Question

A package of mass 19.81 kg is released from rest at a warehouse loading dock and slides down a 4.37-m-high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 42.89 kg, from the bottom of the chute. Suppose the packages stick together.

(a)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 19.81 kg rebound?

Answer 1

P.E(top)K.E(bottom) mgh-12ugh 2*9.8*4.37-85.652-9.2548ml s(letit beu) a)bylawof conservation of momentum mu+mvmV 19.81 *9.2548+ 42.89*0 = (19.81+42.89)びーびー2.92404m / s b)let velocities of twobodies after collision bevand v, Byconservation of energy, 92548() bylawof conservation of momentum, 19.819.2548+0 19.81v +42.89v2 18333759-19.8M +4289y (2) by solving ()and (2) we get, 3407mls Thus againbylawof conservationof energy,letthe package of mass 19.81kg gain a height h meter due to the velocityof 3.407mi s PS(top)=KB(bottom)→ 1981*98*h=112 *19.81*(-3407): on solving the above equation we get 0.5902m.