Conservation of linear momentum

In summary, the problem presented is to calculate the minimum speed a car must travel at in order to propel a cow 30 feet down a road. After considering the weight of the cow and car, as well as the coefficient of friction between the cow and the road, it is determined that the cow's velocity after impact will be roughly 3/2 of the car's initial velocity. Using this information, the minimum speed needed for the car to successfully propel the cow 30 feet is calculated to be approximately 16 mph. However, this solution may vary depending on the specifics of the cow's movement and the coefficient of friction.
  • #1
DomStone
6
0
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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  • #2
Do you know what the conservation of linear momentum is?

- Warren
 
  • #3
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.
 
  • #4
wouldn't the weight of the cow be needed?
 
  • #5
average weight of the cow is 700kg
 
  • #6
and the car weighs 1500kg if that helps
 
  • #7
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
 
  • #8
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
 
  • #9
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: )
 
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  • #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!
 

FAQ: Conservation of linear momentum

What is the conservation of linear momentum?

The conservation of linear momentum is a fundamental law of physics that states that the total momentum of a system remains constant if there are no external forces acting on it. This means that the total momentum before an event or interaction is equal to the total momentum after the event or interaction.

How is linear momentum calculated?

Linear momentum is calculated by multiplying an object's mass by its velocity. It is a vector quantity, meaning it has both magnitude and direction. The SI unit for linear momentum is kilogram meters per second (kg·m/s).

What are examples of conservation of linear momentum in everyday life?

A common example of conservation of linear momentum is when a ball is thrown against a wall. The ball has a certain momentum as it approaches the wall, and after it bounces off the wall, it has the same momentum in the opposite direction. Another example is when two billiard balls collide, they will exchange momentum and continue moving in different directions.

How is conservation of linear momentum related to Newton's third law of motion?

Newton's third law states that for every action, there is an equal and opposite reaction. This means that when two objects interact, their forces on each other are equal and in opposite directions. The conservation of linear momentum is related to this law because the total momentum of the system remains constant, even though the individual objects may experience equal and opposite changes in momentum.

What are the implications of the conservation of linear momentum in terms of collisions?

The conservation of linear momentum is crucial in understanding collisions between objects. It allows us to predict the final velocities of objects after a collision, based on their initial velocities and masses. It also helps us understand the transfer of energy and the behavior of objects during collisions. Furthermore, the conservation of linear momentum is essential in designing safe and efficient transportation systems and in studying the dynamics of celestial bodies.

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