Question

Two masses m1 and m2 are connected by a spring of spring constant k and are placed on a frictionless horizontal surface. Initially, the spring is stretched through a distance x0 when the system is released from rest. Find the distance moved by the two masses before they again come to rest.

Answer 1

Distance stretched= x and x=x1+x2

Let the ratio of the distance be moved by m1 to that m2 be such that x1/x2=m2/m1

Also, x1/(x1+x2)=m2/(m1+m2)

~x1/x=m2/(m1+m2)

x1=m2x/(m1+m2)

m1 again comes to rest when it moves a distance of 2x1

~ 2x1 = m2*2x/(m1+m2)

Similarly for m2 , 2x2= m1*2x/(m1+m2)

Answer 2

Thanks for asking the question!

ANSWER::

Mass of the 1st block = m₁

Mass of the second block = m₂

Initially the spring is stretched by x₀

Spring constant = K

For both the blocks to come to rest again ,

Let the distance travelled by m₁ and m₂ be x₁ and x₂ respectively.

As x₀ external force acts in horizontal direction ,

m₁x₁ = m₂x₂ [Equation 1]

Again , the energy would be conserved in the spring.

(1/2) K x x² = (1/2) K ( x₁ + x₂ - x₀ )²

x₀ = x₁ + x₂ - x₀

x₁ + x₂ = 2x₀ [Equation 2]

x₁ = 2x₀ - x₂

Similarly , x₁ = 2m₂x₀ / (m₁ + m₂)

m₁( 2x₀ - x₂) = m₂x₂

2m₁x₀ - m₁x₂ = m₂x₂

x₂ = 2m₁x₀ / (m₁ + m₂)

Hope it helps!