Speed needed to knock a cow 10 metres down a road

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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.
 
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
 
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
 
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: )
 
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The most favourable situation! The minimum speed of the car, able to throw the cow at the distance b from the impact point!