Two blocks of masses $10 k g$ and $4 k g$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulsive force gives a velocity of $14 {m s}^{- 1}$ to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that moment is

30 {m s}^{- 1}

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20 {m s}^{- 1}

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10 {m s}^{- 1}

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5 {m s}^{- 1}

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Solution

The correct option is C $10 {m s}^{- 1}$
$v_{c m} =$ velocity of center of mass
$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{c m} = \frac{(10 \times 14) + (4 \times 0)}{10 + 4}$
$v_{c m} = 10 m / s$

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Two blocks of the masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block. The velocity of centre of mass is

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Two blocks of masses 10kg and 4kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulsive force gives a velocity of 14ms−1 to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that moment is

The correct option is C 10ms−1vcm=velocity of center of massvcm=m1v1+m2v2m1+m2vcm=(10×14)+(4×0)10+4vcm=10m/s

The correct option is C $10 {m s}^{- 1}$
$v_{c m} =$ velocity of center of mass
$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
$v_{c m} = \frac{(10 \times 14) + (4 \times 0)}{10 + 4}$
$v_{c m} = 10 m / s$