Re: Velocity and Kinetic Energy


"Tom" <poliquin@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:49e3a939$0$25031$b86e1d2@xxxxxxxxxxxxxxxxxxxxxxxx

I tutor a 16 year old student in computer science
but he asked me this physics question which has
me baffled (making me think I should stick
to CS :-) ) I hope this is the right venue.

A man is in spacesuit floating in space. There
is a fixed reference point 'A' nearby. He has
a small hand held rocket motor (or gun that fires
projectiles).

His initial velocity relative to the fixed point
'A' is zero.

He energizes the rocket for a moment, enough to add
200 joules worth of velocity to himself. This means
that his velocity (relative to the fixed point 'A')
from the basic kinetic energy relation, is now,

V1 = sqrt ((2 * e) / m)

where e = 200 joules
m = 100 Kg

or,
v1 = sqrt ((2 * 200) / 100)
V1 = 2 M/S


So far so good,

Now let's do it again. Aiming
the rocket in the same direction as
before the spaceman adds another 200
joules of energy to his velocity.

The spaceman's new frame is moving
at 2 M/s relative to the previous one.

Unfortunately the spaceman
forgot about the brick wall
in the original fixed reference
frame, and smashes into it.

So, let's calculate the energy he
expended against the brick wall,
assuming that he doesn't bounce off.

His velocity appears to be,

Vtot = 2M/s + 2M/s = 4 M/s

Thus the energy the brick wall gets is,

Ew = 0.5 * m * Vtot^2

or,

Ew = 0.5 * 100 * (4 M/s * 4 M/s)
Ew = 800 joules.

Oops, we just violated something as
we only put 400 joules (two blasts
of our rocket) into the system.

But wait.

Since we know that we only put 400 joules
into the system, let's calculate the
his final velocity, again using the
basic kinetic energy relation.

Vf = sqrt ((2 * e)/ m)

or,

Vf = sqrt ((2 * 400)/100)
Vf = 2.8 M/s

Hmmmm ....

Something is wrong somewhere.

I know at several items are being ignored.

- The spaceman's decreased mass as rocket fuel (or bullets) are
expended is ignored, but it seems as the conflict is so
egregious that this might not matter.

- The conservation of momentum details enabling us to add
200 joules worth of energy to our velocity is also being
ignored, but it seems as this could be replaced by
kicking off a pole in the fixed frame as we go by.

I did some looking on the web and couldn't find much.

The only thing I can think of is that combining energy
across reference frames is problematic (at least to me).

Any help greatly appreciated.

Tom


When firing a handgun most people are more concerned with the
energy of the projectile than the energy of the recoil.
After all, getting shot and shooting someone are very different events.

Since momentum is conserved, the mass of the bullet times the velocity
of the bullet equals the mass of the shooter times the velocity of the
shooter, as per Newton's third law. i.e. Mv + mV = 0.

Having completely ignored the energy of the two projectiles, the second
of which has a lower velocity (relative to the fixed point 'A') than the
first,
it cannot be surprising that you find the result perplexing. I suggest you
recalculate (using a spread*** since this is a computer "science" project)
without leaving anything out. There are no short cuts to doing it properly
and guessing as you are doing has led to failure.
With a spread*** you can easily change the initial conditions and fully
understand what is going on.

That's the best help I can give your student.




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