# Complicated Catapult/Conservation of Energy Test Question

 Sci Advisor P: 2,470 The displacement would depend on where the other end of the spring is attached. If it's right bellow the point where it's attached to arm at equilibrium, your value of .3m is correct. I would go with that given no other information. Do you happen to remember what the spring constant is? Also, is the equilibrium position equilibrium for the spring, equilibrium for unloaded catapult, or equilibrium for loaded catapult? Each of these would give you different results. In the final case, the speed will be maximum through equilibrium. But in the first two cases the kinetic energy might work out to be negative if the spring coefficient was too low. The gravitational potential energy as function of angle looks like this. $$U = sin(\theta)\frac{1}{2}mgL + sin(\theta)MgL$$ Kinetic energy as function of velocity. $$T = \frac{1}{6}mv^2 + \frac{1}{2}Mv^2$$ Where L is the length of the arm, m is mass of the arm, M is mass of the boulder, and v is velocity of the boulder. The angle is measured from horizontal. Assuming equilibrium is equilibrium for the spring, the rest is easy. Add spring potential to gravitational potential at lower point, subtract of gravitational potential at the equilibrium point, and that's your kinetic energy. Solve for v, and you're done. If some other equilibrium condition is meant, you'll have to correct for it.