Simple problem
Matt Pyne
This should be a simple problem to solve, but I can't think of the
answer and it is annoying me. And yes it's all about throwing balls on
trains.....
You are standing on a train and throw a ball at the wall in the
direction of the train's motion. It has a completely elastic collision
with the wall (it's one of those special physist balls that the general
public aren't allowed to know about), and comes back to you.
First of all you determine the speed of the ball during the experiment,
when it is half-way to the wall and half-way back again. These too
measurement determine the ball is travelling at the same speed (albeit
in opposite directions) to and from the wall. Call this speed v.
I stand on the platform. I measure the speed of the train, call it u.
I measure the speed of the ball at the same two points that you did. I
measure the speed to be (v+u) towards the wall and (v-u) away from the
wall, a difference of 2u.
So, far so good, now my problem, which I'm sure you lovely people can
solve for me. We both determine the kinetic energy of the ball. You
observe no change in energy of the ball, but I observe that the ball
loses kinetic energy to the train, who is right? What is the change in
energy of the ball. Moreover, you conclude that the collision was
elastic and I conclude the collision was not, who is right?
I pretty sure I'm missing something obvious here, presumably you are
going to tell me I have no right to expect the measured energies to be
the same, but why?
Matt.
Bryan W. Reed
Matt Pyne <mat...@softsim.com> wrote:
>Hi all,
>
>This should be a simple problem to solve, but I can't think of the
>answer and it is annoying me. And yes it's all about throwing balls on
>trains.....
>
>You are standing on a train and throw a ball at the wall in the
>direction of the train's motion. It has a completely elastic collision
>with the wall (it's one of those special physist balls that the general
>public aren't allowed to know about), and comes back to you.
>
>First of all you determine the speed of the ball during the experiment,
>when it is half-way to the wall and half-way back again. These too
>measurement determine the ball is travelling at the same speed (albeit
>in opposite directions) to and from the wall. Call this speed v.
>
>I stand on the platform. I measure the speed of the train, call it u.
>I measure the speed of the ball at the same two points that you did. I
>measure the speed to be (v+u) towards the wall and (v-u) away from the
>wall, a difference of 2u.
>
>So, far so good, now my problem, which I'm sure you lovely people can
>solve for me. We both determine the kinetic energy of the ball. You
>observe no change in energy of the ball, but I observe that the ball
>loses kinetic energy to the train, who is right?
We're both right, in our frames of reference. Kinetic energy is not something
inherent in a particle--it depends on your frame of reference. Likewise
the transfer of kinetic energy.
The train is much more massive than the ball, so the momentum transfer to
the train is extremely small as the ball bounces back. But it's not zero.
Let's call the tiny amount by which the train speeds up as the ball hits
the wall dv, and the mass of the train M. Mdv is the momentum transfer--equal
to 2mv (m is the mass of the ball). In the reference frame of the person
on the train, the train gains a kinetic energy (1/2)M(dv)^2. It involves the
square of a very small value dv, so it's really small. Negligible, really. On
the other hand, in the frame of the person standing along the tracks, the train
gains a kinetic energy (1/2)M((u+dv)^2 - u^2), or Mu*dv (to first order in dv).
Which is the momentum transfer times the speed of the train--not so small
an amount, compared to the kinetic energy of the ball. And in my frame, the
ball loses a significant amount of energy to the train.
Makes sense, since in my frame the ball does work on the train (the train
went a fair distance while the ball was applying force to it). In your frame,
the work done is negligible, since the wall of the train is basically standing
still relative to you.
> What is the change in
>energy of the ball. Moreover, you conclude that the collision was
>elastic and I conclude the collision was not, who is right?
>
The collision is elastic, since kinetic energy is conserved in both cases.
Remember, it's a two-body problem. You're looking at a collision between
a ball and a train, and it'll look different in two reference frames. Don't
let the fact that one of the bodies is much more massive than the other
confuse you--it's just like a billiard-ball problem.
>I pretty sure I'm missing something obvious here, presumably you are
>going to tell me I have no right to expect the measured energies to be
>the same, but why?
