Note: DO NOT ROUND OFF when it is not yet the final answer.Tip: The collision part is not considered. Please consider that stage and use conservation of angular momentum to analyze the collision. Note that the collision is perfectly inelastic.The heights of the rod and box above the ground are not the same in general. You need to find theheight of the center of mass of the rod at the very end of the swing in order to write down the gravitational potential energy associated with its center of mass.The total (initial or final) potential energy should be equal to the sum of potential energies associatedwith the position of the block above the ground, and the position of the center of mass of the rod above the ground.Another tip: In solving this item, you will need the height h' of center of mass of the rod at the end of the motion. This can be solved by applying methods of basic geometry and trigonometry. Note that the rod is uniform so that its center of mass is located at its geometric center. Slides and Sticks!m = 0.250 kgM = 0.500 kgA small block with mass m = 0.250 [kg] slidesdown a frictionless inclined plane that has a heighth = 1.50 [m]. The block collides and sticks to auniform rod of mass M = 0. 500 [kg] pivoted atpoint 0. Find the maximum height H that the smallblock will reach.h = 1.50 m

Question

Accepted Answer

Given that:
m=0.250 kgh=1.50 mM=0.500 kgL=h=1.50 m

A small block with mass m = 0. 250 [kg] slides down a frictionless inclined plane that has a height h = 1.50 [m]. The block collides and sticks to a uniform rod of mass M = 0.500 [kg] pivoted at point 0. Find the maximum height H that the small block will reach.