Center of mass problem involving shell

[mmattson07]

Homework Statement

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?

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c09/fig09_46new.gif

Homework Equations

M v = m1 u1 + m2 u2

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.

 

[mmattson07]

just realized I posted this twice. My apologies.

 

[mmattson07]

Apparently the new speed is 2(13.52m) ?

 

[PhanthomJay]

Homework Statement

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?

http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c09/fig09_46new.gif

Homework Equations

M v = m1 u1 + m2 u2

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?

Apparently the new speed is 2(13.52m) ?

Please explain why and indicate its direction

 

[mmattson07]

Nobody could point this out I guess but
m*v=0.5m*u
2v=u

So the new speed is twice the initial

 

[PhanthomJay]

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?

 

[mmattson07]

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) ;)

 

[PhanthomJay]

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) ;)

I don't know where you luckily found it, but you should try to find it on your own. Otherwise the solution is of no meaning to you. You have the right conservation of momentum equation; you should apply it at the top of the trajectory immediately before and after the explosion. If you do not know the direction of the initial speed of the 2nd fragment immediately following the collision, you cannot solve the problem. Momentum is a vector quantity, and as such, it has direction.