Two blocks of masses 10 kg and 4 kg and 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

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Solution

velocity of center of mass $v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = 10 \frac{m}{s}$

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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 10 kg and 4 kg and 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

velocity of center of massvcm=m1v1+m2v2m1+m2=10×14+4×010+4=10ms

velocity of center of mass $v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = 10 \frac{m}{s}$