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 velocities of blocks (in m/s) in case of maximum compression are

5, 5

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10, 10

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8, 4

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7, 7

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Solution

The correct option is B 10, 10

At maximum compression, both block must have same speed.
$V_{c m} = \frac{10 \times 14 + 4 \times 0}{14} = 10 m / s$
At max compression, $V_{10} = V_{4} = V_{C M} = 10 m / s$

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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 velocities of blocks (in m/s) in case of maximum compression are

The correct option is B 10, 10 At maximum compression, both block must have same speed. Vcm=10×14+4×014=10 m/s At max compression, V10=V4=VCM=10 m/s

The correct option is B 10, 10

At maximum compression, both block must have same speed.
$V_{c m} = \frac{10 \times 14 + 4 \times 0}{14} = 10 m / s$
At max compression, $V_{10} = V_{4} = V_{C M} = 10 m / s$