Need help with a Conservation of Energy and Momentum problem

Question

A spring with a 850 N/m spring constant is compressed 20 cm between a 0.5 kg box and a 2.6 kg box. The spring and boxes are initially at rest on a frictionless floor. The spring is then allowed to expand launching both boxes outward.

a) Calculate the speed of the 0.5 kg box after it leaves the spring.

b) Calculate the speed of the 2.6 kg box after it leaves the spring.

Answer 1

There must be conservation of momentum for the two block system m₂v₂ = - m₁v₁ or v₁ = -m₂v₂/m₁

The energy balance is 1/2 kx² = 1/2 m₁v₁² + 1/2 m₂v₂²

Now, sub in for v₁ using momentum conservation equation

1/2 kx² = 1/2 m₁(m₂v₂/m₁)² + 1/2 m₂v₂²

solving for v₂:

v₂ = sqrt( kx²/(m₂ + m₂²/m₁)) note that kx² = 850 N/m (.2 m)² = 34 J

I get 1.45 m/s for m/s and 7.55 m/s for v₁

Please consider a tutor. Take care.

Answer 2

PE = ½kx² where k = 850 N/m and x = 20 cm = 0.2 m

PE= ½(850)(0.2)² = 17 J = total KE after each box leaves the spring.

Conservation of Energy

½m₁v₁² + ½M₂v₂² = 17 which means that

(1) m₁v₁² + M₂v₂² = 34

Conservation of Momentum

m₁v₁ = M₂v₂ which means that

(2) v₁ = (M₂/m₁)v₂

Solving for v₂ ( the speed of the heavier box)

Substituting (2) into (1) gives

(3) m₁(M₂/m₁)²v₂² + M₂v₂² = 34 which means that (M₂²/m₁ + M₂)v₂² = 34

plugging in m₁ = 0.5 and M₂ = 2.6 gives

((2.6)²/(0.5) + 2.6)v₂² = (13.52 + 2.6)v₂² = 16.12v₂² = 34

v₂ = √(34/16.12) ≈ 1.45 m/s

Solving for v₁ ( the speed of the lighter box)

v₁ = (M₂/m₁)v₂ = (2.6/0.5)(1.45) = 7.55 m/s

Answers

a) The speed of the 0.5 kg box after it leaves the spring is 7.55 m/s

b) The speed of the 2.6 kg box after it leaves the spring is 1.45 m/s