# Kinetic energy and momentum puzzler

Discussion in 'Physics' started by Mark44, Mar 29, 2008.

1. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
While I was looking up definitions of kinetic energy and momentum for a question on another thread, I ran across this question in my physics text.

A light mass and a heavy mass have equal kinetic energies of translation. Which one has the larger momentum?​

Mark

2. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
No takers, huh?
Maybe if I get you started...

Let m and M be the masses of the light and heavy masses.
Let v and V be the velocities of the light and heavy masses.

Let M = k^2 * m, where k > 1 (and hence, k^2 > k)

We know that Mv^2 = mV^2, since the kinetic energies are equal.
Can you find a relationship between the momenta?
Mark

3. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Similar problem: Two rectangular solids of different height have the same volume. If the bases of both solids are square, which rectangular solid has the larger non-square faces?

Rephrased: If M$\propto$H and V$\propto$W or D, and m$\propto$h and v$\propto$w or d, and w = d and W = D, (and therefore kinetic energy $\propto$ volume) which non square face (momentum $\propto$ surface of non-square face) is larger? By how much?

p$\propto\sqrt{P}$

4. ### hsvjap New Member

Apr 1, 2008
1
0
if p = momentum, E = kinetic energy
E = 2P/mass.,

hence, larger mass, greater momentum

hsvjap

5. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
I agree with your conclusion, but I don't follow your reasoning, and in particular how you get E = 2p/m.

We know that E (kinetic energy) = 1/2 m*v^2, and that p = mv.
If I solve the 2nd equation for v, I get v = p/m.
Substituting p/m for v in the KE equation, I get
E = 1/2 p^2/m, not 2p/m as you show.
Mark

6. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
It looks like people have lost interest in this one, so I won't leave you in suspense any longer.

The problem statement was this: A heavy body and a light body have the same kinetic energy. Which one has the greater momentum?

(1) Mv$^2$ = mV$^2$
Here M = mass of larger body, m = mass of smaller body, v = velocity of larger body, and V = velocity of smaller body.

(2) Let M = k$^2$m, where k > 1, and hence k$^2$ > k

From (1) and (2), Mv$^2$ = k$^2$mv$^2$ = mV$^2$
So k$^2$v$^2$ = V$^2$
Or V = kv

Working with the momentum of the larger body,
Mv = k$^2$mv = mk$^2$v > mkv = mV, by the assumption that k$^2$ > k in (2)

Therefore, Mv > mV, so the more massive body has greater momentum.

7. ### Caveman Active Member

Apr 15, 2008
471
0
I know we've done this one, but I wanted to show another logical way of doing this.
If we assume that we have two equal objects with the same mass and velocity and therefore momentum and kinetic energy to start.

K.E. = m*v*v
Momentum = m*v

Now let's increase the mass of one while adjusting the velocity so that the kinetic energies stay the same. So we will multiply one by the number k > 1 to get the new bigger mass.
New K.E. = (m*k)*(v*v)/k.
We have to divide by a total of k to make it stay the same. But actually each velocity will decrease by a factor of SQRT(k).

Now the new momentum will be
(m*k)*(v/SQRT(k) = m*v*SQRT(k).
Since k > 1, then SQRT(k) > 1. (Remember we are talking about magnitudes here.)

Therefore, the new momentum is larger than the old momentum. So more mass with the same kinetic energy will increase momentum.

8. ### triggernum5 Active Member

May 4, 2008
216
0
Your algebra is wrong.. Your k multiplication would cancel out, doing nothing, there would be no square root..
Also, don't forget the 1/2 coefficient on the KE..
The simplest way to envision it is that the lighter particle's velocity advantage 'squared' merely matches the energy of the heavier particle.. When that velocity advantage is forced to compete in an equation where it doesn't get to be squared in order to compensate for its mass deficiency, it comes up short..

9. ### Caveman Active Member

Apr 15, 2008
471
0
I will admit that it wasn't as clear as I would have like, but the algebra is correct. The k's are supposed to cancel out. Otherwise, you wouldn't maintain a constant K.E. What I'm showing is that if you increase the mass by a factor of k, in order to maintain the same kinetic energy, the velocity must be divided by SQRT(k).

10. ### triggernum5 Active Member

May 4, 2008
216
0
No, you started with
K.E. = (m)*(v*v)
Then you multiplied m by k, and divided v*v by k.. But this is one term since everything is multiplied..
you essentially did K.E. = m*v*v*(k/k).. You multiplied (2x) the kinetic energy by 1.. I think what you were aiming at doing is multiplying both sides by K.E. in an attempt to make the actual K.E. eqv to one unit, but that isn't apropriate either..

