relativity

bdbafh said:

Sam the Bam said:

A bicyclist is climbing a steep hill, 5 mph. Another
is heading down, 30 mph. Same size, same bikes.
They collide.

Who gets hurt worse?

Sam

The center of mass for the descending rider will be higher, as its "up
the slope".

The riders are at the same hight at the point of collision.
The riders have the same relative velocity with respect to each other.
The riders have the same relative momentum with respect to each other.
The riders have the same relative kinetic energy wrt each other.

if a model is considered in which the bicyclists are replaced with two
perfect spherical balls of the same size, the balls will experience
equal damage and will experience an equal change of speed, as
described in the conservation of momentum post.

Since the bicycles' shapes are very complex, luck will be more of a
factor here than physics.

But this is an experiment that can be easily done. Make 100
colliisions and see what the statistics shows. Can dummies ride
bicycles?

My answer is - nobody knows.

<[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

if a model is considered in which the bicyclists are replaced with two
perfect spherical balls of the same size, the balls will experience
equal damage and will experience an equal change of speed, as
described in the conservation of momentum post.

Since the bicycles' shapes are very complex, luck will be more of a
factor here than physics.

But this is an experiment that can be easily done. Make 100
colliisions and see what the statistics shows. Can dummies ride
bicycles?

My answer is - nobody knows.

Hal has proved they can post on rec.martial-arts so I guess anything is
possible.

Fraser

Sam the Bam said:

A bicyclist is climbing a steep hill, 5 mph. Another
is heading down, 30 mph. Same size, same bikes.
They collide.

Who gets hurt worse?

Sam

IIRC

There's a lot of ways to work this out. Long story short: the person
who experiences the largest change in velcocity is likely to be hurt
the most. Velocity relates to force (F=ma where a can be derived to
(v2-v1)/t), which relates to 'hurt' (Impulse - Force x Time).

However, in the case provided, you've provided insufficient
information. At a minimum, need to provide mass of cyclists. Then you
can calculate collison momentum and in which direction (of course, the
collison would likely tend to favour the cyclist with higher mass).

Quoted message said:

From this you can derive force at point of impact (and in which

direction). Of course, to an extent, that assumes rigid body
mechanics. This you are left to decide where the impact is along the
perfectly plastic : perfectly elastic scale (ie: how much of the
resultant force goes into deforming the cyclist). Not sure what the
value is for a human body - certainly less than 1 (no momentum lost,
bounce off each other) but obviously more than 0 (all momentum lost in
one cyclist, velocity turned into deformation energy, heat, sound etc)

Simplified version (again, IIRC), assuming a more plastic type
collision:

Eg:
Cyclist A = 80kg
Cylist B = 70 kg
Velocity A = 8.05 km/hr (5mph)
Velocity B = 48.3 km/hr (30mph)
Time in impact = 1 second

=(m1v1)+(m2v2)=(m1+m2)Vfinal
=(80*8.05) + (70*-48.3)=(80+70)Vfinal
=644 + (-3381) = (150)Vf
-2737 = 150Vf
Resultant velocity is 18.24ms in the direction of the heavier rider

F=ma
F=(80+70)a
=(80+70)x(v2-v1)/1
=(150) x (18.24-8.05)/1
=1528.5N

I'd be happier with the answer if I knew the height of the hill
cyclist A was coming down from. It gives you a way to double check.

Again - IIRC. It's been a while....

Quoted message said:

Simplified version (again, IIRC), assuming a more plastic type
collision:

Eg:
Cyclist A = 80kg
Cyclist B = 70 kg
Velocity A = 8.05 km/hr (5mph)
Velocity B = 48.3 km/hr (30mph)
Time in impact = 1 second

=(m1v1)+(m2v2)=(m1+m2)Vfinal
=(80*8.05) + (70*-48.3)=(80+70)Vfinal
=644 + (-3381) = (150)Vf
-2737 = 150Vf
Resultant velocity is 18.24ms in the direction of the heavier rider

That should read "In the direction of the lighter rider" / guy coming
down the hill / one with more momentum.

Also to clarify: Impulse = Force x Time

Quoted message said:

F=ma
F=(80+70)a
=(80+70)x(v2-v1)/1
=(150) x (18.24-8.05)/1
=1528.5N

So assuming the above figure is correct (1528.5N), each person would
'feel' 1528.5Ns-1 of impact force. That still leaves you with the
problem of working out constants of collision / coefficient of
restitution.

.... and I'm not so sure about the F=ma calculation either, on
reflection.

