Class 11 Physics Chapter 4: Laws of Motion worksheet with answer key
A 20-mark worksheet on Laws of Motion for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (d) impulse and momentum
Impulse = force × time = change in momentum, so both are measured in N s = kg m s⁻¹, with dimensions [MLT⁻¹]. Force is in N, [MLT⁻²].
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Answer: (c) They act at the same instant, on different bodies.
The two mutual forces are simultaneous, equal and opposite, and act on different bodies; there is no cause and effect between them, and they cannot cancel because they do not act on the same body.
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Answer: (a) zero
The puck moves with constant velocity, so its acceleration is zero; by the first law the net external force on it is zero. Its weight is cancelled by the normal force of the ice.
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Answer: (c) the gravitational pull of the earth on it
Centripetal force is not a new kind of force; it is the name for whatever material force gives the inward acceleration v²/R. For a satellite that is the earth's gravitational pull, just as the sun's pull does for a planet.
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Answer: (b) 0.2 N, directed towards the nail
Force on the magnet by the nail is equal and opposite to the force on the nail by the magnet. The magnet pulls the nail towards itself, so the nail pulls the magnet towards the nail with 0.2 N. The forces act at a distance; contact is not needed.
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Impulse = area under the force-time graph, a triangle: ½ × 600 × 0.02 = 6 N s. Impulse = change in momentum, so 0.1 v = 6, v = 60 m s⁻¹.
- impulse = area = 6 N s (1)
- v = 6/0.1 = 60 m s⁻¹ (1)
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Only the gravitational and electrical forces are relevant in mechanics. Contact forces arise ultimately from the electrical forces between the charged constituents (nuclei and electrons) of the bodies in contact.
- gravitational and electrical (1)
- contact forces are electrical, between charged constituents (1)
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Forces on the girl: weight 500 N down, normal force R = 400 N up (the scale reads R). Taking down as positive: 500 − 400 = 50a, a = 2 m s⁻² downward. The lift is either speeding up going down or slowing down going up.
- mg − R = ma set up with R = 400 N (1)
- a = 2 m s⁻², downward (1)
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(a) The block can stay at rest only if tan θ ≤ μₛ. Here tan 37° = 0.75 > 0.5, so static friction cannot hold it and it slides. (b) Down the plank: mg sin θ − μ_k mg cos θ = ma, so a = 10(0.6 − 0.25 × 0.8) = 10 × 0.4 = 4 m s⁻².
- compares tan 37° = 0.75 with μₛ = 0.5: it slides (1)
- net force mg sin θ − μ_k mg cos θ (1)
- a = 4 m s⁻² (1)
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The change in momentum in stopping is fixed by the mass and the speed (impulse = F × Δt = Δp). (a) The foam bed takes a longer time to stop the jumper, so the average force on the body, Δp/Δt, is smaller and does not injure. (b) Crumpling lengthens the time in which the car and passengers are stopped, so the force on them is reduced.
- Δp fixed; F = Δp/Δt (1)
- foam bed lengthens stopping time, smaller force (1)
- crumple zone lengthens collision time, smaller force (1)
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(a) Momentum is conserved: 4 × 3 = (4 + 2)v, v = 2 m s⁻¹ in the original direction. (b) 2 kg trolley: Δp = 2 × 2 − 0 = +4 N s. 4 kg trolley: Δp = 4 × 2 − 4 × 3 = −4 N s. (c) The impulses are equal and opposite. The two trolleys are in contact for the same time, so the forces on them, F = Δp/Δt, are equal and opposite, as the third law requires.
- common velocity 2 m s⁻¹ (1)
- impulses +4 N s and −4 N s (1)
- equal and opposite impulses over the same time give equal and opposite forces (1)
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