# Horizontal Speed of bullet fragment

1. Oct 25, 2007

### bulldog23

[SOLVED] Horizontal Speed of bullet fragment

1. The problem statement, all variables and given/known data
A 14-kg shell is fired from a gun with a muzzle velocity 125 m/s at 33o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. What is the horizontal speed of the other fragment?

2. Relevant equations

3. The attempt at a solution
I divided the mass by two, because the bullet splits into two equal fragments. I have no idea how to figure out the horizontal speed of the second fragment. Can someone please explain this to me?

2. Oct 25, 2007

### mgb_phys

Are you supposed to assume all the KE goes into the second half or do you assume that the momentum of the explosion puts all the forward momentum into the second half?

I think you are going to have to do it by energy.
You know the KE at the start. Work out the maximum height reached ( v^2 = u^2 + 2as).
Then you know how much KE has gone into PE.
When the shell explodes all the remaining KE must go into the second half so you know it's total speed.

Think about the path of the second half and you should get it's horizontal velocity.

Last edited: Oct 25, 2007
3. Oct 25, 2007

### bulldog23

I'm guessing that you are to assume that all of the KE goes into the second half. Can you help me out, I'm lost?

4. Oct 25, 2007

### bulldog23

I am still confused, I don't understand how to figure out the height.

5. Oct 25, 2007

### mgb_phys

For now just think about the upward velocity.
Draw a triangle with the angle given in the question, the total speed as the hypotonuse and you will get the vertical and horizontal velocities from the other two sides.

You know at the top of the curve it's vertical velocity must be zero.
You know what rate it is slowing down as it goes up ( g = 9.8m/s^2)
And there is a well known equation that says:
V^2 = U^2 = 2 a s where v=final, u=initial velocity, a = acceleration and s=distance.
Be careful, a is -g because it gravity is acting downwards.

6. Oct 25, 2007

### bulldog23

So the initial velocity is 125 m/s. The accel is -9.8 m/s^2. I still need the distance right? And then I solve for the final velocity?

7. Oct 25, 2007

### mgb_phys

The initial velocity is 125m/s at 33deg and the vertical accelaration is -9.8 m/s^2
You have to find the initial vertical velocity to find the height.

8. Oct 25, 2007

### bulldog23

can you explain to me how to do that?

9. Oct 25, 2007

### mgb_phys

10. Oct 25, 2007

### bulldog23

so then I just take 125 and divide by -9.8?

11. Oct 25, 2007

### mgb_phys

The velocity is 125m/s at an angle of 33deg - velocity must always have a speed and direction.
That's equivalent to 125 sin(33) vertically and 125 cos(33) horizontally. You can see this if you draw a triangle.

A nice feature about physics is that you can treat forces at right angles independantly so we can ignore the horizontall bit fro now and just look at the vertical.
It starts off going up at 125sin(33) and slows down at 9.8m/s^2
v^2 = u^2 + 2 a s At the top when u=0 and a = -9.8
125sin(33) ^2 = 2 * 9.8 * s where s (the height) =236.5m

The kinetic energy is = 1/2 m v^2 and potential energy is = m g h
At the top of the curve you have used up some of the initial KE as PE - so we now work out how much KE is left.
KE = 1/2m v^2 - mgh = 1/2* M * 125^2 - M * 9.8 * 236.5 = M ( 1/2*125^2 - 9.8*236.5).

Now the shell is only moving horizontally at the top of the curve so all this kinetic energy is going to go into the horizontal velocity.
But the mass has just halved - so:

M * those numbers = 1/2 1/2 M V^2 where V is the new velocity.

12. Oct 25, 2007

### chocokat

Would this have been solved a bit more easily as a conservation of momentum problem? You just figure out the horizontal velocity, then

$$m_1v_1 = m_2v_2$$

fill in the $$m_1\ \ v_1 \ \ m_2$$ and solve for $$v_2$$

13. Oct 25, 2007

### HallsofIvy

Staff Emeritus
I think you HAVE to use conservation of momentum. Since the bullet "explodes" you cannot assume that total energy is conserved.

14. Oct 25, 2007

### mgb_phys

Doh - I was forgetting that the vertical momentum component was zero when it separates!

15. Oct 25, 2007

### bulldog23

What do I put in for m1 and m2?

16. Oct 25, 2007

### bulldog23

never mind I got it, thanks