e =0 for perfectly inelastic bodies e = 1 for perfectly elastic bodies 5. Measure the angle of the graph and determine the square root of the tangent of the angle that represents the coefficient of restitution between the two colliding bodies. Momentum is a vector quantity, so that the required sign must be observed in the equation for conservation of momentum. For instance, the algebraic signs of the e= yz = Stane momenta in Fig. I and Fig. 3 would be positive (+) for miv1 , negative (-) for myvz , negative (-) for myuj , and positive (+) for myuz - If the velocity is unknown and it comes out positive in the result of the solution., Data and Observations the assumed direction is correct; if negative, the direction of the velocity is opposite to that arbitrarily assumed. The magnitude of the velocity in each case is the correct value. Trial Height of Fall Height of Rebound Coefficient of Procedures: Y, (cm ) Ya (cm Restitution ( e) 1. Place a meter stick vertically on top of a solid metal block and drop a steel ball from a 80 N N 70 certain height. Practice doing this procedure and try to determine the height of 60 rebound of the steel ball. 50 40 2. When you have developed the skil in determining the height of rebound, begin with an initial height of fall at the 80-cm mark of the meter stick. Do this five times by decreasing the height of the fall by 10 cm and determine again the corresponding Computation and Analysis: height of rebound. CH TRIAL 1 3. Compute the coefficient of restitution between the steel ball and the metal block using 80- D the equation e = =? ", FLOOR Since the metal block is stationary, then vz and uz are 0., v,= 12gy, , and u, = /2872 . Hence, we have #2 - 31 2872 1 1 - 2 where: y1 is the height of fall of the steel ball, and yz is the height of rebound of the steel ball Conclusion: 4. From the data obtained in step 3, plot the height of the fall along the abscissa scale (x- axis) and the height of rebound along the ordinate scale (y-axis)