Surface area of each cyclist is also going to play a role re: pressure
(higher pressure is going to be felt as more pain)

My gut tells me it's along these lines, but I'm not a physics whizz.
Corrections from sci.physics gladly accepted, as I'm curious too.

....and I've just realised I failed to convert km/h into m/sec, so I'll
stop here and bow out.

The climbing rider gets it worse. Given all of those assumptions,
both riders fare about the same in the primary collision. In the
secondary collision (with the ground), the climbing rider is probably
going to land flat on his back. Unless the two get tangled up, the
descending rider is likely to get his arms up and roll when he lands.
It doesn't matter how much momentum he transfers to the climbing
rider. Hitting the road back first is almost always going to do more
damage than landing arms first.

Quoted message said:

"Sam the Bam" <[email hidden]> wrote in message news:
[email hidden]...

Quoted message said:

A bicyclist is climbing a steep hill, 5 mph. Another
is heading down, 30 mph. Same size, same bikes.
They collide.

Who gets hurt worse?

Sam

All you need is the Law of Conservation of Linear Momentum.

You didn't specify how steep the hill is, but that is not important in
this regard. We can simply ignore that, and assume it is level.

The momentum in the 5mph cyclist is (Mass x 5), while the momentum is
the 30mph cyclist is (-Mass x 30). After they collide, the total
momentum is (-Mass x 25). The combined twisted wreck will be
travelling at a speed of (-Mass x 25)/(2 x Mass)=(-12.5 mph).

That is, after they have collided, the combined twisted wreck will be
travelling at 12.5 mph in the downhill direction. Since the downhill
cyclist suffers a reduction in speed of 17.5 mph, and the uphill
cyclist suffers a change in speed of (5+12.5=17.5)mph, both of them
have a change in speed of 17.5mph. And because Force is proportional
to dV/dt, they both will experience the same traumatic force due to
the collision.

Do you agree?

No.

Which rider would YOU rather be?

Bill "creamer, not the creamee" S.

Quoted message said:

Sam the Bam said:

A bicyclist is climbing a steep hill, 5 mph. Another
is heading down, 30 mph. Same size, same bikes.
They collide.

Who gets hurt worse?

Sam

IIRC

There's a lot of ways to work this out. Long story short: the person
who experiences the largest change in velcocity is likely to be hurt
the most. Velocity relates to force (F=ma where a can be derived to
(v2-v1)/t), which relates to 'hurt' (Impulse - Force x Time).

Equal mass, equal momentum ==> each will experience same change in
velocity.

Sam Wormley said:

[email hidden] wrote:
Equal mass, equal momentum ==> each will experience same change in
velocity.

Right.

I think I must have read something different when I saw 'same size'.
In retrospect, I'm guessing he meant 'same mass', as opposed to where
my head was at (re: same physical size but different masses).

I think the other points I made (re: coefficient of restitution, time
taken in impacting, surface area / pressure) are valid when it comes
to deciding who 'hurts' more? Or not?

I don't think it's just down to 'velocity' (impulse) alone. Else, the
simple out is to say they both 'hurt equally', when gut instinct would
indicate otherwise.

Sam Wormley said:

Quoted message said:

Quoted message said:

A bicyclist is climbing a steep hill, 5 mph. Another is heading
down, 30 mph. Same size, same bikes. They collide.
Who gets hurt worse?

Quoted message said:

Center of mass calculations. Obviously the guy going fast is worse
off. I notice that when a train hits a standing car. It destroys the
train... doesn't it?

In this case the train and car have equal mass.
Recite Newton's third law. And remember the motion is relative.

Yes, that was the point. (see the subject header)
Newton's Third says it's equal. (Newton invented
relativity, long before that screwball pacifist!) But
intuition says the slower rider, going uphill, gets creamed.

We know intuition is often wrong, in questions
of science. Is it also off base here? Is there
some other factor, such that intuition is correct?

More generally, we are interested in striking,
and damage. On impact, per Newton's Third,
both players feel it equally. Why does the punchee
ouch more than the puncher?

Sam

Sam the Bam said:

Sam Wormley said:

Quoted message said:

> A bicyclist is climbing a steep hill, 5 mph. Another is heading
> down, 30 mph. Same size, same bikes. They collide.
> Who gets hurt worse?

Quoted message said:

Quoted message said:

Center of mass calculations. Obviously the guy going fast is worse
off. I notice that when a train hits a standing car. It destroys the
train... doesn't it?

Quoted message said:

In this case the train and car have equal mass.
Recite Newton's third law. And remember the motion is relative.

Yes, that was the point. (see the subject header)
Newton's Third says it's equal. (Newton invented
relativity, long before that screwball pacifist!) But
intuition says the slower rider, going uphill, gets creamed.

