How Fast Does the Lighter Block Travel When the Spring Relaxes?

[naianator]

Homework Statement

A system consists of two blocks, of masses m and 2m, attached to the ends of a massless spring with a force constant k. The system is placed on a horizontal frictionless surface. Initially, the spring is relaxed. The blocks are then pulled apart an “extra” distance x and simultaneously released from the state of rest.

Find the speed v1 of the block of mass m at the instant the spring is relaxed again. Answer in terms of m, k, and x.

Homework Equations

U_spring = 1/2*k*x^2
K = 1/2*m*v^2
m_1*v_1+m_2*v_2 = m_1*v_1' + m_2*v_2'

The Attempt at a Solution

Momentum is conserved and initially 0 so

m*v_1 = 2m*v_2

and

v_1 = 2v_2

Potential energy in the compressed spring is equal to 1/2*k*x^2 and when the spring reaches equilibrium the potential is 0 and the kinetic energy is equal to 1/2*m*v_1^2+1/2*(2m)*(2v_1)^2 so

1/2*k*x^2 = 1/2*m*v_1^2+4*m*v_1^2

= k*x^2 = m*v_1^2+8*m*v_1^2

= k*x^2 = 9*m*v_1^2

and finally

v_1 = sqrt(k*x^2/(9*m))

Where am I going wrong?

 

[olivermsun]

If $v_1 = 2v_2$, then why is $\frac{1}{2}kx^2 = \frac{1}{2} mv_1^2+4mv_1^2$?

 

[AlephNumbers]

v_1 = 2v_2

1/2*m*v_1^2+1/2*(2m)*(2v_1)^2

You substituted 2v₁ for v₂.

 

[naianator]

You substituted 2v₁ for v₂.

Yes! Thank you