Melonie+H.


 * Week 10 (April 16th, 2012) **

01**. A steel ball of mass 0.5 kg is fastened to a cord that is 70 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of it's path the ball strikes, a 2.5 kg steel block initially at rest on a frictionless surface. The collision is elastic.**
 * Find, **
 * a) the speed of the ball **
 * b) the speed of the block, both just after the collision **


 * Hint: first draw a diagram. **
 * find the velocity of the ball, just before it hits the block. **


 * To find the Velocity of the ball before collision, conservation of energy theorem can be applied. PE=KE. mgh=.5mv^2. m1=2.5 g=9.8 h=0.7m. m2=.5 and we solve for v. **

**Using this value of v, conservation of momentum theorem can be applied. m1v1i +m2v2i=m1vf1 + m2v2f. This equation has to unknowns. Vf1 and Vf2.** **You can solve for one of these unknowns in this equation since we found V1i and we know V2i=0. We can set Vf1 to equal to 3.7-5Vf2. This is equation 1.**

**Since it is an elastic collision, the kinetic energy is the same before and after the collision. The resulting equation would be .5m1vi1^2+.5m2v2i^2=.5m1vf1^2+.5m2vf2^2.The .5m2v2i^2 term goes away so you're left with .5m1v1i^2 = .5m1vf1^2 + .5m2vf2^2. The equation turns out to be 3.42=.25Vf1^2 +1.25Vf2^2.This is equation 2. Substitute 3.7-5Vf2 for Vf1 and solve for Vf2. Once you solve for Vf2, substitute that into the equation Vf1 = 3.7-5Vf2.**