We know intuition is often wrong, in questions
of science. Is it also off base here? Is there
some other factor, such that intuition is correct?

No. The intuition is incorrect.

When a mosquito hits the windshield of a truck, there is as much
impact on the mosquito as there is on the truck, physically speaking.

Now, practically, the windshield can absorb MUCH more impact than the
mosquito can absorb before exhibiting effects. But the force exerted
by the mosquito on the truck is EXACTLY equal to the force the truck
exerts in the mosquito.

Quoted message said:

More generally, we are interested in striking,
and damage. On impact, per Newton's Third,
both players feel it equally. Why does the punchee
ouch more than the puncher?

Because the fist can endure more force with fewer ill effects than the
parts of the body the fist is aimed at. If you want to see a more
balanced example, then have somebody punch someone else's clenched
fist as hard as they can, and THEN see whether the punchee hurts more
than the puncher.

Quoted message said:

Sam

Quoted post said:

Quoted message said:

Since the downhill cyclist suffers a reduction in speed of
17.5 mph, and the uphill cyclist suffers a change in speed of
(5+12.5=17.5) mph, both of them have a change in speed of
17.5mph. And because Force is proportional
to dV/dt, they both will experience the same traumatic force
due to the collision.
Do you agree?

No. The guy going uphill gets way more whiplash.

huh?

Sam

Quoted message said:

Quoted message said:

A bicyclist is climbing a steep hill, 5 mph. Another is heading
down, 30 mph. Same size, same bikes. They collide.
Who gets hurt worse?

Center of mass calculations. Obviously the guy going
fast is worse off. I notice that when a train hits a standing car.
It destroys the train... doesn't it?

Practically speaking, the fast rider is worse off because it isn't a
mirror image collision so there is some sort of glancing blow. After
colliding both riders may be equally injured but the fast guy still
has a hard collision with the ground or other obstacle coming before
he stops.

I was thinking about the collision only, not hitting the
pavement or whatever. It wasn't a trick question.

Sam

Quoted message said:

Who gets hurt worse?

Is this a helmet thread?

"Sam the Bam" <[email hidden]> wrote in message news:
[email hidden]...

Quoted message said:

More generally, we are interested in striking,
and damage. On impact, per Newton's Third,
both players feel it equally. Why does the punchee
ouch more than the puncher?

Sam

Your cyclists are not a good example to compare the punchee and the
puncher.

The puncher packs the kinetic energy of the travelling arm and fist,
plus the rotational energy of his torso into his punching target via
the front end of his fist.

The punchee uses a small area of his body (equivalent to the surface
area of the front end of the puncher's fist) to absorb the puncher's
kinetic energy.

If the puncher's fist lands on the punchee's head, the head is too
hard and light in weight to absorb the kinetic energy and will fly off
but the neck is trying to hold it in place.

If the puncher's fist lands on the punchee's torso, the part of the
torso at the landing site will absorb the incoming kinetic energy
which will result in structural damage because the energy needed to
cancel out (decelerate) the incoming fist is equal to the integral of
(Force) x (Distance travelled by the incoming fist after impact), but
the contact area is so small that the pressure, which is (Force/Area),
will break the ribs or bruise the flesh and internal organs.

Wannabe
=======

Quoted message said:

"Sam the Bam" <[email hidden]> wrote in message news:
[email hidden]...

Quoted message said:

A bicyclist is climbing a steep hill, 5 mph. Another
is heading down, 30 mph. Same size, same bikes.
They collide.

Quoted message said:

Who gets hurt worse?

Quoted message said:

Sam

All you need is the Law of Conservation of Linear Momentum.

You didn't specify how steep the hill is, but that is not important in
this regard. We can simply ignore that, and assume it is level.

The momentum in the 5mph cyclist is (Mass x 5), while the momentum is
the 30mph cyclist is (-Mass x 30). After they collide, the total
momentum is (-Mass x 25). The combined twisted wreck will be
travelling at a speed of (-Mass x 25)/(2 x Mass)=(-12.5 mph).

That is, after they have collided, the combined twisted wreck will be
travelling at 12.5 mph in the downhill direction. Since the downhill
cyclist suffers a reduction in speed of 17.5 mph, and the uphill
cyclist suffers a change in speed of (5+12.5=17.5)mph, both of them
have a change in speed of 17.5mph. And because Force is proportional
to dV/dt, they both will experience the same traumatic force due to
the collision.

Do you agree?

Wannabe
=======

What if one of the cyclists was your mom!

http://www.yikers.com/picture_gallery_funny_pictures_fattest_ass.html

Scary.