# Kinetic Energy

1. May 12, 2006

### zbobet2012

A quick look through the FAQ and search didn't reveal anything so...

I know that K.e. = ½MV², however I am looking to find the energy of an impact when φ != 90°.

Or to the more specific problem, if I have a Spaceship shaped like a needle moving through space at .9c(Vnet) and a micrometor hits it with a mass of 1g(M) and the angle between the metorite and the hull is 10°, how much energy would the ship absorb.

This is not a homework question, which is why I didn't post it there...sorry should have posted that before the moderator moved it.

Last edited: May 12, 2006
2. May 12, 2006

### Andrew Mason

Without knowing the speed of the meteorite in the rest frame of the observer, this is not answerable. Also are we to assume that the meteorite sticks to the ship?

AM

3. May 12, 2006

### zbobet2012

The meteorite is assumed to be "at rest" (thus the Vnet next to the .9c). And no the meteorite does not stick to the spaceship, I am assuming the material used on the outer hull is hard enough to prevent elasticity. I am simply asking what the force exerted on the hull will be.

I don't think we will have to use the relativistic version here.

Last edited: May 12, 2006
4. May 12, 2006

### Andrew Mason

Well, you were asking what the energy would be absorbed by the ship. Which is it? Force or energy? You will need to give us the mass of the spaceship to work out the energy absorbed by the spaceship if the meteorite bounces off.

In analysing a collision you have to apply conservation of momentum. Think of the spaceship as being at rest and the meteorite travelling at a velocity of .9c 0° (S) and deflecting at a direction of 20° (SW, say). Use conservation of momentum to determine the speed and direction of the spaceship.

AM

Last edited: May 12, 2006
5. May 12, 2006

### zbobet2012

I intended to say energy there, I was being a bit loose with my terminology. Sorry.

6. May 13, 2006

### zbobet2012

After much work I have arrived at what should have been obvious , velocity is a vector quantity. And in a 2d plane that means it is composed of two parts, x and y. Without delving to deeply into any kind of (if basic) vector calculus, the x = vcosφ and y = vsinφ. Therefore the energy actually transferred to the ship would only be the sin componet, eg K.E.absorbed = 1/2M(vsinφ)^2.

7. May 14, 2006

### Curious3141

Basically, this sort of problem (object striking ship) should be analysed with conservation of momentum. After the final velocities of the ship and the object are worked out, then the kinetic energy of each can be computed to find out how much KE got "transferred".

Remember that unless the collision is perfectly elastic, Kinetic Energy is not conserved. There are unaccountable losses in heat and light (flash or spark) energy. Sound does not play a role here in a vaccuum but there are solid matter vibrations which dissipate energy also.

So it makes little sense to talk about figuring out how much KE was transferred without considering momentum transfer first.

BTW, a meteorite is a fragment of a meteor that has burned through the atmosphere to land on the ground. Similarly a meteor (even a "micrometeor") only exists within the context of the atmosphere. For your scenario (in space), I suppose you should call it an asteroid.

8. May 14, 2006

### Andrew Mason

As Curious and I have been saying, you have to use conservation of momentum. So you need to know the mass of the spaceship (M) as well as the mass of the meteor (m) and you also have to know the final velocity of the meteor:

$$m\vec v_{mf} + M\vec v_{ss} = m\vec v_{mi}$$

Work out the magnitude and direction of the space ship's velocity from the change in momentum of the meteor. Then use KE = $.5mv^2$ to determine the energy of the spaceship, which is the amount transferred.

AM

9. May 14, 2006

### zbobet2012

Curious: I wasn't trying to derive how much energy would be transferred in a given circumstance, but how much the ships outer hull would have to withstand to make the collison perfectly elastic. For example we can say that on a pool table momentum is conserved(within the limits of reason) because the pool balls are hard enough and strong enough that they don't break or vibrate etc. I am not saying this wouldn't happen on the ship, it would, but I am looking at the maximum amount of energy (in whatever form) that will be directed at the hull.

Last edited: May 14, 2006