Physics · System of particles and rigid body · chapter 6 of 14

Class 11 Physics Chapter 6: Systems of Particles and Rotational Motion case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

On the smooth ice of a frozen lake in Gulmarg, Arjun (60 kg) and Meera (40 kg) stand at rest 10 m apart, each holding one end of a light rope. Arjun pulls the rope in hand over hand, while Meera only holds on. Friction between their skates and the ice is negligible, and the rope stays horizontal. The two slide towards each other and meet. Treat each skater as a particle.

Part 1 · 1 mark

How far from Arjun's starting position do they meet?

(a) 6.0 m(b) 5.0 m(c) 4.0 m(d) 2.4 m

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Answer: (c) 4.0 m

They meet at their centre of mass, which stays at rest: X = (60 × 0 + 40 × 10)/100 = 4.0 m from Arjun. 6.0 m is the distance from Meera.

Part 2 · 1 mark

At an instant when Arjun is moving at 1.2 m s⁻¹, Meera's speed is

(a) 0.8 m s⁻¹(b) 1.8 m s⁻¹(c) 1.2 m s⁻¹(d) 2.7 m s⁻¹

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Answer: (b) 1.8 m s⁻¹

Total momentum stays zero: 60 × 1.2 = 40 × v, so v = 1.8 m s⁻¹, towards Arjun. 0.8 m s⁻¹ inverts the mass ratio.

Part 3 · 2 marks

Explain why the meeting point would be the same if Meera pulled the rope and Arjun only held on.

Answer and marking scheme

Whoever pulls, the rope's forces on the two skaters are internal to the system and there is no external horizontal force. So the centre of mass, initially at rest, stays at rest (MA = F_ext = 0). They always meet at the centre of mass, 4.0 m from Arjun, whatever internal forces act.

rope forces are internal; no external horizontal force (1)
centre of mass stays at rest, so they meet there either way (1)

Find the total linear momentum of the two skaters at any instant during the pull, and the velocity of their centre of mass. Give reasons.

Answer and marking scheme

With no external horizontal force, dP/dt = F_ext = 0, so P stays equal to its initial value, zero (both started at rest). Since P = MV, the velocity of the centre of mass is V = P/M = 0 throughout.

no external force, P conserved, P = 0 (1)
V = P/M = 0 (1)

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