1clearly momentum won't be conserved because there is an impulse acting on the system(here ball)
so only saving is energy conservation!
A ball is bouncing vertically up and down. It has a velocity v0 when it strikes the ground. The acceleration due to gravity is slowly reduced by 10% during a very long interval of time. Assuming no air resistance and perfectly elastic collisions, find the corresponding change in v0.
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18 Answers
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66Yes that's correct. However 0.1 is still quite small compared to 1. The final v0 is 3√0.9 times v0 which gives a fraction change of 3.45% approximately which is not very far from 3.33%.
However, for the sake of exactness, we indeed cannot what I did.
1Sir , just for the sake of completeness , your last sentence is not fully correct I believe . That statement is only true for very small changes in " g " , i . e , like " 1 % " , " 2 % " at most . But " 10 % " is not an allowable value of change in " g " to calculate " Δ V " in that way I suppose : )
1Thanks for your appreciation Anant Sir . As far as your solution goes , I ' ll consider it as a masterpiece . Also , that question mark " ( ? ) " in your question is left for students to realize , isn't it ?
My answer : - Since , this is a one - dimensional translational motion , hence there is only one degree of freedom .
66Sorry for the delayed response.
@Ricky, such a flurry of activity from you in the recent days is indeed commendable. [1] I am quite happy for you. Even more satisfactory has been your transformation in the physics dept. There has hardly been a problem left unsolved by you and what is more important is that you have solved them correctly. Nice work.
As far the current problem goes, definitely finding out the adiabatic invariant of the process solves it in a minute. What I had in mind (and that will kind of justify the name adiabatic) was to make an analogy of this bouncing ball system with a one dimensional gas. We can think of the relation of g with pressure and the h to which is bouncing to the volume. So changing slowly g will amount to a change in pressure adiabatically. So if we denote the pressure as p and the "volume" as V then this slow variation in g should give a process like
pV^\gamma=\text{ constant}
What about \gamma. Since \gamma - 1 = 2f where f is the number of degrees of freedom, and in this case there is only one degree of freedom(?), so we have γ=3 and so the process becomes pV3 = constant.
Now, pV is proportional to the internal energy which in turn will be proportional to v02 and V is proportional to h. So we get
v02 h2 = constant
But since h is proportional to v02/g, so we get
v06g2 = constant
So we ultimately get
v03g = constant
for the process.
That means the fractional change in v0 = 13 X fractional change in g
1Another solution using Adiabatic Invariants : - Entirely based upon my knowledge about them collected from Wikipedia only -
Let the momentum at any moment of time be " p " , the other variables from my previous solution remaining unchanged . Then , conserving energy : -
p 2 = p02 - 2 m 2 g x
Now , since the given process can be assumed to be an adiabatic process , hence , the quantity : -
A = 0 ∫ H p dx = A constant = K ............ ( let )
Carrying out the necessary integration and remembering that , " p02 = 2 m 2 g H " , we obtain -
V03 = g K ' ............... ( where K ' is another constant )
So , the final answer remains same , that is " VFinal = . 969 V0 " .
1@Anant sir,
The gravity is changing very slowly (very long interval of time). I read in some mechanics book that such problems can be solved by employing the concept of inadiabatic invariants (I am no expert at it though)
1@kaymant sir , isn't the ang. mom. of the ball remains conserved about the center . can we use that fact
71awesome thought of your friend ricky bhaiya... LOOKS NICE.
1Here goes my somewhat - faulty looking solution -
Applying " Energy " conservation and letting the initial height reached by the ball " h " , I got -
v 2 = v02 - 2 g x
where " x " is the disatance of the ball from the ground .
Let us look at the " average " velocity , which in this case , can either be found out using " x " or " time " , but as Anant Sir has pointed out , " g " is also a function of time . Hence , I ' ll average the velocity using " x " as the independant variable .
Vav = 1H 0∫H ( v02 - 2 g x ) 1 / 2 dx
where - 12 m v02 = m g H ..................... ( H - The height of the highest point initially )
Vav = v033 g H = 23 v0 = a constant
Since " H " is also constant ,
Hence , v03g = constant
We have found out a relation between " v0 " and " g " which does not involve time .
Hence , if " V " is the final velocity and " G " , the final accelaration due to gravity , then -
v03g = V3G
Or , V = . 969 v0 .
Note -
1 . The idea of averaging is not my own , a close friend of mine had read this method of solving these type of problems somewhere .
2 . I am pretty sure that this problem is well suited to be set in IIT .
66Well said Ricky[1]. And swordfish, why exactly do you think that this problem requires the use of adiabatic invariants?
1Let me assure you , Swordfish , Anant Sir is not known to give problems to us which are far above JEE level . He only gives such problems which require , say , in essence , " Momentum " and " Energy " conservation and a bit of understanding of the situation at hand .
The concept of " Adiabatic Invariants " is pretty interesting though , as I just found out by " Wiki - ing" it .
1sir, if I understand the question correctly, doesn't this require the use of adiabatic invariants which is out of our syllabus?
66Try to find some expression involving the velocity v0 and g that does not depend on time. (Remember, here g is not constant; its changing with time)
66Its a one dimensional problem, so there is no necessity of considering angular momentum. In any case, what center are you referring to?