§ 4 ✓NCERT Class 11 · Physics · Chapter 4

Chapter test: Laws of Motion

30 questions · Solve each question on paper first, then open "Show answer" to check your working.

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026

Q1.
Find the net force needed to accelerate a 5 kg body at 3 m s⁻², in newtons.
Show answer
15 N — F = ma = 5 × 3
Q2.
A 0.5 kg ball moving at 10 m/s is caught and stopped. Find the magnitude of the impulse delivered to it, in N·s.
Show answer
5 N·s — J = Δp = 0.5 × (10 − 0) = 5 kg m/s = 5 N·s
Q3.
A 20 g bullet is fired horizontally at 400 m/s from a 5 kg gun free to recoil. Find the recoil speed in m/s.
Show answer
1.6 m/s — 0 = 0.02 × 400 + 5V → V = −8/5 (speed 1.6, opposite the bullet)
Q4.
A 60 kg passenger stands on a scale in a lift accelerating upward at 2 m s⁻². Find the scale reading in newtons.
Show answer
720 N — N = m(g + a) = 60 × 12 (accelerating up → heavier)
Q5.
A 10 kg block on a floor (μ_s = 0.4, μ_k = 0.3) is pulled horizontally with 50 N. Find its acceleration in m s⁻².
Show answer
2 m s⁻² — 50 N > μ_smg = 40 → slides; a = (50 − μ_kmg)/m = (50 − 30)/10
Q6.
Find the maximum speed (m/s) with which a vehicle can round a level curve of radius 80 m if μ_s = 0.5.
Show answer
20 m/s — v_max = √(μ_srg) = √(0.5 × 80 × 10) = √400
Q7.
A 24 N force pushes a 2 kg block pressed against a 4 kg block on a frictionless floor. Find the contact force between the blocks in newtons.
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16 N — a = 24/6 = 4; contact = mass ahead (being pushed) × a = 4 × 4
Q8.
A frictionless road is banked for a vehicle moving at 10 m/s on a curve of radius 10 m. Find the banking angle in degrees.
Show answer
45° — tan θ = v²/rg = 100/100 = 1
Q9.
Action and reaction forces never cancel each other because they: (a) have unequal magnitudes (b) act on different bodies (c) are not simultaneous (d) act in the same direction
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(b) — Different bodies → they cannot enter one ΣF equation
Q10.
The inertia of a body is measured by its: (a) velocity (b) acceleration (c) mass (d) momentum
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(c) — Mass is the measure of inertia (first law)
Q11.
A book rests on a table. The normal force N on the book and its weight mg: (a) form an action–reaction pair (b) balance each other but are not a third-law pair (c) are both contact forces (d) would remain equal even in a freely falling lift
Show answer
(b) — N and mg act on the same body and are different force types — balance, not pair
Q12.
A 10 N horizontal push acts on a block whose limiting static friction is 15 N. The friction force on the block is: (a) 15 N (b) 10 N (c) 5 N (d) zero
Show answer
(b) — 10 < 15 → no sliding; static friction self-adjusts to 10 N
Q13.
Two bodies couple and move together after colliding on a frictionless track. Which quantity is definitely conserved? (a) momentum only (b) kinetic energy only (c) both (d) neither
Show answer
(a) — Coupling is inelastic: momentum survives, KE converts to heat/deformation
Q14.
A man stands on a weighing scale inside a lift whose cable has snapped (free fall). The scale reads: (a) mg (b) more than mg (c) less than mg but not zero (d) zero
Show answer
(d) — N = m(g − g) = 0; weightless but gravity still acts
Q15.
A car rounds a level circular track at constant speed. The "centripetal force" on the car is supplied by: (a) the engine's drive (b) friction on the tyres from the road (c) the normal reaction (d) a special centripetal force of nature
Show answer
(b) — Level road: friction is the only inward force; "centripetal" is the job, not a force
Q16.
A block remains at rest on a rough incline whose angle is below the angle of repose. The friction on it equals: (a) μ_smg cos θ (b) mg sin θ (c) zero (d) μ_kmg cos θ
Show answer
(b) — Equilibrium along incline: f_s = mg sin θ (self-adjusted below the ceiling)
Q17.
