Failed rocket problem (momentum conservation, Kleppner 4-4)

In summary, the "Failed rocket problem" illustrates the principle of momentum conservation by examining a rocket that does not achieve the desired propulsion due to fuel expulsion issues. The problem highlights that even if the rocket's mass decreases, the momentum can remain constant if the expulsion of fuel is not directed properly. It emphasizes the importance of both the conservation of momentum and the correct application of thrust to achieve motion, as discussed in Kleppner's physics text.
  • #1
Michael Korobov
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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.
Relevant Equations
Momentum conserved
Kinematics equations
Hi,
Can anyone hint me if there is issue in the problem statement?
Consistent answer can be obtained if one presumes that the trajectory of center mass is parabolic.
Assuming this, the CM will land at distance L right to the axis of symmetry of parabola.
But the problem tells about a rocket, therefore first part of rocket movement is not free fall and thus the horizontal distance of falling after the top of trajectory is not necessarily L.
Is it deduction correct?
 
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  • #2
Michael Korobov said:
But the problem tells about a rocket, therefore first part of rocket movement is not free fall and thus the horizontal distance of falling after the top of trajectory is not necessarily L.
Is it deduction correct?
Yes, it is not necessarily L. That's why the scientist in charge was surprised when it landed there.
 
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  • #3
Michael Korobov said:
But the problem tells about a rocket, therefore first part of rocket movement is not free fall and thus the horizontal distance of falling after the top of trajectory is not necessarily L.
Is it deduction correct?
If we assume powered flight all the way to the top of the trajectory then we have, in principle, no way to know the horizontal velocity of the rocket just prior to the explosion.

It is possible, for instance, that the rocket reverses course, that the "explosion" is a mere fizzle and that both rocket pieces land together at the launch site.

Accordingly, one assumes, as you did in your solution, that the rocket burns out almost immediately after launch and then carries on in a parabolic free fall trajectory. Otherwise, there is no way to a solution.
 
  • #4
jbriggs444 said:
If we assume powered flight all the way to the top of the trajectory then we have, in principle, no way to know the horizontal velocity of the rocket just prior to the explosion.

It is possible, for instance, that the rocket reverses course, that the "explosion" is a mere fizzle and that both rocket pieces land together at the launch site.

Accordingly, one assumes, as you did in your solution, that the rocket burns out almost immediately after launch and then carries on in a parabolic free fall trajectory. Otherwise, there is no way to a solution.
Indeed, this was the only way to get consistent solution.
Thanks!
 

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