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NEET PHYSICSMOTION IN A PLANEMedium

Question

A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a speed of 1 m/s, then after 4 s his displacement relative to the ground will be:

A

5 m

B

4.8 m

C

3.2 m

D

3.0 m

Step-by-Step Solution

Since the system (man + trolley) is initially at rest and there is no external force in the horizontal direction, the total linear momentum of the system is conserved .

  1. Define Variables: Mass of man (mm) = 80 kg80\text{ kg} Mass of trolley (MM) = 320 kg320\text{ kg} Velocity of man relative to trolley (vrelv_{rel}) = 1 m/s1\text{ m/s} Let vTv_T be the velocity of the trolley relative to the ground.
  • Let vmv_m be the velocity of the man relative to the ground.
  1. Relative Velocity Relation: vrel=vmvT    vm=vT+vrel=vT+1v_{rel} = v_m - v_T \implies v_m = v_T + v_{rel} = v_T + 1.

  2. Conservation of Momentum: Pinitial=0P_{initial} = 0 Pfinal=mvm+MvT=0P_{final} = m v_m + M v_T = 0 Substitute vmv_m: 80(vT+1)+320vT=080(v_T + 1) + 320 v_T = 0 80vT+80+320vT=080 v_T + 80 + 320 v_T = 0 400vT=80    vT=0.2 m/s400 v_T = -80 \implies v_T = -0.2\text{ m/s}.

  3. Calculate Man's Velocity w.r.t. Ground: vm=0.2+1=0.8 m/sv_m = -0.2 + 1 = 0.8\text{ m/s}.

  4. Calculate Displacement: Displacement = vm×t=0.8 m/s×4 s=3.2 mv_m \times t = 0.8\text{ m/s} \times 4\text{ s} = 3.2\text{ m}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOTION IN A PLANE. 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.

PHYSICSMOTION IN A PLANEweighingstandingtrolleyweighingtrolley

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