# Kinetic energy

1. Jul 11, 2009

### leena19

1. The problem statement, all variables and given/known data
Consider the foll. statements
(A) If the Kinetic Energy of a particle is constant with time,its momentum also should be constant with time
(B) If the momentum of a particle is constant with time,its kinetic energy should also be constant with time
(C) If the momentum of a particle varies linearly with time,its Kinetic Energy should also vary linearly with time
Which of the above statements is/are true?
(1)only A (2)B only(3) Conly (4)A&Bonly (5)A&C only
2. Relevant equations

KE = 1/2mv^2
momentum=mv

3. The attempt at a solution
At first,i thought all A,B,C were true,(since they were talking of the same particle of mass m,so both would only depend on the velocities?),but,
I don't think A is necessarily true,cause momentum is conserved only if there's no external force acting on the system,but then again,doesn't an external force affect the KE as well?I don't think so,cause the conservation of energy depends on whether a collision is elastic or not,right?but is this even relevant here?
I'm just so confused.
for(B) I think if B was true,then C would also probably be true?
Once again,I'm just guessing here. :(
I hope my problem's clear
Thank you

Last edited: Jul 11, 2009
2. Jul 11, 2009

### Hootenanny

Staff Emeritus
In terms of (A) & (B) consider which quantities are conserved in a (i) elastic collision and (ii) an inelastic collision in the absence of external forces. You are correct in the assumption that both KE and momentum are only conserved if there are no external forces acting.

In terms of (C), consider a change in velocity from v = v0 to v = 2v0. How does the kinetic energy change? How does the momentum change?

3. Jul 11, 2009

### leena19

Sorry,Hootenanny,I didn't realise you had responded before editing my post,
and I'm sorry i could't reply earlier,i had to go for my prayers.

KE and momentum is conserved during an elastic collision,so if KE is a constant with time,then momentum would also be constant?so I think A would be true?

For B,momentum is conserved even during an inelastic collision,so I guess B would be wrong?

For C,when the velocity is increased to 2v,the momentum would be doubled,but KE would be 4times,but their individual variation with time is proportionate ,right?

4. Jul 11, 2009

### Fenn

Well, I think you're over-complicating by talking about collisions. In a collision, there is more than one particle, and conservation of energy or momentum describes the total quantity in the system.

This question asks about a single particle, and so you don't need to consider collisions or conservation. You're on the right track with the relevant equations, and particularly your assessment of (c).

5. Jul 11, 2009

### leena19

Thanks for replying,Fenn,Although I'm not very sure how i could determine the answer by just considering the equations.I feel we should use the conservation laws too,but i don't know.
As for C) i'm confused with the terms proportionate variation and linear variation,are they both the same?
I know if something is proprtionate,y/x=m,a constant,and the graph goes through the origin,but is it the same for linear variation as well?I know this is a very basic question,but i'd like to clear this up as well.

THANK YOU

6. Jul 11, 2009

Hello leena .Picking up on Hootenanny's advice think of say an elastic collision with a wall where the particle bounces back in the opposite direction with the same speed.Has the K.E changed?Has the momentum changed?Could it be that one or both of them is a vector?As for part C write out the equations and see what you get.It might help you understand it more if you sketched graphs.

7. Jul 11, 2009

### Fenn

Ok, well consider what (C) is saying. If the momentum varies linearly with time, then

$$p(t) = A \cdot t,\quad\text{where }A\text{ is some constant}$$

You have an expression for momentum that describes it in terms of mass and velocity. It's reasonable to assume the mass of the particle is constant, so the velocity must be changing as some function of time. If the momentum is changing linearly with time, then you can show that the velocity is changing linearly with time.

$$p(t) = A \cdot t = m \cdot v(t)$$

or, solving for $$v(t)$$,

$$v(t) = \frac{p(t)}{m} = \frac{A}{m}t$$

With this, plug it into your equation for kinetic energy. Does this expression depend linearly on time?

After re-reading the question, I see the reasoning behind talking about collisions. In particular, note what happens to the momentum and kinetic energy when $$v\rightarrow -v$$.

8. Jul 11, 2009

### Dr. Mirrage

You shouldn't need any conservation laws, because the problem deals with a single particle, so if momentum is conserved then there is no change in v. But, the problem never says that v can't change, in fact, part C requires it to change.

Also, note that momentum is a vector, since p = mv and mass is not a vector, velocity is a vector, so a change in momentum is not necessarily a change in KE, if the magnitude is the same because KE is not a vector, it's a scalar.

9. Jul 11, 2009

### leena19

Oh I see now!Thank you so much,sir.

No,cause it'a scalar quantity
YES,cause it's a vector.
which would mean A is wrong?and B is right?

I drew the graph,and for KE,I get a curve with an increasing gradient ,which would mean,
KE doesn't vary linearly with time(I don't know why i didn't see it earlier,it's so easy)
and,
momentum varies linearly with time,but not proportionately(cause it's a y=mx+c graph)?
Am I right?
Everything seems so clear to me right now,i feel i am correct. :)
Now,I also (think?) I get what Fenn referred to when she asked me to consider the equations,cause the equations tell me whether KE and momentum are vector or scalar quantities?

Thank you

10. Jul 11, 2009

### Fenn

Glad to know you understand. But, "she"? Last I checked I'm a guy :).

11. Jul 11, 2009

### leena19

Thank you,Fenn.
Thank you Dr.Mirrage.Thank you both for trying to help me,I see what you mean(i think?to some extent),but I prefer using the conservation laws to deal with the problem,and I think I finally understood it ,so thanks again.