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NEET PHYSICSLAWS OF MOTIONMedium

Question

Two carts of masses 200 kg and 300 kg on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are the same. If the 200 kg cart travels a distance of 36 m and stops, then the distance travelled by the cart weighing 300 kg is:

A

32 m

B

24 m

C

16 m

D

12 m

Step-by-Step Solution

  1. Conservation of Momentum: When the carts are pushed apart, the total momentum of the system remains zero. Thus, the magnitudes of their momenta are equal: p1=p2p_1 = p_2 m1v1=m2v2m_1 v_1 = m_2 v_2 [Source 61].
  2. Work-Energy Theorem with Friction: The kinetic energy acquired by each cart is dissipated by the work done against friction to bring it to a stop. K=Work=f⋅sK = \text{Work} = f \cdot s Since friction f=μN=μmgf = \mu N = \mu mg, we have: 12mv2=μmgs\frac{1}{2}mv^2 = \mu m g s Alternatively, expressed using momentum (K=p2/2mK = p^2/2m): p22m=μmgs\frac{p^2}{2m} = \mu m g s s=p22μgm2s = \frac{p^2}{2 \mu g m^2} [Source 77, 78].
  3. Proportionality: Since pp, μ\mu, and gg are constant for both carts: s∝1m2s \propto \frac{1}{m^2} s2s1=(m1m2)2\frac{s_2}{s_1} = \left( \frac{m_1}{m_2} \right)^2
  4. Calculation: Given m1=200 kgm_1 = 200 \text{ kg}, s1=36 ms_1 = 36 \text{ m}, and m2=300 kgm_2 = 300 \text{ kg}: s2=36×(200300)2s_2 = 36 \times \left( \frac{200}{300} \right)^2 s2=36×(23)2=36×49=16 ms_2 = 36 \times \left( \frac{2}{3} \right)^2 = 36 \times \frac{4}{9} = 16 \text{ m}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONmasseshorizontalpushedsupposecoefficient

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