# Find momentum of pieces of shrapnel

1. Oct 28, 2008

### Chrisleo13

An artillery shell of mass 10.00 kg, lying at rest on the ground, accidentally detonates, exploding into three pieces of shrapnel. One piece of mass 1.70 kg is flung in the -x direction at 21.0 m/s, while a second piece of mass 3.40 kg is propelled in the -y direction at 13.9 m/s. Determine the angle at which the remaining piece of shrapnel travels, relative to the +x axis.

b. Calculate the speed of the third piece of shrapnel.

I found the momentum for 1 and 2 and I tried to invert those to try to find the third. I'm sort of lost on how to find the angle and such.

2. Oct 28, 2008

### LowlyPion

Re: Momentum

You almost have it then. Since there is a conservation of momentum and since momentum is a vector (you know like V is a vector) then the 2 momentum vectors must add to the third to give you 0.

$$\vec 3 = - (\vec 1 +\vec 2) = - ( m_1v_1 * \hat x + m_2v_2 * \hat y)$$

To calculate the speed then divide the resultant momentum magnitude by the remainder of the mass from the 10 kg. (I think you are supposed to assume negligible mass of actual explosive is exploded.)

3. Oct 29, 2008

### Chrisleo13

Re: Momentum

Ok, I understand that but what is x and y in that equation?

m1v1(x) + m2v2(y)

what are the x and y variables here?

4. Oct 29, 2008

### LowlyPion

Re: Momentum

Those are the unit vectors in the x and y direction, because the resultant vector necessarily has both an x and y component to balance the x direction of vector 1 and the y direction of vector 2.

5. Oct 29, 2008

### Chrisleo13

Re: Momentum

Hmm, I think I understand, but the unit vectors aren't given. Do I have to find them using the speeds?

6. Oct 29, 2008

### LowlyPion

Re: Momentum

I used the unit vectors as a device to emphasize the direction. In actuality they are length 1, in the x,y direction. The magnitudes are the scalar quantities represented by the momentums.

The important concept is that you must treat the momentums in the orthogonal directions separately and they must balance to 0. Momentums in x = 0 and momentums in y = 0. The resulting components define then the resultant vector.

7. Oct 29, 2008

### Chrisleo13

Re: Momentum

Hmm, sorry for keep bothering you but I seem to miss the problem when I work it like that.

I got 82.96 when I added m1v1 to m2v2

then I divided it by 4.9 to find the speed. But that's not correct.

8. Oct 29, 2008

### LowlyPion

Re: Momentum

No that can't be correct. These components are at right angles to each other.

Think vectors.

Think Pythagoras.

9. Oct 29, 2008

### Chrisleo13

Re: Momentum

Ahhhh, why didn't I think of that before. I got it. The angle still is unclear. Can I just take the tan of the two velocities?

10. Oct 29, 2008

### LowlyPion

Re: Momentum

That should work.

Remember though the direction is opposite the III quadrant as the x,y directions of the other pieces are negative.

11. Oct 29, 2008

### Chrisleo13

Re: Momentum

I did

13.9 m/s / 21.0 m/s

tan-1()

then I should subtract that angle from 270 or 180?

12. Oct 29, 2008

### LowlyPion

Re: Momentum

They want the angle with respect to the positive X axis. Draw a picture.

13. Oct 29, 2008

### Chrisleo13

Re: Momentum

Well, I may have it. I have one try left on my homework.

I drew it out and found that it would be the opposite of what I said previously.