Conservation of linear momentum

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Discussion Overview

The discussion revolves around the application of conservation of linear momentum to determine the speed a car must travel to propel a cow a certain distance down a road after an impact. The scenario includes various assumptions about the cow's weight, the car's weight, and the conditions of the impact, including whether the cow flies through the air or skids along the road.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks for help in calculating the speed of a car needed to propel a cow 30 feet down a road, indicating a lack of confidence in their physics knowledge.
  • Another participant introduces the concept of conservation of linear momentum and questions whether the weight of the cow is necessary for the calculations.
  • Participants provide estimates of the cow's weight (700 kg) and the car's weight (1500 kg) to facilitate calculations.
  • One participant suggests that the cow's velocity after impact could be roughly 3/2 of the car's initial velocity, depending on the weight ratio of the cow to the car.
  • Another participant proposes that if the cow is considered to weigh half as much as the car, its velocity after impact would be about 4/3 of the car's initial velocity.
  • Participants discuss the specifics of the cow's trajectory, including whether it flies through the air or skids along the road, and the implications of these conditions on the calculations.
  • One participant calculates the friction force acting on the cow and estimates the distance it would travel after impact, leading to a proposed car speed of about 7.42 m/s (approximately 16 mph) to achieve the desired distance.
  • A later reply introduces a different approach using equations related to momentum conservation and projectile motion, suggesting a minimum speed for the car based on specific conditions.

Areas of Agreement / Disagreement

Participants express various viewpoints on the calculations and assumptions involved, with no consensus reached on a definitive answer. Different models and approaches are proposed, leading to ongoing debate about the conditions and parameters necessary for the calculations.

Contextual Notes

The discussion includes assumptions about the cow's trajectory, the coefficient of friction, and the specifics of the impact, all of which remain unresolved and may affect the calculations.

DomStone
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I've stupidly told a friend that I am good at physics, but I am really not. He's set me a question that he wants answering, with an equation. The question is

How fast does a car have to be traveling to hurtle a cow 30feet (10metres) down a road?!

Sounds stupid I know, but any help you could give me would be greatly appreciated!

Thanks
 
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Do you know what the conservation of linear momentum is?

- Warren
 
I'm afraid I am a complete novice. I left school with no qualifications and I work selling pizzas.. Thats why I need your help. Please help.
 
wouldn't the weight of the cow be needed?
 
average weight of the cow is 700kg
 
and the car weighs 1500kg if that helps
 
Well, apparently you can think of a cow as being roughly a third of the weight of a car. (A cow's about 1,000 lbs, and a car's about 3,000 lbs).

This means that the cow's velocity after impact is going to be roughly 3/2 of the car's initial velocity. If the car is going 30 mph, the cow will start off going about 45 mph. You can see this by typing some numbers into the calculator here:

http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html#c5

The difficulty with this problem is that it's not well-specified. Does the cow fly through the air for those 30 feet, or does it scrape along the road? What is the coefficient of friction between cow and asphalt?

This isn't really a question that can be met with a precise, specific answer, but you can explain to your friend the thought process involved.

- Warren
 
Or, if you'd prefer a heavier cow (half the weight of a car), then the cow's velocity after impact would be about 4/3 the car's initial velocity. If the car was going 30 mph, the cow would start off going 40 mph.

What happens to the cow afterwards depends highly on the specifics of the cow's, um, adventure.

- Warren
 
lets just say it went 15feet through the air and skidded 15ft on the floor
 
  • #10
During the time the cow is in the air, there is some small amount of friction due to wind resistance, but it's almost certainly small enough to be ignored. Therefore, while the cow is in the air, it is not slowing down at all.

The problem comes down, again, to the coefficient of friction between the cow and the road.

If you want to just make an assumption that the coefficient is 0.5, then you can solve the problem -- the friction slowing the cow has a magnitude that's half that of the cow's weight. That'd be about 3,430 N:

http://www.google.com/search?hl=en&lr=&q=(9.8+m/s^2)+*+700+kg+*+0.5&btnG=Search

The cow, with an initial velocity of 40 mph, would have kinetic energy of about 112,000 joules.

http://www.google.com/search?hl=en&lr=&q=0.5+*+700+kg+*+(40+mph)^2&btnG=Search

When subjected to a friction force of 3,430 N, it would take a distance of about 30 meters:

http://www.google.com/search?hl=en&lr=&q=(0.5+*+700+kg+*+(40+mph)^2)+/+(3430+N)&btnG=Search

That's quite a bit too long, of course -- almost 90 feet.

The entire equation would simplify to

[tex]\frac{\frac{4}{3} v_\textrm{car}^2}{g} == 10[/tex]

Solving this equation, the car would have to be going about 7.42 m/s

http://www.google.com/search?hl=en&lr=&q=((4/3)+*+7.42+m/s)^2/(9.8+m/s^2)&btnG=Search

or about 16 mph:

http://www.google.com/search?hl=en&lr=&q=7.42+m/s+in+mph&btnG=Search

- Warren
 
  • #11
I propose a solution:

We consider that the cow's velocity after impact is at [tex]45^o[/tex] (that is the most convenable situation). From the momentum conservation we have
(1) [tex]Mv_0=mvcos 45^o[/tex]
and the "gunshot" of the cow
(2) [tex]b=\frac{v^2}{g}sin 2 \alpha[/tex]
where alpha=45 degree.

From these two equations you will get the minimum speed of the car
[tex]v_0=\frac{m}{M} \sqrt{\frac{bg}{2}}[/tex]

(sometime I would like to eat a pizza at your restaurant :smile: )
 
Last edited:
  • #12
and what is the minimum speed?! 16mph as stated above?!
 
  • #13
The most favourable situation! The minimum speed of the car, able to throw the cow at the distance b from the impact point!
 

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