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Center of mass problem involving shell 
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#1
Nov2210, 05:05 PM

P: 31

1. The problem statement, all variables and given/known data
A shell is shot with an initial velocity 0 of 23 m/s, at an angle of θ0 = 54° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 942). 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 M v = m1 u1 + m2 u2 3. The attempt at a solution This is what I tried Init vertical speed = 23sin54=18.61m/s Init horizontal speed= 23cos54=13.52m/s >Time to reach top of trajectory= 18.61/9.8=1.899s >Horizontal distance to top= 13.52*1.899=25.674m Then the speed of the second shell must be 23 m/s because momentum is conserved and it will take the same 1.899s to fall as it did to rise so it will go another 23*1.899=43.677m and a total of 69.351m...however I am not getting the correct answer. 


#2
Nov2210, 06:57 PM

P: 31

just realized I posted this twice. My apologies.



#3
Nov2210, 07:58 PM

P: 31

Apparently the new speed is 2(13.52m) ?



#4
Nov2210, 09:31 PM

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Center of mass problem involving shell



#5
Nov2210, 09:44 PM

P: 31

Nobody could point this out I guess but
m*v=0.5m*u 2v=u So the new speed is twice the initial 


#6
Nov2210, 10:09 PM

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#7
Nov2210, 10:22 PM

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Yeah. Luckily I found that somewhere else or I'd still be lost. I don't know what the direction is , don't care to find it because the question doesn't ask for it. But it will have something to do with arctan(y/x) ;)



#8
Nov2210, 10:32 PM

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