# When the spring is maximally extended, find v_1f

## Homework Statement

A mass m_1, initially moving at a speed v_0, collides with and sticks to a spring attached to a second, initially stationary mass m_2. The two masses continue to move to the right on a frictionless surface as the length of the spring oscillates. At the instant that the spring is maximally extended, the velocity of the first mass is...?

p_i = p_f
E_i = E_f

## The Attempt at a Solution

Initially, there's only m_1 moving, so p_i = m_1v_0.
When the masses collide and stick together, assuming the spring to be massless, they will travel at a final velocity v_f together (so v_f is the velocity of the center of mass), and p_f = (m_1+m_2)*v_f.

Equating p_i and p_f, we get that v_f = m_1v_0/(m_1+m_2).

However, I'm not sure how to carry on from here, with the spring oscillating. This sounds like m_2 might have a greater velocity and m_1 a little less at some times and other times m_1 might have more velocity than m_2 does, but how can I figure this out definitely?

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-from p_i=p_f you can get the center of mass velocity
-we go into the intertial system that moves with the center of mass velocity
-in this frame, the system does not move to the right anymore, the masses only oscillate. At the point where the spring is maximally extended, both masses have zero kinetic energy. That means, in the absolute inertial system, both will move with the center of mass velocity at this point.

rude man
Homework Helper
Gold Member
I don't think either energy nor momentum conservation will help.
I suggest F = ma for m1 and m2.