# Shell Explosion

1. Sep 28, 2007

### Oblio

A shell traveling with velocity vo explodes into three pieces of equal masses. Just after the explosion, one piece has velocity v1=vo and the other two have velocities v2 and v3 that are equal in magnitude but mutually perpendicular.
Find v2 and v3 and sketch their velocities.

In starting the question, the x and y components of v2 and v3 have to be equal before and after the explosion, and the Net external forces must cancel out?

2. Sep 28, 2007

### learningphysics

v2 and v3 don't exist before the explosion. What is the total momentum before the collision?

If vo is in the horizontal direction... that means that before the collision, the momentum is 0 vertically. So what does that tell you about the vertical components of v2 and v3?

what is the angle v2 and v3 make with the horizontal?

3. Sep 28, 2007

### FedEx

Law of conservation of momentum. The initial total momentum shouldbe equal to the total final momentum now as the masses are same for the three parts we can note them as m/3 each wherem is the mass of the whole shell. So we can elimate m from both the sides.

And now we can proceed easily.But beware of the vector sums.

4. Sep 28, 2007

### Oblio

So:

mvo = (m/3)v1 + (m/3)v2 + (m/3)v3

If v2 and v3 are mutually perpendicular, and if the vertical component of the y velocity is zero before the explosion, the y components must be in opposite directions?

45 degrees above horizontal and 45 degrees below horizontal, in the +x direction?

5. Sep 28, 2007

### learningphysics

Yes, this is true in vector form... in other words:

$$m\vec{vo} = (m/3)\vec{v1} + (m/3)\vec{v2} + (m/3)\vec{v3}$$

but when you are looking at only the horizontal direction, what equation do you get? call the magnitude of v2 and v3, vk... so what's the conservation of momentum equation in the x-direction? also you know that $$\vec{v1} = \vec{vo}$$, so substitute that in also.

Yes, that's true. The angle is 45.