Two blocks of masses 10kg and 4kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14ms^–1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass is

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10 ms^–1

30 ms^–1

20 ms^–1

5 ms^–1

answer is C.

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Detailed Solution

Given that the velocity of the heavier block of 10 kg is = 14m/s.

We know that the velocity of COM is = $\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

Substituting the values,

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

Hence the correct answer is 10m/s.

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