A couple of physics questions

[Dr Jekyll]

kiseca said:

Square of its speed relative to what? The ground, or the object it hits?

Relative to the speed after the collision.

[kiseca]

Dr Jekyll said:

kiseca said:

Square of its speed relative to what? The ground, or the object it hits?

Relative to the speed after the collision.

Still, speed compared to what?

Two cars hit eachother head on, both travelling at 60mph. How much kinetic energy is shared in the collision?

Ground speed is arbitary. The ground is moving too. Nothing is absolutely stationary. IMO the only time you'd use speed compared to ground in a kinetic energy equation is when the object is striking the ground.

[4Q]

Maybe I’m missing something obvious but if someone runs into my outstretched fist it would cause far less damage than if I punched them as they ran at me?

[thebraketester]

Surely it’s velocity relative to the point where KE of “the system” =0

Edited by thebraketester on Friday 31st July 09:23

[thebraketester]

4Q said:

Maybe I’m missing something obvious but if someone runs into my outstretched fist it would cause far less damage than if I punched them as they ran at me?

Quite. But that’s a different proposition to what the original post describes.

[Kawasicki]

That this is still questioned baffles me.

[Dr Jekyll]

kiseca said:

Dr Jekyll said:

kiseca said:

Square of its speed relative to what? The ground, or the object it hits?

Relative to the speed after the collision.

Still, speed compared to what?

Two cars hit eachother head on, both travelling at 60mph. How much kinetic energy is shared in the collision?

Ground speed is arbitary. The ground is moving too. Nothing is absolutely stationary. IMO the only time you'd use speed compared to ground in a kinetic energy equation is when the object is striking the ground.

It doesn't matter. We are trying to measure the kinetic energy converted in the collision, providing we measure the speed on the same basis before and after we can work it out.

[Monty Python]

kiseca said:

Still, speed compared to what?

Two cars hit eachother head on, both travelling at 60mph. How much kinetic energy is shared in the collision?

Ground speed is arbitary. The ground is moving too. Nothing is absolutely stationary. IMO the only time you'd use speed compared to ground in a kinetic energy equation is when the object is striking the ground.

"The ground is moving too"? So if you stand still, the ground will suddenly move under your feet?

[kiseca]

Dr Jekyll said:

kiseca said:

Dr Jekyll said:

kiseca said:

Square of its speed relative to what? The ground, or the object it hits?

Relative to the speed after the collision.

Still, speed compared to what?

Two cars hit eachother head on, both travelling at 60mph. How much kinetic energy is shared in the collision?

Ground speed is arbitary. The ground is moving too. Nothing is absolutely stationary. IMO the only time you'd use speed compared to ground in a kinetic energy equation is when the object is striking the ground.

It doesn't matter. We are trying to measure the kinetic energy converted in the collision, providing we measure the speed on the same basis before and after we can work it out.

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

[thebraketester]

kiseca said:

Dr Jekyll said:

kiseca said:

Dr Jekyll said:

kiseca said:

Square of its speed relative to what? The ground, or the object it hits?

Relative to the speed after the collision.

Still, speed compared to what?

Two cars hit eachother head on, both travelling at 60mph. How much kinetic energy is shared in the collision?

Ground speed is arbitary. The ground is moving too. Nothing is absolutely stationary. IMO the only time you'd use speed compared to ground in a kinetic energy equation is when the object is striking the ground.

It doesn't matter. We are trying to measure the kinetic energy converted in the collision, providing we measure the speed on the same basis before and after we can work it out.

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

Post collision you now have double the mass.

You are getting into conservation of momentum territory now.

[kiseca]

Where did the extra mass come from?

[thebraketester]

Post collision in your example you have two cars now with half the initial velocity but double the mass.

[Monty Python]

I think the problem is that you're treating the two cars as somehow being linked together prior to the collision - they're not. They are independent bodies that each possess the same amount of energy that's a consequence of their velocity and mass. When they collide, each of them has the same amount of energy to dissipate.

It's logical when you think about it - if you're in a car doing 120mph, I suspect you wouldn't survive a head-on collision with another car.

[kiseca]

kiseca said:

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

How about someone answers this? I'm sure it will clear up my question.

Monty Python said:

"The ground is moving too"? So if you stand still, the ground will suddenly move under your feet?

Only if the ground starts moving relative to you. My problem is everyone seems to be treating velocity along the ground as absolute zero. I'm saying if you're talking about two moving objects, then the ground isn't the correct reference point. What if it were two aircraft instead? Or two satellites? When two spaceships dock, they're both doing maybe 15,000mph, but because they are doing maybe 1mph relative to eachother, the docking doesn't end in a catastrophic fireball. Why does having wheels on the ground make that different? If a car doing 120 rear ends a car doing 119 in the same direction, you might not even have a mark to show for it.

