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Conservation of linear momentum

  1. Jan 17, 2005 #1
    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 travelling to hurtle a cow 30feet (10metres) down a road?!

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

    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Jan 17, 2005 #2


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    Do you know what the conservation of linear momentum is?

    - Warren
  4. Jan 17, 2005 #3
    I'm afraid Im a complete novice. I left school with no qualifications and I work selling pizzas.. Thats why I need your help. Please help.
  5. Jan 17, 2005 #4
    wouldn't the weight of the cow be needed?
  6. Jan 17, 2005 #5
    average weight of the cow is 700kg
  7. Jan 17, 2005 #6
    and the car weighs 1500kg if that helps
  8. Jan 17, 2005 #7


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    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:


    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
  9. Jan 17, 2005 #8


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    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
  10. Jan 17, 2005 #9
    lets just say it went 15feet through the air and skidded 15ft on the floor
  11. Jan 17, 2005 #10


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    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:


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


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


    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


    or about 16 mph:


    - Warren
  12. Jan 17, 2005 #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: Jan 17, 2005
  13. Jan 17, 2005 #12
    and what is the minimum speed?! 16mph as stated above?!
  14. Jan 17, 2005 #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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