How Does Igniting Fuel at the Trajectory's Apex Affect Missile Debris Dispersal?

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SUMMARY

This discussion focuses on the physics of missile debris dispersal when fuel ignites at the apex of its trajectory. The missile, initially with a mass of 3m, disintegrates into two pieces: one with mass m and the other with mass 2m. The conservation of momentum is applied to determine the velocities of the debris pieces after ignition, revealing that the smaller piece's change in velocity is twice that of the larger piece but in the opposite direction. The analysis concludes that the timing of the ignition at the trajectory's apex is crucial for understanding the motion of the debris.

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Homework Statement


In testing a missile defense system, a missile is red from the ground on a
trajectory that would directly hit a bunker some distance away. When the missile is
at the top of the trajectory, a laser light from the bunker ignites fuel in the missile
and the missile disintegrates into two pieces, one twice as massive as the other. The
pieces reach the ground nearly simultaneously, 60m apart from each other.

(a) By how much does the larger piece miss the bunker? Hint: Consider motion of
the center of mass.
(b) By how much does the smaller piece miss the bunker?
(c) How important is the information that the fuel ignited at the top of the trajectory?

Homework Equations



The Attempt at a Solution




so far I figured out this:
http://imgur.com/MdLrl
http://imgur.com/YqBet

I know that I need one more equation to get all the unknowns.
the homework is due in 5 hours if anyone could help I would love you forever.
 
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This looks like a conservation of momentum problem, more than a kinematics problem.

For part (c), the rocket is tangent to the x-axis (i.e., going horizontally) when the laser zaps it, so there should be no change in the y components of the velocity or the acceleration when it blows. Both pieces should have the same velocity and acceleration in that dimension. At max height, the y component of velocity is zero, so all of the momentum is in the x direction.

If we use your definition of the initial mass as 3m and the two pieces as m and 2m, and define the initial velocity as v0, then conservation of momentum is

p0 = p1 + p2

where

p0 = m0v0 = (3m)v0

Let's define the change in velocity of the small piece as v1 and the change in velocity of the larger piece as v2. This is a bit different than the usual notation, but there's a method to my madness.

Define (v0 + v1) as the velocity of the small piece and (v0 + v2) as the velocity of the big piece. v1 and v2 are the changes in velocity from the fuel exploding. Intuitively, one of these should be the opposite of the other as they fly apart, but let's leave the signs as positive for now.

So, p1 = m (v0 + v1) and p2 = 2m (v0 + v2)

The conservation law is then p0 = p1 + p2 or

3m(v0) = m (v0 + v1) + 2m (v0 + v2)

Multiply it out:

3m(v0) = m(v0) + m(v1) + 2m(v0) + 2m(v2)

Collect similar terms:

3m(v0) = 3m(v0) + m(v1) + 2m(v2)

Subtract 3m(v0) from both sides:

0 = m(v1) + 2m(v2)

Divide by m:

0 = v1 + 2(v2)

Subtract 2(v2) from both sides

- 2(v2) = v1, or v1 = -2(v2)

So, intuition and our signs match up. The little piece's change in velocity is opposite in direction to the big one and its velocity changes twice as much as the big piece changes.

That should be a big enough hint for you.
 
Last edited:

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