## A 18.1-kg block and a 32.6-kg block are resting on a horizontal frictionless surface.

**A 18.1-kg block and a 32.6-kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant = 1580 N/m). The spring is compressed by 0.160 m from its unstrained length and is not attached permanently to either block.**

**With what speeds do (a) the 18.1-kg block and (b) the 32.6-kg block move away after the mechanism keeping the spring squeezed is released and the spring falls away?**

## Answer ( 1 )

The conservation of momentum :

0 = m1v1 + m1v1

v1= -(m2/m1) * v2 = -1.80*v2 =

The conservation of the energy :

(Potential Energy spring = Kinetic Energy )

1/2 k * x2 = 1/2 m1 v12 + 1/2 m2 v22

1580 * (0.16)2 = 18.1*(-1.80*v2)2 + 32.6 * v22

—> v2 = 0.67 m/s

—-> v1 = -1.80 * 0.67 = -1.21 m/s