Class 11 Physics Chapter 3: Motion in a Plane worksheet with answer key
A 20-mark worksheet on Motion in a Plane 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: (c) can have any value from 0° to 180°
Only in one dimension are velocity and acceleration along one line. In two or three dimensions the angle between them may be anything from 0° to 180°.
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Answer: (a) of constant magnitude, always directed towards the centre, so it is not a constant vector
The velocity keeps changing direction, so there is an acceleration, v²/R, towards the centre. Its magnitude is constant but its direction turns with her, so it is not a constant vector.
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Answer: (d) 1 : 3
Every point of the disc has the same ω, and v = Rω, so vA : vB = 5 : 15 = 1 : 3. 1 : 1 would be the ratio of angular speeds; 1 : 9 wrongly squares the ratio of the radii.
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Answer: (c) 5
Rx = 2 + (−5) = −3, Ry = 3 + 1 = 4, so |R| = √(9 + 16) = 5. √53 is |A − B|, and 7 adds the magnitudes of the components.
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Answer: (b) P ≠ Q, since their directions differ
Two vectors are equal only if they have the same magnitude and the same direction. P and Q have equal magnitudes but different directions, so P ≠ Q.
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Tminute = 1 h and Thour = 12 h, and ω = 2π/T, so ω_m : ω_h = 12 : 1. v = Rω, so vm : vh = (6 × 12) : (4 × 1) = 72 : 4 = 18 : 1.
- ω ratio 12 : 1 from the periods (1)
- v = Rω gives 18 : 1 (1)
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The equation holds only for a constant acceleration vector. In uniform circular motion the acceleration has constant magnitude v²/R but always points to the centre, so its direction changes continuously. The acceleration is not constant, and the equation does not apply.
- equation needs a constant acceleration vector (1)
- centripetal acceleration changes direction, so it is not constant (1)
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Multiplying a vector by a positive scalar changes its magnitude by that factor and keeps its direction: 5 m s⁻¹ × 4 s = 20 m, due north. Its dimension is [LT⁻¹][T] = [L]: it is the displacement of the scooter in 4 s.
- 20 m, due north (1)
- identifies it as a displacement, dimension [L] (1)
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x = ½ × 2 × t² = t², y = 3t. Eliminating t: x = y²/9, a parabola (its axis along x). At t = 2 s, v = 2t î + 3ĵ = 4î + 3ĵ, speed √(16 + 9) = 5 m s⁻¹.
- x = t², y = 3t (1)
- x = y²/9, a parabola (1)
- speed 5 m s⁻¹ at t = 2 s (1)
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Same direction: P + Q = 17. Right angles: P² + Q² = 169. Then 2PQ = 17² − 169 = 120, PQ = 60; P and Q are the roots of x² − 17x + 60 = 0, so P = 12 N and Q = 5 N. At 60°: R = √(144 + 25 + 2 × 12 × 5 × 0.5) = √229 ≈ 15.1 N.
- P + Q = 17 and P² + Q² = 169 (1)
- P = 12 N, Q = 5 N (1)
- R at 60° = √229 ≈ 15.1 N (1)
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At rest, F₁ + F₂ + F₃ = 0, so F₃ = −(F₁ + F₂). F₁ + F₂ = (3 − 5)î + (4 + 2)ĵ = −2î + 6ĵ, so F₃ = (2î − 6ĵ) N. |F₃| = √(4 + 36) = √40 ≈ 6.32 N. tan θ = −6/2 = −3, θ ≈ 71.6° below the +x-axis (in the fourth quadrant).
- F₃ = −(F₁ + F₂) = 2î − 6ĵ N (1)
- |F₃| = √40 ≈ 6.3 N (1)
- ≈ 71.6° below the +x-axis (1)
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