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 velocity of the center of mass $$is____m/s$$

Question

Infinity Learn · Accepted Answer

The velocity vCM of the centre of mass can be obtained by using the principle of conservation of linear momentum, $$MV=(M+m)vCMvCM=MV(M+m)=10kg$$×$$14ms$$−$$1(10+4)kg$$                           $$=10ms$$−$$1$$

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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 velocity of the center of mass is____m/s

The velocity v_CM of the centre of mass can be obtained by using the principle of conservation of linear momentum, $MV = (M + m) v_{CM}$
$\begin{array}{l} v_{CM} = \frac{MV}{(M + m)} = \frac{10 kg \times 14 {ms}^{- 1}}{(10 + 4) kg} \\ = 10 {ms}^{- 1} \end{array}$

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Centre of mass(com)

Remember concepts with our Masterclasses.

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 velocity of the center of mass is____m/s

The velocity vCM of the centre of mass can be obtained by using the principle of conservation of linear momentum, MV=(M+m)vCMvCM=MV(M+m)=10kg×14ms−1(10+4)kg =10ms−1

Ready to Test Your Skills?

free study material

The velocity v_CM of the centre of mass can be obtained by using the principle of conservation of linear momentum, $MV = (M + m) v_{CM}$
$\begin{array}{l} v_{CM} = \frac{MV}{(M + m)} = \frac{10 kg \times 14 {ms}^{- 1}}{(10 + 4) kg} \\ = 10 {ms}^{- 1} \end{array}$