Two blocks of masses $8$ kg are connected by a spring of negligible mass and placed on a frictions less horizontal surface. An impulse gives a velocity of $12$ m/s to the heavier block in the direction of lighter block. The velocity of the center of mass is:-

12

m/s

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10

m/s

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8

m/s

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6

m/s

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Solution

The correct option is D $6$ m/s
Velocity of center of mass is $= \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$= \frac{8 \times 12 + 8 \times 0}{8 + 8}$

$= \frac{96 + 0}{16}$

$= 6 m / s$
Hence, the answer is $6 m / s .$

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Two blocks of masses 8kg are connected by a spring of negligible mass and placed on a frictions less horizontal surface. An impulse gives a velocity of 12m/s to the heavier block in the direction of lighter block. The velocity of the center of mass is:-

The correct option is D 6m/sVelocity of center of mass is =m1v1+m2v2m1+m2 =8×12+8×08+8 =96+016 =6m/sHence, the answer is 6m/s.

Two blocks of masses $8$ kg are connected by a spring of negligible mass and placed on a frictions less horizontal surface. An impulse gives a velocity of $12$ m/s to the heavier block in the direction of lighter block. The velocity of the center of mass is:-

The correct option is D $6$ m/s
Velocity of center of mass is $= \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$= \frac{8 \times 12 + 8 \times 0}{8 + 8}$

$= \frac{96 + 0}{16}$

$= 6 m / s$
Hence, the answer is $6 m / s .$