Class 11 Physics Chapter 6: Systems of Particles and Rotational Motion worksheet with answer key
A 20-mark worksheet on Systems of Particles and Rotational 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) ĵ
In cyclic order î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ. −ĵ is î × k̂, the reversed order; the null vector is a × a.
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Answer: (a) 300 m
The explosion forces are internal, so the centre of mass follows the original parabola and lands at 240 m. Both fragments land together, so 240 = (180 + x)/2 and x = 300 m.
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Answer: (b) 3/2
Total momentum stays zero, so the fragments have equal and opposite momenta p. K = p²/2m, so K₁/K₂ = m₂/m₁ = 3/2. The answer 1 assumes equal energies; 9/4 squares the mass ratio.
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Answer: (b) the acceleration due to gravity is the same at every point of the body
With g the same for all particles it comes out of the sum Σrᵢ × mᵢg = 0, giving Σmᵢrᵢ = 0 about the CG, which is the condition for the centre of mass. Symmetry, rest or rigidity is not what makes the two points coincide.
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Answer: (b) 3
The centre of mass is nearer the heavier ball, and mP × 15 = mQ × 45, so mP/mQ = 45/15 = 3. The answer 1/3 puts the centre of mass nearer the lighter ball.
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The centre of mass depends only on how the mass is distributed (R = Σmᵢrᵢ/M); the centre of gravity is the point about which the total gravitational torque is zero. They coincide only when g is the same at every point. For the very tall tower g is smaller at the top, so the upper parts weigh relatively less and the centre of gravity lies below the centre of mass.
- CM from mass distribution only; CG is where the gravitational torque is zero (1)
- not coincident: g varies, CG lies lower (1)
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P = τω, so τ = P/ω = 1500/300 = 5.0 N m. At the rim τ = rF⊥, so F = τ/r = 5.0/0.10 = 50 N.
- τ = P/ω = 5.0 N m (1)
- F = τ/r = 50 N (1)
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The two forces are equal in magnitude, opposite in direction and have different lines of action: they form a couple. Their vector sum is zero, so there is no translation; their moments about the axle add in the same sense, so the net torque is non-zero and the wheel rotates.
- names a couple: equal, opposite forces with different lines of action (1)
- zero net force so no translation; non-zero torque so rotation (1)
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(1) The vector sum of the external forces is zero, ΣFᵢ = 0; (2) the vector sum of the external torques is zero, Στᵢ = 0. For coplanar forces (say in the x-y plane) only three remain: ΣFx = 0, ΣFy = 0 and Στ_z = 0. Example: a light rod acted on at its two ends by equal and opposite forces perpendicular to it (a couple) has zero net force but turns.
- both conditions, ΣF = 0 and Στ = 0 (1)
- three for coplanar forces (1)
- a valid example such as a couple on a rod (1)
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a × b = | î ĵ k̂ ; 1 1 0 ; 0 1 1 | = (1·1 − 0·1)î − (1·1 − 0·0)ĵ + (1·1 − 1·0)k̂ = î − ĵ + k̂, which is perpendicular to both. Its magnitude is √3 ≈ 1.73, so n̂ = (î − ĵ + k̂)/√3 (or its negative). Check: (î − ĵ + k̂)•(î + ĵ) = 1 − 1 + 0 = 0.
- a × b = î − ĵ + k̂ (1)
- divides by √3 to get the unit vector (either sign) (1)
- scalar product with a is zero (1)
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I = Σmᵢrᵢ². (i) About the end: 1.0(0)² + 1.0(0.5)² + 1.0(1.0)² = 1.25 kg m²; K = (1/2)Iω² = 0.5 × 1.25 × 16 = 10 J. (ii) About the middle: 1.0(0.5)² + 0 + 1.0(0.5)² = 0.5 kg m²; K = 0.5 × 0.5 × 16 = 4.0 J. The same body has a different I for a different axis.
- I = 1.25 kg m² about the end (1)
- I = 0.5 kg m² about the middle (1)
- K = 10 J and 4.0 J (1)