23for 2nd, MvM=mvmcosa [ (vcom)x = 0]
mgh = 1/2mvm2+1/2 MvM2 = M2vM22mcosa +MvM22
mgh = vM2(2M2+2mMcosa4mcosa)
1. A uniform chain of mass M and length L is held vertically in such a way that its lower end just touch the horizontal floor.The chain is released from rest in position. Any portion that strikes the floor comes to rest .Assuming that the chain does not form a heap on the floor calculate the force exerted by it on the floor when the length x has reached the floor.
2.A block of mass m is placed on a wedge of mass M, which in turn placed on a horizontal surface. Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom.Angle of inclination of wedge is \alpha and height is h.
3.A ball of mass m is dropped onto the floor from a certain height .The collision is perfectly elastic and ball rebounds to the same height and again falls. Find the average force exerted by the ball on the floor during the long time interval.
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7 Answers
23Ndt=(dm)v
dm = Mdx/L , so N =Mv/L (dx/dt) = Mv2/L
v = √2gx, so N = 2Mgx/L
23
1i don't know that's why i asked all of u, by the way it's HC verma question.
1oops i missed a 'time' factor in FORCE.....:(
i guess the question has a meaning with 'a along time interval' and not 'the'......
taking the 1st impact.....
avg F = change in momentum/ total time
= m.2√(2gh)2√(2h/g)
= mg
In each impact an avg force of 'mg' will be exerted by the ball..
now during a long time interval it exerts the mean of such avg forces( each mg) which eventually gives
Favg = mg