>
Because kinetic energy is a measure related to the motion of an object
relative to an observer. It's not inherent in the object--it depends on
choice of reference frame. Energy will be conserved in all frames, but
the precise accounting for it will be different from frame to frame.
Have fun,
Bryan
Bunyips
mat...@softsim.com wrote:
> You are standing on a train and throw a ball at the wall in the
> direction of the train's motion. It has a completely elastic collision
> with the wall (it's one of those special physist balls that the general
> public aren't allowed to know about), and comes back to you.
>
> First of all you determine the speed of the ball during the experiment,
> when it is half-way to the wall and half-way back again. These too
> measurement determine the ball is travelling at the same speed (albeit
> in opposite directions) to and from the wall. Call this speed v.
>
> I stand on the platform. I measure the speed of the train, call it u.
> I measure the speed of the ball at the same two points that you did. I
> measure the speed to be (v+u) towards the wall and (v-u) away from the
> wall, a difference of 2u.
to me, it's more simple to picture if you express the velocity in the first
case to be v and -v and the second case to be u+v and u-v.
> So, far so good, now my problem, which I'm sure you lovely people can
> solve for me. We both determine the kinetic energy of the ball. You
> observe no change in energy of the ball, but I observe that the ball
> loses kinetic energy to the train, who is right?
Both.
If the person on the train is holding the ball in his hand, the person on
the platform will say that it has a KE=1/2mu^2 and the person on the train
will say it's 0.
If the Roadrunner and the Coyote are on a train, which is not acclelerating,
and the Roadrunner shoots an Acme cannon at the Coyote, it makes no difference
what direction the cannon is pointed. If the Roadrunner is on the train and
shoots the Coyote standing on the platform, the cannonball would squish the
Coyote more if it were shot forwards than it would if it were shot backwards
after passing the Coyote.
> I pretty sure I'm missing something obvious here, presumably you are
> going to tell me I have no right to expect the measured energies to be
> the same, but why?
Because the velocity that is measured depends on your reference frame and the
kinetic energy depends on the velocity and the mass.
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
John T. Lowry
aviation. When turning from heading upwind to heading downwind, the airplane
apparently has a large gain of kinetic energy (as measured with respect to
the Earth underneath). Even though the forces acting on the airplane during
the turn (aerodynamic + weight) are the same in both the Air Mass- and the
Earth-based systems (uniform motion between them), the kinetic energies are
not. It's because Delta_KE is the integral of Force dot dr and the space
increments dr are NOT the same in the two systems.
Rather than go into this with pilots I usually simplify to a flight
attendant handing a passenger a magazine, then turning on her pumps and
walking back the way she came from. Similar to the bouncing ball.
John.
John T. Lowry, PhD
Flight Physics; Box 20919; Billings MT 59104
Voice: 406-248-2606
Bryan W. Reed wrote in message <79sg5h$p...@newsstand.cit.cornell.edu>...
>In article <36C180E8...@softsim.com>,
>Matt Pyne <mat...@softsim.com> wrote:
>>Hi all,
>>
>>This should be a simple problem to solve, but I can't think of the
>>answer and it is annoying me. And yes it's all about throwing balls on
>>trains.....
>>
>>You are standing on a train and throw a ball at the wall in the
>>direction of the train's motion. It has a completely elastic collision
>>with the wall (it's one of those special physist balls that the general
>>public aren't allowed to know about), and comes back to you.
>>
>>First of all you determine the speed of the ball during the experiment,
>>when it is half-way to the wall and half-way back again. These too
>>measurement determine the ball is travelling at the same speed (albeit
>>in opposite directions) to and from the wall. Call this speed v.
>>
>>I stand on the platform. I measure the speed of the train, call it u.
>>I measure the speed of the ball at the same two points that you did. I
>>measure the speed to be (v+u) towards the wall and (v-u) away from the
>>wall, a difference of 2u.
>>
>>So, far so good, now my problem, which I'm sure you lovely people can
>>solve for me. We both determine the kinetic energy of the ball. You
>>observe no change in energy of the ball, but I observe that the ball
>>loses kinetic energy to the train, who is right?
>
>>I pretty sure I'm missing something obvious here, presumably you are
>>going to tell me I have no right to expect the measured energies to be
>>the same, but why?
>>
>