Concerning Elastic collisions and Kinetic energy

[neeraj93]

Homework Statement

A ball of mass 2m is moving with a velocity V_o collides elastically and head-on with another ball of mass m at rest.
Find the minimum kinetic energy of the system.

Homework Equations

K = 1/2 mv²

The Attempt at a Solution

I personally do not understand what is expected of me here. The answer expected is
(2mV_o²)/3.
But collision is elastic. And I have to find the Kinetic energy of the system. Won't the Kinetic energy of the system remain constant (mV_o²) throughout?

 

[ideasrule]

Yes, it should be constant. I suspect the question meant something else, or the answer is wrong.

 

[troy611]

Yes, the kinetic energy remains constant, but a more close look tells you that, the 'initial' and 'final' kinetic energies are same, i.e. during a collision (for the split second when they are in contact), the kinetic energy is not conserved. The reason is as follows:

Consider a spring attached to the body initially at rest.
During collision, the spring is deformed, and at the same time, the body at rest gains speed, while the body which was moving initially is slowed down, also the spring gets compressed.
Now, a time will come, when both the bodies will travel with same velocity, and the compression in the spring would be max. or the energy stored in the spring will be max. Later on, the body with spring will gain much more speed than the other and speed away.

So, you can see that some part of the kinetic energy was stored in the spring as its potential energy. But, later it again converted to their kinetic energies.

Similarly, in bodies, there's some deformation (similar to spring, although for a short while), and as a result, some potential energy is gained(at the cost of kinetic energy).
Momentum can be conserved anytime, as the spring(or contact) force acts on both the bodies equally in opposite directions, and as a system, there's no external force.
For your answer,
Initial momentum = 2mv₀
Final mometum(when they are traveling with common velocities (v_c))
= 3mv_c

So, v_c = (2/3)v_o

KE of system (sum of KEs of two mases) = (2/3)mv_o²

Since you are an Indian (so am I)...I suggest you read up HC Verma part 1 Pg. 145 for better understanding...good luck...btw, u in 11th now, preparing for IIT, right?

Well, I am in 12th now, giving IIT this year...

 

[neeraj93]

Oh, It isn't there in Resnick, so no wonder. I, personally, don't prefer using Verma.. I'll have a look, but. thanks.
I'm in 12th too, btw.

 

[troy611]

oh...ok...all the best for your boards n all !

 

[neeraj93]

Thanks, good luck to you too!