IIT JEE , AIEEE Forum - Physics , Mechanics

ruhi

hi

plz anyone solve my question:

"a particle of mass 'm' is released from a certain height 'h' with zero initial velocity. it strikes the ground elastically. plot the graph between its kinetic energy and time till it returns to its initial position."

plz tell me the steps how to deal with such type of questions

ruhi

prakhar

i faced ditto problem
but u got to spend seom nickel to ask the expert.

i'm posting it on the expert zone, for some tips.

yo

prabhu

prakhar

see it here.... http://www.goiit.com/posts/list/0/54.htm#90

rahul.a

Hi

seems like a simple problem

As the collision will be elastice ( e =1 ), so in the entire journey the energy will be constant ( E = mgh )

Now at t =0 body starts movind downwards

at time 't'
V = gt

and Kinetic Energy
E_k = 1/2 mv²= 1/2 m g²t²So the final curve between E_k and t will be a parabola and at t =

2h/g ( at this time body will hit the ground) E_k will have its maximum value E_k = mgh and at time t = 2

2h/g body will gain its original height h and E_k will be zero. In this fashion body can move endlessly.

A simple twist can be given to the problem, say body hit the ground with elastic cofficient e ... here e < 1 .

try to plot the same curve in this situation.

Cheers
Rahul

ayush007

lets , start from the equations----

when the partical is coming down,

v=gt ---------------1

due to elasticcollision ,its velocity changes from +ve to -ve,but ,k.e not changes , before & after collision,means the magnitude of velocity remains unchanged even after collision....

kinetic energy=1/2(mv^2)=1/2m(gt)^2.

now as, we know , m & g are the constants, we take 1/2mg^2=x;
then kinetic energy = xt^2.
therefore,,
K.E.=xt^2

IT IS A UPWARD PARABOLA,which is like, y=kx^2 curve.

the max. k.e.(1/2m(gt)^2) . (corresponding to t=(2h/g)^1/2 and symmetrically ,it goes down then after ,to attend value of k.e=0.

please ,think about it carefully, you 'll surely benefit from this....

vasanth

thanx guyzzzzzzzz

ayu

thanks ayush007 and rahul.a ,i really benefitted with your explanations ,so pls, accept my salutes.

ayush007

thanks, ayu, i feel happy that you benefitted from my answers,
and one more thing, you are the person who has voted me most , thanks for all that....

i would like you to answer my doubt on gravitation,
its posted entitiled with "THINK ,YOU CAN CRACK THIS!" its a little wierd but please give it ,at least ,a thougth.
once again thanks for rating me..

keep smiling......

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