3. A mass M is dropped from a height H onto one end of a stick of mass M, length L,pivoted about the opposite end. Assume the moment of inertia of the stick about thepivot is ML. Upon collision, the mass adheres to the stick. Respond to the followingin terms of M, L, H, and g.(a) Find the speed of the mass just before impact.(b) Find the angular speed of the system immediately after impact.(c) Find the linear speed of the mass M at its lowest point.(d) Determine the mechanical energy lost as a result of the collision.

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered Given, A mass M is dropped from height H onto one end of a stick of mass M length L pivoted about the opposite . (a.) Just before the impact the mass M travelled a distance H. So by equation of kinematics , v2=u2+2as⇒v2=0+2gH⇒v=2gH (b.) By conservation of angular momentum we have, Initial angular momentum=Final angular momentum MvL=Iω⇒M2gHL=13ML2ω⇒ω=32gHL (c.) By conservation of energy we have, The lowest point of the mass will be the position when it is complete Total energy before the impact= Total energy after the impact 12Mv2+MgH=12Iω2+12M+Mv'2-MgL⇒12M×2gH+MgH=1213ML232gHL2+Mv'2-MgL⇒Mv'2=2MgH+MgL-3MgH⇒v'2=MgL-H⇒v'=MgL-H Answer: (a.) Speed of the mass just before the impact= 2gH (b.) Angular speed of the system= 32gHL (c.) Linear speed of mass M at the lowest position= MgL-H

Answered: A mass M is dropped from a height H…