Physics · Laws of motion · chapter 4 of 14

Class 11 Physics Chapter 4: Laws of Motion case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, 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

In a school laboratory, Ananya pulls a 0.5 kg cart along a smooth horizontal track with a force sensor and records its velocity with a motion sensor. After 1.5 s she lets go of the sensor and the cart rolls freely. Her readings are given in the table. Friction and air resistance are negligible.

time (s)	0	0.5	1.0	1.5	2.0	2.5
velocity (m s⁻¹)	0	1.0	2.0	3.0	3.0	3.0

Part 1 · 1 mark

The force on the cart during the first 1.5 s is

(a) 1.5 N(b) 2 N(c) 1 N(d) 0.5 N

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Answer: (c) 1 N

a = (3.0 − 0)/1.5 = 2 m s⁻², F = ma = 0.5 × 2 = 1 N.

Part 2 · 1 mark

The net force on the cart between 1.5 s and 2.5 s is

(a) 1 N, forward(b) zero(c) 1 N, backward(d) 0.5 N, forward

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Answer: (b) zero

The velocity stays at 3.0 m s⁻¹, so the acceleration and, by the first law, the net force are zero.

Part 3 · 2 marks

Find the impulse on the cart in the first 1.5 s in two ways, and show that they agree.

Answer and marking scheme

From the force: impulse = F × t = 1 × 1.5 = 1.5 N s. From the momentum: Δp = 0.5 × 3.0 − 0 = 1.5 N s. They agree: impulse = change in momentum.

F × t = 1.5 N s (1)
Δp = 1.5 N s; equal (1)

Ananya repeats the run with a 1.0 kg cart and the same force for the same 1.5 s. Find its final velocity, and state what is the same for both carts.

Answer and marking scheme

Impulse = 1 × 1.5 = 1.5 N s = 1.0 × v, so v = 1.5 m s⁻¹. Both carts receive the same impulse, so they end with the same momentum, 1.5 N s; the heavier cart moves more slowly.

v = 1.5 m s⁻¹ (1)
same momentum (1.5 N s) for both (1)

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