Conservation of Momentum and Energy

[Tetra]

Don't really get what to do...

 

A hunter shoots a 500g arrow at a 2kg bird perched on a tall tree growing on a flat ground. The arrow is launched from ground level with a speed of 40m/s at an angle of 30 degrees above the horizon. It is traveling horizontally when it strikes and embeds into the bird. How far from the base of the tree do the bird and arrow land?

 

So I found that (using the law of conservation of momentum) the momentum of the bird + arrow is 20kgm/s, and the velocity of the arrow + bird is 20m/s. And I know we can form a triangle thing out of the velocities or momentums but...I can't seems to find distance! Which is the thing we're looking for! Help?

[darkenlighten]

If you can find the height of the tree (where the bird is at) that means you can find how long it takes for them to fall. And if you know that you can find how far away from the tree they went.

Edited July 21, 2010 by darkenlighten

[Tetra]

So...what does it have to do with conserving momentum?

And for the height, we just use one of the kinematic equations?

[darkenlighten]

You use conservation of momentum to get the horizontal velocity after collision.

 

And yea since you know that the arrow is horizontal when hitting the bird, I believe it is safe to assume that is when the arrows vertical velocity is 0.

[Tetra]

Um...I muse be stupid or something...

 

So say I used that equation v2^2 = v1^2 + 2(a)(d) and I find d.

Then I find the horizontal velocity.Then i can find time, which is the amount of time it takes for the bird to drop. But since that's a vertical component, I'm supposed to use the vertical velocity to calculate for time. And, since time should be the same for both horizontal and vertical components, then I can use the horizontal component and time to find distance...is that it?

 

I thought there was another way of doing it...I guess I must have imagined it....

[swansont]

momentum of the bird + arrow is 20kgm/s.

 

You need to look at this again. That's the initial momentum of the arrow (.5 kg* 40 m/s) but it is launched at a 30º angle. You want the horizontal component; the vertical component becomes zero, owing to the force of gravity, at the time of the collision.