Failed Rocket Momentum Problem (Kleppner 4-4)

[TimSon]

Homework Statement

An instrument carrying rocket accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is L. The rocket breaks into two pieces that fly apart horizontally. The larger piece has three times the mass of the smaller piece. To the surprise of the scientist in charge, the smaller piece returns to Earth at the launching station. How far away does the larger piece land? Neglect air resistance and effects due to the Earth's curvature.

Homework Equations

Momentum conserved
Kinematics equations

The Attempt at a Solution

Let M4 = initial mass
Let 3m = mass that flies away from the launch site.
Let m = mass that flies towards the launch site.

4m * V(M4) = - m * V(m) + 3m * V(3m)

Since velocity at the top is 0:

M4(0) = - m * V(m) + 3m * V(3m)

m * V(m) = 3m * V(3m)

Therefore V(3m) = V(m)/ 3

Since m travels L horizontally,

L = V(m) * t

therefore V(m) = L/ t

If we plug in V(m) = V(3m) * 3

we get V(3m) = L / (t*3) (since velocities are all horizontal, both take the same amount of time to fall down)

Plugging this into the Kinematics Eqn for distance, we get

D = V(3m) * t - .5 * 9.8 * t^2

D = (L/(3*t)) * t - .5 *9.8 * (L/(3*(v(3m)))^2

I don't know how to get rid of the v(3m).

thanks in advance.

 

[haruspex]

You seem to be confusing vertical velocities and horizontal velocities.

 

[TimSon]

thanks for the hint. it helped me realize that the initial mass's velocity in the x-direction is not zero at the top and then I solved the problem.

 

[almarpa]

May the solution be D=(5/3)L?

 

[Baybora]

did you do by this logic?

Momentum of small piece + Larger piece = total mass * speed before explosion

 

[almarpa]

Yes, and assuming that the time that the total mass, the small piece, and the large piece are flying through the air is the same, as movement in the x direction is concerned.

Is it all right?