Exploding projectile problem

[Noob.com]

Homework Statement

Fire at angle 45 degree.
Initial kinetic energy = Eo
At top of projectile explode with addtional energy Eo into 2 fragment.
1. Mass = m1, go straight down.
2. Mass = m2, direction: unknown.

Question:
1. Velocity (mag and dir) of m2
2. Velocity of m1
3. What's ratio m1/m2 when m1 is maximun?

Homework Equations

Conservation of momentum:
mo x vo = m1 x v1 + m2 x v2
Eo = 1/2 mo x vo^2
mo = m1 + m2
Center of Mass:
No clue

The Attempt at a Solution

Ok, so I set up the conservation of momentum formula. Now what? How can I find the direction of m2? How can I use center of mass formula? Thanks!

 

[Delphi51]

"At top of projectile" must mean the top of the trajectory where vertical speed is zero.
I think the first step is to find the horizontal speed at that point so you know how fast the combined mass is going just before the explosion. You can't find a number for it, but you can get an expression involving the initial energy Eo.

 

[Noob.com]

huh? And why would I want to find that? Cos(45) x vo? For what?

Note that this problem is not a simple x and y direction! It's the conservation of center of mass, conservation of momentum and conservation of energy! I just don't know how to use it correctly!

 

[Delphi51]

Well, if you are going to use conservation of momentum during the explosion, you will need to know the momentum before. That is m x Cos(45) x vo.

 

[Noob.com]

Sorry but I don't see how can you start the problem with that! It's not like the right hand side of conservation of momentum have the variable vo. Beside I know exactly that this problem does not start from conservation of momentum formula. It got to start from conservation of center of mass formula which is mass x distance. I just don't know how to use it. Tell me how to use that center of mass formula not like telling me you should try this or try that! It seems like you just look at the problem and oh just try this, see if it will work or not! This is not an introduction Physics problem. It's classical Physics problem! Thank you!

 

[Delphi51]

Sorry, I don't see how conservation of center of mass applies. The thing is, the center of mass is moving both horizontally and vertically. The vertical movement stops at the top of the trajectory, but the horizontal motion continues. When the parts split apart, their center of mass is still moving with this same horizontal speed. Perhaps you could write this in terms of center of mass but it will amount to the same thing as writing it in terms of momentum before = momentum after (in each direction).

You will also need to make use of the extra energy contributed by the explosion in order to find the velocity of m2. Yes, quite a complicated problem!

 

[Noob.com]

Exactly. because it's moving. That's why the formula is NOT a static problem. Setting it up is HARD because of that! You have to set it up as a derivative formula so it can change at different time. For all my classes at this level, there is no such thing as horizontal or vertical anymore. It's x', x'' and such. And how can you find the direction with that energy conservation? it's Eo + Eo = 1/2m1(v1)^2 + 1/2 m2(v2)^2. So what? How can you find direction from that? Beside you know NOTHING about the mass nor velocity of BOTH of the exploding object. It's not simple like that!

 

[Delphi51]

you know NOTHING about the mass nor velocity

m1 and m2 will have to appear in the final answer!
As for the velocities, you have one equation about the energy relating them, and you can get a second equation relating them from conservation of momentum. Two equations, two unknowns, not so bad.

Careful with the first Eo; it is partly gone or converted to mgh potential energy. Better, I think, to just use the Ek due to speed Cos(45) x vo at the time of the explosion.

Direction. Well you've got the upward/downward (do you call it z or x'''?) momentum due to m1 going straight down and also the horizontal momentum (from before the explosion) in the direction of the original launch. That should pin down the directions after you do conservation of momentum in those two perpendicular directions.