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 impulsive force gives a velocity of 14 m s- 1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass of the system at that very moment is

Question

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 impulsive force gives a velocity of 14 m s- 1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass of the system at that very moment is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-nzehu7nmdpu8.png" /> (A) 30 m s- 1 (B) 20 m s- 1 (C) 10 m s- 1 (D) 5 m s- 1

Tardigrade Admin · Accepted Answer

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest. Hence, vC M=(m1 v1 + m2 v2/m1 + m2)=(10 × 14 + 4 × 0/10 + 4)=(140/14)=10 m s- 1

$30 m s^{- 1}$

$20 m s^{- 1}$

$10 m s^{- 1}$

$5 m s^{- 1}$

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,
$v_{CM} = \frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 m s^{- 1}$

30ms−1

20ms−1

10ms−1

5ms−1

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest. Hence, vCM​=m1​+m2​m1​v1​+m2​v2​​=10+410×14+4×0​=14140​=10ms−1

$30 m s^{- 1}$

$20 m s^{- 1}$

$10 m s^{- 1}$

$5 m s^{- 1}$

At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,
$v_{CM} = \frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 m s^{- 1}$