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Two blocks of masses 6 kg and 4 kg are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of 14 m/s in the direction of lighter one, then the velocity gained by the centre of mass will be
  • a)
    7.4 m/s
  • b)
    14 m/s
  • c)
    8.4 m/s
  • d)
    10 m/s
Correct answer is option 'C'. Can you explain this answer?
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Initial Conditions:
- Mass 1 (m1) = 6 kg
- Mass 2 (m2) = 4 kg
- Initial velocity of m1 = 14 m/s
- Initial velocity of m2 = 0 m/s

Calculating Centre of Mass:
- Centre of mass (C) = (m1 * v1 + m2 * v2) / (m1 + m2)
- Substitute the values to get initial velocity of centre of mass
- C = (6 kg * 14 m/s + 4 kg * 0 m/s) / (6 kg + 4 kg) = (84 + 0) / 10 = 8.4 m/s

Final Conditions:
- After collision, the spring will compress and the masses will move together
- The final velocity of both masses together will be the same as the initial velocity of the centre of mass

Therefore, the velocity gained by the centre of mass will be 8.4 m/s.
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Two blocks of masses 6 kg and 4 kg are placed on a frictionless surface and connected by a spring. If the heavier mass is given a velocity of 14 m/s in the direction of lighter one, then the velocity gained by the centre of mass will bea)7.4 m/sb)14 m/sc)8.4 m/sd)10 m/sCorrect answer is option 'C'. Can you explain this answer?
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