Two masses m₁ and m₂ are initially at rest with a spring compressed between them. What is the ratio of their velocities after spring has been released.

(Ans: `\frac {v_1^'}{v_2^'}` = `\frac {m_2}{m_1}`)


Given:

Two masses m₁ and m₂

Velocity of mass m₁ = v₁ = 0 m s⁻¹

Velocity of mass m₂ = v₂ = 0 m s⁻¹


To Find:

Let `\v_1^'` and `\v_2^'`  are the velocities after the spring released then to find

Ratio of the velocities = `\frac {v_1^'}{v_2^'}` = ?


Solution:

According to the law of conservation of momentum

Initial momentum = final momentum  

m₁ v₁ + m₂ v₂ = m₂ v₂' - m₁ v₁'  

As after the released of the spring, the masses will move in opposite direction.

by putting the corresponding values

0 = m₂ v₂' - m₁ v₁'

or

m₁ v₁' = m₂ v₂' 

or

`\frac {v₁'}{v₂'}` = `\frac {m₂}{m₁}` ------------Ans

Hence the required velocities ration after the spring released is equal to the inverse ratio of masses.

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