Two blocks of masses 2 kg and 0.5 kg attached at two ends of a compressed spring. The elastic P.E. stored in the spring is 10 J. Find the velocities of the blocks if the spring delivers its energy to the blocks when released.  

(Ans: 1.4 m s⁻¹,  5.6 m s⁻¹)



Given:

Mass of 1st block  = m₁ = 2 kg

Mass of 2nd block = m₂ = 0.5 kg

Total energy stored =  10 J  


To Find:

Velocity of 1st block  = v₁ = ?

Velocity of 2nd block = v₂ = ?


Solution:


According to the law of conservation of energy

Energy of the blocks before released  = Energy of the blocks before released 

Elastic P.E stored in spring = K.E. of the both blocks

Elastic P.E stored in spring = `\frac {1}{2}` m₁ v₁² + `\frac {1}{2}` m₂ v₂²


by putting the corresponding values

10 J = `\frac {1}{2}` (2 kg) v₁² + `\frac {1}{2}` (0.5 kg) v₂²

or 

20 J = (2 kg) v₁² + (0.5 kg) v₂² ---------------(1)


Now according to the law of conservation of momentum

Momentum of the blocks before released  = Momentum of the blocks after released 

0 =  m₁ v₁ + m₂ v₂


by putting the corresponding values

0 = (2 kg) v₁ + (0.5 kg) v₂ 

or

(2 kg) v₁ = - (0.5 kg) v₂ 

or

 v₁ = - `\frac {0.5 kg v₂}{2 kg}`

 v₁ = - 0.25 v₂   ------------(2)

putting the value of v₁ in equation (1)

20 J = (2 kg) (- 0.25 v₂)² + (0.5 kg) v₂²

20 J = (0.125 kg) v₂² + (0.5 kg) v₂²

20 J = (0.125 kg + 0.5 kg) v₂²

20 J = (0.625 kg) v₂²

or

v₂² = 32 m² s⁻²

or (taking square root both side)

v₂ = 5.65 m s⁻¹  ------------------Ans (1)

Putting in v₂ = 5.65 m s⁻¹ in equation (2)

 v₁ = - 0.25 (5.65 m s⁻¹)

 v₁ = - 1.42 m s⁻¹ -------------Ans.(2)

-ve sign shows the opposite direction  

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