Assertion: Passengers standing in a bus jerk forward when the driver brakes sharply. Reason: The upper part of the body continues in its state of motion due to inertia while the feet stop with the bus.
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(a) — Classic inertia of motion; R is the mechanism behind A
Q18.
Assertion: A satellite in a circular orbit moves with constant speed. Reason: The gravitational force supplying the centripetal resultant is always perpendicular to the satellite's velocity and does no work.
Show answer
(a) — ⊥ force ⇒ W = 0 ⇒ KE, hence speed, constant — R explains A
Q19.
Statement I: The total momentum of an isolated system of interacting bodies remains constant, whatever interactions occur inside. Statement II: Internal forces occur in action–reaction pairs that cancel when the system's net force is summed.
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(a) — II is precisely the derivation step that proves I
Q20.
Statement I: Kinetic friction on a sliding block is approximately independent of its speed. Statement II: Static friction on a resting block always equals μ_sN, where N is the normal reaction.
Show answer
(d) — I true (kinetic regime is flat); II false — static friction is ≤ μ_sN, self-adjusting
B1.
[Pattern mirror: NEET impulse style] A 0.15 kg ball strikes a wall perpendicularly at 10 m/s and rebounds at 10 m/s. If contact lasted 0.01 s, the average force on the wall is about: (a) 150 N (b) 75 N (c) 300 N (d) 600 N
Show answer
(c) — Δv = 10 − (−10) = 20 → Δp = 3 kg m/s; F = 3/0.01 = 300 N (third law: same on wall)
B2.
[Pattern mirror: JEE Main pulley style] A 3 kg block on a frictionless table connects over an ideal pulley to a 2 kg hanging block. The system's acceleration is: (a) 2 m s⁻² (b) 5 m s⁻² (c) 10 m s⁻² (d) 4 m s⁻²
Show answer
(d) — a = m₂g/(m₁+m₂) = 20/5 = 4 m s⁻²
B3.
[Pattern mirror: NEET boat-recoil style] A 50 kg girl jumps horizontally at 4 m/s from a 200 kg boat at rest. The boat's recoil speed is: (a) 0.5 m/s (b) 2 m/s (c) 1 m/s (d) 4 m/s
Show answer
(c) — 0 = 50×4 + 200V → V = −1 m/s (speed 1 m/s, opposite the jump)
B4.
[Pattern mirror: JEE Main rough-incline style] A block slides down a 37° incline with μ_k = 0.5. Its acceleration is: (a) 6 m s⁻² (b) 2 m s⁻² (c) 4 m s⁻² (d) 10 m s⁻²
Show answer
(b) — a = g(sin 37° − μ_kcos 37°) = 10(0.6 − 0.4)
B5.
[Pattern mirror: NEET banking style] A frictionless road curve has radius 40 m and banking angle 37°. Its optimum speed is: (a) 10 m/s (b) 10√2 m/s (c) 20 m/s (d) 10√3 m/s
Show answer
(d) — v₀ = √(rg tan 37°) = √(40×10×0.75) = √300 = 10√3 ≈ 17.3 m/s
B6.
[Pattern mirror: JEE Main contact-force style] A 16 N force pushes a 3 kg block pressed against a 5 kg block on a frictionless floor. The contact force between them is: (a) 6 N (b) 16 N (c) 10 N (d) 8 N
Show answer
(c) — a = 16/8 = 2; contact = mass ahead (being pushed) × a = 5 × 2 = 10 N
B7.
[Pattern mirror: NEET lift style] A 60 kg man rides a lift accelerating downward at 5 m s⁻². The floor's normal force on him is: (a) 600 N (b) 300 N (c) 900 N (d) 0
Show answer
(b) — N = m(g − a) = 60 × 5 = 300 N (downward acceleration → lighter)
B8.
[Pattern mirror: JEE Main F–t pulse style] A force rises linearly from 0 to a 20 N peak over 0.2 s and then drops instantly to zero. The impulse delivered is: (a) 1 N·s (b) 2 N·s (c) 4 N·s (d) 0.4 N·s
Show answer
(b) — Impulse = area of triangle = ½ × 0.2 × 20 = 2 N·s
B9.
[Pattern mirror: NEET explosion style] A 4 kg body at rest explodes into a 3 kg fragment and a 1 kg fragment. If the 3 kg fragment moves at 3 m/s, the 1 kg fragment's speed is: (a) 3 m/s (b) 6 m/s (c) 12 m/s (d) 9 m/s
Show answer
(d) — 0 = 3×3 + 1×v → v = −9 (speed 9 m/s, opposite the heavy fragment)
B10.
[Pattern mirror: JEE Main spring style] A 2 kg block hangs at rest from a vertical spring of constant 400 N m⁻¹. The extension is: (a) 5 cm (b) 2 cm (c) 10 cm (d) 20 cm
Show answer
(a) — kx = mg → x = 20/400 = 0.05 m = 5 cm

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