For all that momentum to turn into a destructive situation, it matters what the two soon-to-collide items are doing relative to eachother leading up to the impact.

Which brings me back to this:

kiseca said:

one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

Edited by kiseca on Friday 31st July 13:24

[Kawasicki]

kiseca said:

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

The energy involved in the crash deformation in both scenarios is the same.

What's your point?

[kiseca]

Kawasicki said:

kiseca said:

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

The energy involved in the crash deformation in both scenarios is the same.

What's your point?

So 1 car doing 120 hitting an identical stationary car is equal in severity to the same two cars each doing 60mph and having a head on collision?

If that's correct then you're agreeing with my original point.

[Big-Bo-Beep]

Fascinating discussion.

It's a good job that science is 100% fact based and not just wishy-washy interpretation, opinion and speculation.

So, have we reached an opinion that all are agreed on ?

[Dr Jekyll]

kiseca said:

kiseca said:

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

How about someone answers this? I'm sure it will clear up my question.

Monty Python said:

"The ground is moving too"? So if you stand still, the ground will suddenly move under your feet?

Only if the ground starts moving relative to you. My problem is everyone seems to be treating velocity along the ground as absolute zero. I'm saying if you're talking about two moving objects, then the ground isn't the correct reference point. What if it were two aircraft instead? Or two satellites? When two spaceships dock, they're both doing maybe 15,000mph, but because they are doing maybe 1mph relative to eachother, the docking doesn't end in a catastrophic fireball. Why does having wheels on the ground make that different? If a car doing 120 rear ends a car doing 119 in the same direction, you might not even have a mark to show for it.

For all that momentum to turn into a destructive situation, it matters what the two soon-to-collide items are doing relative to eachother leading up to the impact.

Which brings me back to this:

kiseca said:

one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

What matters is the difference between the what they are doing before, and what they are doing after, If a car doing 120 slows to 119 suddenly it has almost as much kinetic energy after as before. If it slows to zero suddenly that energy that's a lot of energy to be released. If it slows to 60 because it's hit another car and both are now doing 60, then the energy is absorbed by accelerating the other car rather than by heating up brake pads or destroying the crumple zones and making a loud bang.

Edited by kiseca on Friday 31st July 13:24

[kiseca]

Dr Jekyll said:

kiseca said:

kiseca said:

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

How about someone answers this? I'm sure it will clear up my question.

Monty Python said:

"The ground is moving too"? So if you stand still, the ground will suddenly move under your feet?

Only if the ground starts moving relative to you. My problem is everyone seems to be treating velocity along the ground as absolute zero. I'm saying if you're talking about two moving objects, then the ground isn't the correct reference point. What if it were two aircraft instead? Or two satellites? When two spaceships dock, they're both doing maybe 15,000mph, but because they are doing maybe 1mph relative to eachother, the docking doesn't end in a catastrophic fireball. Why does having wheels on the ground make that different? If a car doing 120 rear ends a car doing 119 in the same direction, you might not even have a mark to show for it.

For all that momentum to turn into a destructive situation, it matters what the two soon-to-collide items are doing relative to eachother leading up to the impact.

Which brings me back to this:

kiseca said:

one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

What matters is the difference between the what they are doing before, and what they are doing after, If a car doing 120 slows to 119 suddenly it has almost as much kinetic energy after as before. If it slows to zero suddenly that energy that's a lot of energy to be released. If it slows to 60 because it's hit another car and both are now doing 60, then the energy is absorbed by accelerating the other car rather than by heating up brake pads or destroying the crumple zones and making a loud bang.

Edited by kiseca on Friday 31st July 13:24

So, I read that, I think I understand it, and I conclude that the two scenarios in my question would give the same results. Do you conclude the same?

Edited by kiseca on Friday 31st July 13:50

[Kawasicki]

kiseca said:

Kawasicki said:

kiseca said:

OK, so start with one car doing 120mph and the other doing 0. They hit eachother. Immediately after the collision, both cars are now doing 60mph in the same direction. Then, compare that to two cars doing 60mph. After the collision, they're both doing 0mph.

What values do you put in the equation?

To me, the results will be the same. If that's wrong, that's where my understanding needs to be fixed.

Edited by kiseca on Friday 31st July 10:00

The energy involved in the crash deformation in both scenarios is the same.

What's your point?

So 1 car doing 120 hitting an identical stationary car is equal in severity to the same two cars each doing 60mph and having a head on collision?

If that's correct then you're agreeing with my original point.

Yes, I agree with you.

Edited to add that your earlier explanation was quite smart.