Posted by **Anonymous** on Thursday, November 5, 2009 at 11:43pm.

A mass, m1 = 9.36 kg, is in equilibrium while connected to a light spring of constant k = 112 N/m that is fastened to a wall.

A second mass, m2 = 6.71 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.201 m

The system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. Determine the value of v.

- phsyics -
**drwls**, Friday, November 6, 2009 at 5:58am
At the spring's equilbrium point, all of the stored potential energy of the compressed spring becomes kinetic energy of both masses. No further force is applied by the spring at that instant, and then the masses separate as the spring force reverses direction.

The V (of both masses) at that instant is given by

(1/2) k A^2 = (1/2)(m1 + m2) V^2

Solve for V

## Answer this Question

## Related Questions

- physics - A mass, m1 = 5.5 kg, is in equilibrium while connected to a light ...
- Physics Problem - A 3.00-kg mass is fastened to a light spring that passes over ...
- Physics - A body of mass m is suspended from a spring with spring constant k and...
- physics - A light spring of constant k = 170 N/m rests vertically on the bottom ...
- physics help please... - A light spring of constant k = 170 N/m rests vertically...
- Physics - A spring, (spring constant of 400 N/m) is displaced .75 m from its ...
- physics - A light spring with a spring constant of 15.2 N/m rests vertically on ...
- Physics - A light spring of constant k = 158 N/m rests vertically on the bottom ...
- Physics - A mass m = 13 kg is connected to two different springs, one on the ...
- physics - A steel ball of mass m=5 g is moving at a speed of 250 m/s toward a ...

More Related Questions