11. ### Caveman Active Member

Apr 15, 2008
471
0
I'm trying to make K.E. stay the same despite the fact that I am increasing mass by a factor of k. To do this, the K.E. equation must have 1/k somewhere else. Since there are only two variables, m and v, the 1/k defines the change needed in velocity. Actually it defines the change needed in v^2 so it must be a 1/SQRT(k) decrease for v.

Stated another way. We start with m0 and v0, with K.E. = m0*v0^2.
Now I define m1 = k*m0 and v1 = v0*n, where k > 1. So I am increasing mass.
If I want K.E. to stay the same in both situations, then
K.E. = m0*v0^2 = m1*v1^2 = k*m0*v0^2 * n^2
This reduces to n = 1/SQRT(k).

However, the equations are obscuring the basic idea here, instead of the original intent, which was to illustrate it better.

What I am trying to explain is how to solve problems like this using the concept of growth rates. We start with a certain situation and in a very abstract simple way, we can determine how things will relate to one another without ever really showing numbers.

In this case. If we start with a known K.E. and we increase mass by a factor of k, the velocity will decrease by a factor of SQRT(k) if we are to maintain the same K.E. Now the question is what happens to momentum. Since momentum = m*v, and m is changing by a factor of k and v is changing by 1/SQRT(k), then momentum is changing by a factor of SQRT(k). Since k > 1, SQRT(k) > 1. So as mass increases in a system maintaining the same K.E., the momentum will also increase.

Using this idea, you can show all kinds of cool things, like why animals are limited in size or why they all jump at about same height. Or you can show that as you increase the size of a boat, how the engine power scales.

12. ### triggernum5 Active Member

May 4, 2008
216
0
Oh, ok then.. In that case:
0) KE0 = (1/2)*m*v^2
1) KE = (1/2)*(km)*(v/SQRT(k))^2
2) v = SQRT( 2*KE/(SQRT(k)m) )
3) v = k^(1/4)/m^(1/2) * (2*KE)^1/2
4) p = k^(1/4)*m^(1/2) * (2*KE)^1/2
5) v0 = 1/m^(1/2) * (2*KE)^1/2
6) p0 = m^(1/2) * (2*KE)^1/2
7) p = k^(1/4)*p0
Sorry, its getting too late to follow all these brackets.. I'll double check for errors tomorrow, but from (7) if k > 1 then p > p0..
QED

13. ### Ratch New Member

Mar 20, 2007
1,068
4
To the ineffable all,

Sorry I am late to this thread. Here is my solution.

K.E heavy = K.E. light

The velocity of the light mass MUST be higher so that it can have the same K.E. as the heavy mass.

(M*V^2)/2 = (m*v^2)/2

By simple algebraic manipulation, MV/mv = v/V

So the heavier mass has more momentum by the ratio v/V .

Ratch

Last edited: May 23, 2008
14. ### Ratch New Member

Mar 20, 2007
1,068
4
Let the dimensions of the higher rectangle be designated in upper case, and the lower rectangle in lower case.

Then: (B^2)*H = (b^2)*h because the volumes are the same

Now the base of the lower rectangle MUST be larger in order for the volume to be the same, so b > B.

By simple algebraic manipulation, (B*H)/(b*h) = b/B

Therefore the non-square surface area of the high rectangle (B*H) is larger that the non-square surface area of the low rectangle (b*h) by the ratio b/B.

Ratch

15. ### Ratch New Member

Mar 20, 2007
1,068
4
To the ineffable ALL,

OK, so if a light and heavy mass have the same momentum, which mass has the higher kinetic energy, and by how much? Ratch

May 16, 2005
5,072
6
..........

17. ### triggernum5 Active Member

May 4, 2008
216
0
What weighs more, a pound of feathers or a pound of rocks?

18. ### recca02 Senior Member

Apr 2, 2007
1,211
0
{(M*V^2)/2}/{(m*v^2)/2}
= {(M*V^2)/2}/{(m*v^2)/2}
= {(M*V^2)}/{(m*v^2)}
= {(p.V)}/{(p.v)}
= V/v
E/e = m/M.

Which one hurts more, a 'pound' of feather or a 'pound' of rocks?

19. ### Ratch New Member

Mar 20, 2007
1,068
4
recca02,

Correct and succinct solution. Ratch