1. The problem statement, all variables and given/known data A package of mass m is released from rest at a warehouse loading dock and slides down a h = 3.7 m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. (a) Suppose the packages stick together. What is their common speed after the collision? (b) Suppose the collision between the packages is elastic. To what height does the package of mass m rebound? 2. Relevant equations energy of conservation eqn 3. The attempt at a solution the velocity acquired that mass m when it reaches the bottom of the path v = √2gh h = height of fall So I got v= 7.668 m/s (I went sqrt(2*9.8*3) ) the kinetic energy of the mass by the time it collides with the big mass 2m will be E = mv^2 / 2 after collsion the kientic energy of the combined mass will be equal to the initil kinetic energy of the mass m ( m +2m) * v'^2 / 2 = mv^2 / 2 v ' = is the final velocity of the combined mass . So I got v= 4.4 m/s Does that look right??? b) velocity of the mass m after collision v' = (m2 - m1) v / (m1+m2) v = √gh m2 = 2m m1 = m v = √gh . the height reached by teh mass after collsion h = v'^2 / g ARE THE STEPS TAKEN IN QUESTION B CORRECT? or should it be like this: where vf= final velocity of package with mass m right after collision vi= the velocity from part a (7.668 m/s) vf= ((m1 - m2)*vi) / (m1 + m2) = ((m - 2m)*vi) / (m + 2m) = (-m*vi)/ (3m) = -vi/3 = -7.668/3 = -2.556 m/s So for mass m: where Kf= final kinetic energy Ki= initial kin energy Ugf= final gravitational potential energy Ugi = initial grav. pot. energy yf = final height vi = initial velocity (THIS time I used the 2.556 m/s I just found not 7.668 m/s) Kf + Ugf = Ki + Ugi Kf and Ugi = 0 so, solving for yf I get: yf = vi^2 / 2g = ((-2.556)^2) / (2*9.8) = 0.33 m IS THIS RIGHT? OR ARE NEITHER CORRECT? need some quick help :/ thanks a lot in advance!