Accidental explosion (Kleppner 3.4)

  • #1
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Homework Statement


An instrument-carrying projectile accidentally explodes at the top of its trajectory. The horizontal distance between the launch point and the point of explosion is ##L##. The projectile breaks into two pieces which 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 curvature.

Homework Equations



Center of mass

The Attempt at a Solution



The center of mass of the two pieces is ##\vec R = \frac{1}{4m} (m \vec r_1 + 3m \vec r_2)##,
where ##\vec r_1## is the position of the small piece and ##\vec r_2## the position of the large piece.

The question is to find ##r_{2,x}(t_f)##, where ##t_f## is the time both pieces are back to the ground.

At explosion time ##t_e##, the external forces are ##\vec f_{ext} = ( 0, -4mg)## as the pieces fly apart horizontally.
The equation of motion for the center of mass is :
## R_x(t) = tV_{0,x} ##
## R_y(t) = t(V_{0,y}-\frac{gt}{2})##
Now we are given that:
1. ## V_{0,x} = \frac{L}{t_e} ##
2. ## R_y(t_f) = 0 \Rightarrow t_f = \frac{2V_{0,y}}{g} ##
3. ## \dot R_y(t_e) = 0 \Rightarrow V_{0,y} = g t_e ## because ##R_y(t_e)## is a maximum of the curve ## t \rightarrow R_y(t) ##
4. ## r_{1,x}(t_f) = 0 ## as piece 1 goes back to origin

Using everything above:
## R_x(t_f) = t_f V_{0,x} = 2t_e\frac{L}{t_e} = 2L ##.

Using the definition of ##\vec R##, we get ## r_{2,x}(t_f) = \frac{8}{3}L ##

I found this one hard, is it correct?
 

Answers and Replies

  • #2
BvU
Science Advisor
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There is no fext!
The ##R_x(t_f) = 2L## follows directly from symmetry, then using the definition of R gives r2. Looks fine to me.
That's not all that hard :)
 
  • #3
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72
Hi, thanks for replying.
Why do you say there are no external forces since the pieces are in motion ?
 
  • #4
BvU
Science Advisor
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2019 Award
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You're right, the gravitational force is present and has the same value before and after the explosion. I misinterpreted (= read too fast).
 
  • #5
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Thanks, because if I was wrong, it would mean that I get a right answer with a wrong proof. And that is worse than anything ;-)
 

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