Newton's Second Law for a System of Particles

1. The problem statement, all variables and given/known data
A shell is shot with an initial velocity 0 of 29 m/s, at an angle of θ0 = 56° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?


2. Relevant equations
Equation 1: R=((v0^2)/g)*sin(2θ0)


3. The attempt at a solution
R gives the horizontal range of the projectile, however, due to the change in mass and velocity of the shell, I divided the R by 1/2 to get the distance traveled by the shell before its explosion. After that, I'm not sure what to do.
 

tiny-tim

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A shell is shot with an initial velocity 0 of 29 m/s, at an angle of θ0 = 56° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?
Hi peaceandlove! :smile:

What is the speed of the shell just before it explodes?

What equation can you use to find the speed of the other fragment just after the explosion?
 
Why hello tiny-tim!

I have no clue what the speed of the shell is right before it explodes, so I'm just gonna say 29 m/3.

I dunno if this would be the right equation to use, but according to (m1+m2)(v0)(cos (θ0))=(m1v1)+(m2v2) where v1=0 and (m1+m2)=1, the velocity of the fragment that continues to travel through the air along the y-axis is 32.4332 m/s.
 

tiny-tim

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I have no clue what the speed of the shell is right before it explodes
ok, let's do that first …

what is its vertical component of speed right before it explodes?

and what is its horizontal component of speed right before it explodes? :wink:
 
I figured it out! I found the coordinates of the point where the shell explodes and the velocity of the fragment that did not fall straight down. Since no horizontal forces act, the horizontal component of the momentum is conserved. Since one fragment has a velocity of zero after the explosion, the momentum of the other equals the momentum of the shell before the explosion. I then used this information to determine where the fragment lands by analyzing a projectile launched horizontally at time t=0 and got 119.35124 m.
 

tiny-tim

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Excellent!

:biggrin: Woohoo! :biggrin:
 

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