Bull charging and you charge the bull

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TheWonderer1
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Would you be better off standing your ground? I don't actually know how massive a bull is but let's say about 1,700 kg. It's coming at you with a velocity of 16 m/s. You got a human who is a decent 93 kg and runs in the opposite direction at 12.517 m/s (he's an olympian 100m runner). Would this decrease the impact rather? I'm thinking about conservation of momentum.

The total momentum of the system changes from about 25,600 to 24,600 (compared to standing). It's not that significantly different if he stood there with no momentum but I'm just thinking a bit.
 
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Apologies, I didn't think about kinetic energy and relative speed.
 
TheWonderer1 said:
Would this decrease the impact rather? I'm thinking about conservation of momentum.
One way to think of it is this: by your argument crashing your car into an identical, but stationary, car should trash your car but a head-on collision at speed shouldn't even scratch your bumpers. As Russ says, it's the momentum and energy in your rest frame that matters in determining how much the initial impact will hurt. The Earth frame only becomes important in thinking about how much hitting the ground will hurt.
 
TheWonderer1 said:
The total momentum of the system changes from about 25,600 to 24,600 (compared to standing). It's not that significantly different if he stood there with no momentum but I'm just thinking a bit.
What if the bull stood still in indifference, and the human ran at it?
What would you then say about the impact?
 
I think to answer the question it would hurt far less to run at a bull that was standing still because there is no momentum coming at you. However, it would still hurt.

I'm not sure what the reference point bit is getting at since you just mention that it matters only in your own reference point.
 
TheWonderer1 said:
I think to answer the question it would hurt far less to run at a bull that was standing still because there is no momentum coming at you. However, it would still hurt.
What if the bull was charging at you from the east? The Earth is rotating eastward and by a non-rotating standard there is no momentum coming at you at all.
 
jbriggs444 said:
What if the bull was charging at you from the east? The Earth is rotating eastward and by a non-rotating standard there is no momentum coming at you at all.

Ok but you would still feel that bull pretty badly.

Was I correct in my assumption that you running at bull would hurt less than the opposite?
 
TheWonderer1 said:
I'm not sure what the reference point bit is getting at since you just mention that it matters only in your own reference point.
Fair enough - I didn't say that well. Reference frame doesn't matter to the physics. But it's easier to solve particular problems in some frame, and in this case your rest frame has helpful properties. Particularly because you initially chose to work in it for the first half of the problem statement.
 
TheWonderer1 said:
Ok but you would still feel that bull pretty badly.

Was I correct in my assumption that you running at bull would hurt less than the opposite?
No.
 
jbriggs444 said:
No.
Why not? The overall momentum experienced by you would be less and same as your KE. I understand it's about what you experience so that's the involvement.
 
TheWonderer1 said:
Why not?
Because both are descriptions of the exact same collision. The movement (or not) of the Earth beneath the feet of the person or of the bull is irrelevant.
 
jbriggs444 said:
Because both are descriptions of the exact same collision. The movement (or not) of the Earth beneath the feet of the person or of the bull is irrelevant.
I don't see that since there are different amounts of momentum and energy involved.

Are we comparing the same scenario? If you had a bull charging at you vs. if you charged the bull?
 
Your change in velocity and the associated acceleration and force would be greater if you run toward the bull.
 
Ok there is some important aspect I'm missing. The frame of reference stuff changed my understanding and I get that pretty well. I see why the Earth moving meaning no momentum would be still a hurtful scenario. However, you charging at the bull who is at rest seems different completely. This is still stumping me as I see the bull sitting there in my head. I run at it and the bull feels very little. I feel my momentum bounce me backwards. However, if that bull decided to charge me, I would bounce much further than if I charged or stayed still hurting me a lot more. I'm going to test it :)
 
Force causes damage. F=MA. Acceleration is what counts. Whether you run into the bull at speed v or the bull runs into you at speed v doesn't matter. The acceleration is the same so the damage is the same. If you and the bull are running toward each other, that increases the velocity of the collision and increases the acceleration (=> force => damage) that you would suffer.
 
I would take one step to the side just before estimated time of impact or throw a red rag on the ground and run up a tree.

Fun fact: bulls run faster up hills than down hills.
 
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FactChecker said:
Force causes damage. F=MA. Acceleration is what counts. Whether you run into the bull at speed v or the bull runs into you at speed v doesn't matter. The acceleration is the same so the damage is the same. If you and the bull are running toward each other, that increases the velocity of the collision and increases the acceleration (=> force => damage) that you would suffer.
Ok just to get to the meat of what you are saying, running at a bull and the bull runs at you would mean greater acceleration than the bull at speed v OR the human at speed v. Therefore, you get hurt much worse.
 
Also consider the weight of the human, a big guy will get more impact from the same bull than a lighter guy.
 
Let's do a simple example with simple numbers. You have mass 100kg and the bull 1000kg. You can both run at 10m/s.

1. You are stationary and the bull runs at you at 10m/s. You get stuck on its horns. What is the velocity of the bull+you just after the collision? What is your kinetic energy before and after the collision?

2. The bull is stationary and you run at it at 10m/s. You get stuck on its horns. What is the velocity of the bull+you just after the collision? What is your kinetic energy before and after the collision?
 
It's the same 1100kg mass and the 20 m/s. I think that means you have like 440,000 Joules. That correct, I think the mass are combined because they are attached by the horns?
 
No. Where are you getting 20m/s from?

In the first case, momentum was Mv=1000×10=10,000kg m/s before the collision. Conservation of momentum means that Mv=(m+M)u, so u=9.1m/s for the stuck-together-man-and-bull.

In the second case, mv=100×(-10)=-1,000kg m/s before the collision. Conservation of momentum means that mv=(M+m)u, so this time u=-0.9m/s.

You answer the second part of the questions. What's your kinetic energy before and after the collision?
 
TheWonderer1 said:
This is still stumping me as I see the bull sitting there in my head. I run at it and the bull feels very little. I feel my momentum bounce me backwards. However, if that bull decided to charge me, I would bounce much further than if I charged or stayed still hurting me a lot more.

I think part of the problem is that, according to what I'm getting from you here, you are still thinking in terms of a reference frame of the Earth. Hence why you're saying "bounce much further", you're still thinking in terms of the physics with respect to the ground. You need to forget the ground is even there. Heck, I'd even replace you and the bull with two balls floating in space or some other scenario where all references to the Earth are gone. Then once you understand the physics behind collisions you can apply it to you and a bull.
 
891 J and 91,000 J and I believe it's the same kinetic energy before and after.
 
Are we assuming the bull maintains a constant velocity , highly unlikely.

This problem is so unrealistic in the context of a bull its useless, stick with hard balls in space.
 
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Was in the middle of fixing it :)

Ok before for the bull charging: 1/2mv which means that I think that is 50,000J. Also, I think the change is 45,500 J rather than what I said.
Other case: 5,000 J, same here it's 445.5 J.

I kind of rushed the calculation. It might be off