Center of mass problem involving shell

by mmattson07
Tags: involving, mass, shell
 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. 9-42). 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.
 P: 31 just realized I posted this twice. My apologies.
 P: 31 Apparently the new speed is 2(13.52m) ?
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Center of mass problem involving shell

 Quote by mmattson07 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. 9-42). 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
why 23? use your conservation of momentum equation immediately before and immediately after the explosion
 because momentum is conserved and it will take the same 1.899s to fall as it did to rise
yes
 so it will go another 23*1.899=43.677m and a total of 69.351m...however I am not getting the correct answer.
Correct the velocity...in what direction is it?

 Quote by mmattson07 Apparently the new speed is 2(13.52m) ?
Please explain why and indicate its direction
 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
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 Quote by mmattson07 Nobody could point this out I guess but m*v=0.5m*u 2v=u So the new speed is twice the initial
looks like you pointed it out. The new speed of the 2nd fragment is 27.04 m/s immediately after the collision, and it's direction is?
 P: 31 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) ;)
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