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I can't seem to figure out the relationship between $E_k$ and $p$ or $F$. I understand that the units are pretty different. But for example:

A bullet with a mass of 10.0g is moving at the speed of 1000m/s. A bull with a mass of 400kg is charging at the speed of 5.00m/s Which has the greater kinetic energy?

5000J for the bullet and 5000J for the bull

That worked fine but now if you calculate momentum you get:
$p = 10 kg\cdot m/s$ for the bullet and $p = 2000 kg\cdot m/s$ for the bull.
Which does not make sense!

Does the bull hit harder than the bullet or the same amount as the bullet?

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Why do You think Your result "does not make sense"? I'm quite comfortable with it. –  Georg Jun 2 '11 at 18:42
    
I don't understand it. Kinetic energy, force, and momentum are all similar kind of the same thing, right? I think i'm missing something. –  Dan the Man Jun 2 '11 at 18:47
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"" Kinetic energy, force, and momentum are all similar kind of the same thing, right? "" Not at all. Kinetic energy is mass times velocity squared, momentum is mass times velocity. What makes You to throw in force here, is something I do not understand. –  Georg Jun 2 '11 at 18:50
    
They are closely related, but they are not "kind of the same thing". –  Lagerbaer Jun 2 '11 at 18:50
    
I see. it seems to make more sense now. thanks guys. –  Dan the Man Jun 2 '11 at 18:59
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4 Answers

up vote 15 down vote accepted

If you write down the formulas for kinetic energy, $$E_k = \frac{1}{2} m v^2$$ and momentum $$p = mv$$ you see that you can write the energy in terms of momentum via $$E_k = \frac{p^2}{2m}$$

So, if two objects have the same energy $E_k$, they only have the same momentum if they also have the same mass. Since the bull has a much larger mass than the bullet, it must therefore have a much larger momentum than the bullet to arrive at the same kinetic energy.

Force is change of momentum with time, $F = \dot p$. If we assume that the bullet and the bull hit a target and come to rest in the same time, the bull hits much harder, as it has the higher momentum.

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Thanks a lot! it makes more sense now –  Dan the Man Jun 2 '11 at 19:00
    
Classically, couldn't the momentum of the bullet and the bull also be equal if the bullet is moving really fast? –  Brandon_R Jun 3 '11 at 2:41
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@Brandon_R Momentum depends linearly on mass and velocity. When the bull is $x$ times heavier than the bullet, the bullet needs to be $x$ times faster than the bull. In the example above, they do have the same momentum. But you cannot have two objects have the same momentum and the same energy while having different mass. –  Lagerbaer Jun 3 '11 at 3:36
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If we apply a given force $F$ for a given time $t$, it changes the momentum by $Ft$. If in that time the force has moved whatever it's pushing by a distance $s$, it changes the energy by $Fs$. If we push a light thing and a heavy thing equally hard for the same amount of time --so they both have the same momentum-- the light thing will be pushed farther in the same time because it accelerates more, so it has more energy. Equally, if we push them both equally hard until they have the same energy --that is, for the same distance, remembering that in this kind of Physics the bull puts up no resistance other than its mass (and it may even be a Bull Sphere)-- the light thing will have less momentum, because it will take less time to get there.

When it comes to stopping the bull, you have to keep up a force that's trying to stop it for much more time, the bull keeps on coming, whereas the same force will stop the bullet in the same distance.

There's a fourth concept that also has to be considered, which is pressure. Sorry! Force, Energy, Momentum, and Pressure are all different. To apply a force to the small area that a bullet has is different from applying the same force to the big area that the bull has. It's more difficult, it needs more pressure, to apply the same force to a smaller area.

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They're not the same thing. They have very different implications.

You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc.

As soon as the object hits the target, the E(k) applies (i.e. the damage). The object will transfer the E(k) over time (very short), and as it creates damage (penetrattion) its E(k) is decreasing. The penetration is opposed by friction, which pushes the target backwards with the same Force as it's stopping the bullet.

Simple example: A football and a bullet have the same Momentum, but the bullet has much higher Kinetic Energy. The push (p) is the same, but the ball doesn't have enough E(k) to cause any damage so it just pushes the target. The bullet, however, will first penetrate it "spending" it's E(k) against the toughness, hardness and other factors, and as it's stopping within the target the friction pushes the target backwards resulting in the same push as the one of the ball (unless it pierces though it, in which case just a fraction of the momentum gets transferred)

In short: Energy causes damage. Force and Momentum cause movement.

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I'd argue that it isn't energy or momentum but pressure that makes something deadly. If you put a nearly massless spike on the tip of the football you don't increase it's energy or its momentum but you do make it deadly. –  Brandon Enright Dec 18 '13 at 5:09
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$$K.E.= \frac{1}{2} p v$$ since: $$K.E.=\frac{1}{2} m v^2$$ and $$p=mv$$

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Usually one writes this is $p^2/2m$ as Lagerbaer did, rather than with $pv$. –  Kyle Kanos Mar 20 at 12:35
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protected by Qmechanic Mar 20 at 10:36

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