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 impulse gives a velocity of $14 m / s$ to the heavier block in the direction of the lighter block, The velocity of the centre of mass is:

30 m / s

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20 m / s

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10 m / s

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5 m / s

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Solution

The correct option is C $10 m / s$
Formula,

$V_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$= \frac{10 \times 14 + 4 \times 0}{10 + 4}$

$= 10 m / s$

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Two blocks of 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 in the direction of the lighter block, The velocity of the centre of mass is:

The correct option is C 10 m/sFormula,Vcm=m1v1+m2v2m1+m2=10×14+4×010+4=10m/s

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 impulse gives a velocity of $14 m / s$ to the heavier block in the direction of the lighter block, The velocity of the centre of mass is:

The correct option is C $10 m / s$
Formula,

$V_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$= \frac{10 \times 14 + 4 \times 0}{10 + 4}$

$= 10 m / s$