Question

Two blocks of masses 10 kg and 4kg are connected by a spring of negligible mass and placed on frictionless horizontal surface . an impulse gives a velocity of 14 m/s to the heavier block in the direction of lighter block .the velocity of the centre of mass is??

Answer 1

The velocity of the centre of mass is 10 m/s.

Explanation:

We are given:

Mass of block A = 10 Kg

Mass of block B = 4 Kg

The velocity of the heavier block (10 kg) = 14 m/s.

The velocity of COM is given by:

= m  1   v1 + m  2v / 2 /m  1 +m  2  

= 10 x 14 + 4 x 0 / 10 + 4

= 140 / 14

= 10 m/s

Hence the velocity of the centre of mass is 10 m/s.

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Answer 2

The velocity of the centre of mass is 10 m/s.

Explanation:

We are given the velocity of the block whose mass is 10 kg as 14 m/s.

Then, the velocity with which the centre of mass is moving will have the formula as $\frac{(m1v1)+ (m2v2)}{(m1+m2)}$.

Putting the given values in the above given formula we will get the value of velocity as $\frac{(14 \times 10)+ (0 \times 4)}{(10+4)}$

⇒ $\frac{140}{14}$

= 10 m/s .

So, now we get that on a friction less Surface where two masses are connected with each other and heavier that has velocity of 14 metre per second then the lighter mass will have the velocity of 10 metre per second.

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