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  4. 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- 1 to the heavier block in the direction of the lighter block. The velocity of the centre of mass in m s- 1 is 2n , then n=

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Q. 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 $14 \, m \, s^{- 1}$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass in $m \, s^{- 1}$ is $2n$ , then $n=$

NTA AbhyasNTA Abhyas 2022

Solution:

$\textit{v}_{\text{CM}}=\frac{\textit{m}_{1} \textit{v}_{1} + \textit{m}_{2} \textit{v}_{2}}{\textit{m}_{1} + \textit{m}_{2}}$
$\frac{10 \times 14 + 4 \times 0}{10 + 4}=\frac{10 \times 14}{14}=10 \, m \, s^{- 1}$ $= 2 n \Rightarrow n = 5$
Solution