IIT JEE , AIEEE Forum - Physics , Mechanics

rohith291991

what is the definition of an elastic collision?? one in which kinetic energy is conserved or one in which coefficient of restitution is 1???

karthik2007

rohith291991

nope i think when one is right the other cannot be right...check this post..

http://www.goiit.com/posts/list/mechanics-pleaz-help-too-tough-for-me-40696.htm

gain

the first one

coefficient of resititution is different for perfectly elastic n partially elastic.

kundan_leo

i agree with gain................

but it depends on what the context is..............
in elastic collision the COR is 1 ....
otherwise the objects would just stick to each other which is not the case...........

but then that's a sort of approx becoz actually
ALL COLLISIONS BTW REAL OBJECTS ARE SEMI-ELASTIC.......

WHERE

e- = coeff of restitution = relative velocity of recession / approach......
v2 - v1 / u1 - u2.............

thus ...........both the statements may not necessarily be true..
ke is definitely conserved.........but COR is not always 1......
but in the broader picture yes..........sort of approx u know

akshaybansal

According to defination, collisions in which e=1 , are called elastic collisions.
In elastic collisions, Kinetic energy of the system remains conserved.

sanyogita2023

according to me........in perfectly elastic collision momentum as well kinetic energy is constant after and before collision. Coz in this both the bodies come to their original shape and size after the collision i.e. no fraction of mechanical energy remains stored as deformation pot. energy in the bodies.........and for perfectly elastic collision e is always 1.......and obviously for partially it will b less than 1 and greater than 0.

so for perfectly elastic collision you right........K.E. is const. as well e= 1.

bvsatyaram

The basic definition of Elastic collision is that the energy is conserved. If u review the definition of Coefficient of restitution, u will find that it has nothing to do with energy.

But mathematically, it happens that coefficient of restitution is 1 for perfectly elastic collisions.
So, both the definitions are true for perfectly elastic collisions.
Now coming to the problem, I would like to stress on the exact definition of Coefficient of restitution.

e = (rel. vel. of seperation)/(rel. vel. of approach)
In this definition we talk about two relative velocities. Note that both the relative velocities MUST be taken in the direction of common normal. So, inthis particular proble, when u use the definition of Coefficient of restitution, u need to take the relative velocities along the common normal. ie. normal to the sphere at the point of contact. If u go by this method, u will get the same answer as u get by conservation of energy method.

jasss

by definition in ray&jonson,e=KE(f)/KE(i) only when e=1;now if KE(f)=KE(i),then e has to be equal to one

jasss

this one's very correct n i'm very sure about it.i deal with the basic concepts in physics very thoroughly as its my favourite n i've even confirme it frm my physics teacher

soham406

kinetic energy is conserved , coefficient of restitution is 1

rohith291991

@ bvsatyaram sir.... yes i know that we have to take e along the normal ...thats y i had this doubt in the first place....but anyways i think i have found an answer to my own problem... the definition of an elastic collision has to be modified a bit....it is a collision in which e is always 1...this is always valid.....but based on KE it should be written as.... a collision in which KE is conserved JUST AFTER the collision....here KE is conserved JUST AFTER the collision...but soon normal force acts and some KE is lost... both definitions are now correct.....basically all this confusion arose because i was under the impression that brilliant tutorials cudnt make a mistake....in their solution they used an AMBIGUOUS definition of an elastic collision...in other words they used " a collision in which KE is conserved" ...with "JUST AFTER" removed...hence they considered that the triangle will move back with the SAME velocity as it collides....this is because they took the definition to mean"a collision in which KE is conserved NO MATTER WHAT HAPPENS..." and hence i also got confused.. but this is utter nonsense....if this were true then e wud become greater than 1 and it wud be a super elastic collision.... so they CAN be wrong...everyone is human right??

so i guess everything is clear now...

bvsatyaram

Normal forced never perform work on rigid bodies. Here the sphere is rigid. So, even after the collision, KE is conserved, provided no non-conservative force acts on the body..

rohith291991

but the inclined plane with sphere and triangle is the system... and the system is fixed by some means(say by a nail ) to the ground...so KE of the inclined plane+sphere+triangle system is not conserved... but the KE of inclined plane+sphere+triangle+earth system is conserved....it just depends on the system which u take....btw sir HOW can u say that a normal force does no work on rigid bodies??? contact force is a normal force and it clearly can do work.....

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