Projectile explosion velocity and direction of fragments

[j_namtirach]

Homework Statement

A gun fires a shell with the horizontal component of its velocity equal to 200m/s. At the highest point in its flight, the shell explodes into three fragments. Two of these fragments, which have equal mass, fly off with speeds of 300m/s relative to the ground, one along the flight direction of the shell at the instant of fragmentation and the other perpendicular to it in a horizontal plane. Find the magnitude and direction of the velocity of the third fragment immediatley after the explosion, assuming its mass is three times that of each of the other two fragments. Neglect air resistance.

Homework Equations

ρ = mv

The Attempt at a Solution

I'm afraid I don't know where to start with this question. I was hoping somebody could at least set me off in the right direction. Any help would be appreciated. Thanks

 

[CWatters]

Two of these fragments, which have equal mass, fly off with speeds of 300m/s relative to the ground, one along the flight direction of the shell at the instant of fragmentation and the other perpendicular to it in a horizontal plane

Imagine the shell is fired in the x direction with y being the vertical. The other horizontal direction would be the z direction.

At the top the vertical velocity is zero so the shell is traveling horizontally in the x direction at 200m/s.

The first fragment continues horizontally in the direction of flight so (at least initially) it woud continue in the x direction.

The second fragment heads off perpendicular to the first in a horizontal plane so that would be in the z direction.

So before the explosion the shell has zero momentum in the vertical y axis. Two of the fragments also have zero momentum in the y-axis after the explosion. Applying conservation of momentum...

1)what does that tell you about the momentum of the third part in the y direction?
2)Is this really a 3D problem or a 2D problem?

 

[j_namtirach]

It was less complicated than I at first thought. I've got it now!

Thanks a lot for the very clear explanation.