard · theory
The virtual-work (constraint) method uses the fact that the total work done by the internal string tensions in an ideal system is:
A. Always positive
B. Equal to the change in kinetic energy
C. Zero ✓ Correct
D. Equal to the applied force times distance
Solution: For ideal (massless, inextensible) strings the net work of the tension pairs is zero: Σ T·v = 0, giving the acceleration/velocity constraint relations.
Q24 — Constrained Motion (Kinematic Linkages) · hard · theory
A load hangs from a single movable pulley and the free end of the string is pulled upward. The load moves:
A. At twice the speed of the free end
B. At half the speed of the free end ✓ Correct
C. At the same speed as the free end
D. In the opposite direction to the free end
Solution: A movable pulley is supported by two string segments, so the load moves half as fast (and half as far) as the free end. Pitfall: correspondingly the effort force is half the load.
Q25 — Constrained Motion (Kinematic Linkages) · hard · theory
A block slides on the incline of a wedge that is itself free to move on the floor. The constraint is that:
A. The block's absolute acceleration is along the incline
B. The block cannot move relative to the wedge
C. The block's acceleration relative to the wedge is directed along the incline surface ✓ Correct
D. The wedge must remain stationary
Solution: The block stays in contact with the wedge, so its acceleration relative to the wedge lies along the incline; its absolute acceleration is the vector sum of that and the wedge's acceleration.
Q26 — Constrained Motion (Kinematic Linkages) · hard · theory
For a rigid rod, the constraint linking the velocities of its two ends A and B is:
A. The rod length changes with time
B. The components of v_A and v_B along the rod are equal ✓ Correct
C. v_A = v_B in magnitude and direction
D. v_A and v_B are perpendicular
Solution: A rigid rod cannot stretch, so the velocity components of its two ends along the rod must be equal.
Q27 — Constrained Motion (Kinematic Linkages) · hard · theory
An ideal movable-pulley arrangement in which the load is supported by n string segments gives a mechanical advantage (ideal) of:
A. 1
B. 1/n
C. n²
D. n ✓ Correct
Solution: With n supporting segments, the effort = load/n, so the ideal mechanical advantage is n (velocity ratio is also n).
Q28 — Constrained Motion (Kinematic Linkages) · hard · numerical
A ladder leans against a wall. Its foot (end A) slides along the floor at 5 m/s. When the ladder makes 37° with the floor (sin37° = 0.6, cos37° = 0.8), the top end B slides down at:
A. 5 m/s
B. ≈ 4 m/s
C. ≈ 3.75 m/s
D. ≈ 6.67 m/s ✓ Correct
Solution: Equal velocity components along the rod: v_A cosθ = v_B sinθ ⇒ v_B = v_A cotθ = 5 × (0.8/0.6) ≈ 6.67 m/s.
Q29 — Constrained Motion (Kinematic Linkages) · hard · numerical
A ladder's foot slides at 4 m/s along the floor. When the ladder makes 53° with the floor (cot53° = 0.75), the speed of its top end (on the wall) is:
A. 6 m/s
B. 3 m/s ✓ Correct
C. 4 m/s
D. 5.33 m/s
Solution: v_B = v_A cotθ = 4 × 0.75 = 3 m/s.
Q30 — Constrained Motion (Kinematic Linkages) · hard · numerical
A block on the floor is pulled by a rope passing over an overhead pulley; the rope is drawn in at 5 m/s and makes 60° with the horizontal at the block. The speed of the block is:
A. 10 m/s ✓ Correct
B. 5 m/s
C. ≈ 8.7 m/s
D. 2.5 m/s
Solution: Rate of shortening of the rope = component of block velocity along the rope: 5 = v cos60° ⇒ v = 5/0.5 = 10 m/s. Pitfall: the block moves faster than the rope is pulled.