Shortcut Methods
JEE Main - Motion of the Centre of Mass Integer Type Questions:
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Velocity of the center of mass = (2 kg × 3 m/s - 3 kg × 2 m/s) / (2 kg + 3 kg) = -0.2 m/s (towards the left due to the negative sign).
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Displacement of the body = (v_i + v_f)/2 × t = (5 m/s + 15 m/s)/2 × 2 s = 15 meters.
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Kinetic energy of the body = 1/2 × m × v_f² = 1/2 × 10 kg × (10 m/s)² = 500 Joules.
Multiple Choice Questions:
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The correct answer is b) 3.3 m/s. Explanation: Using the law of conservation of momentum, we have: (10 kg × 5 m/s) + (5 kg × 0 m/s) = (15 kg × v) v = (50 kg m/s + 0 kg m/s) / 15 kg v = 3.3 m/s.
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The correct answer is d) 20 meters. Explanation: Using the equation of motion, we have: s = ut + 1/2 at² s = (5 m/s × 2 s) + 1/2 (-10 m/s²) × (2 s)² s = 20 meters.
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The correct answer is b) 20 Ns. Explanation: Impulse = Force × Time Impulse = 20 N × 2 s = 20 Ns.
CBSE Board - Motion of the Centre of Mass:
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Displacement of the block = (v_i + v_f)/2 × t = (2 m/s + 6 m/s)/2 × 4 s = 16 meters.
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Kinetic energy of the body = 1/2 × m × v_f² = 1/2 × 10 kg × (15 m/s)² = 1125 Joules.
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Impulse imparted to the body = Force × Time = 30 N × 5 s = 150 Ns.
