Register to reply

Center of mass problem involving shell

by mmattson07
Tags: involving, mass, shell
Share this thread:
mmattson07
#1
Nov22-10, 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. 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.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
mmattson07
#2
Nov22-10, 06:57 PM
P: 31
just realized I posted this twice. My apologies.
mmattson07
#3
Nov22-10, 07:58 PM
P: 31
Apparently the new speed is 2(13.52m) ?

PhanthomJay
#4
Nov22-10, 09:31 PM
Sci Advisor
HW Helper
PF Gold
PhanthomJay's Avatar
P: 6,043
Center of mass problem involving shell

Quote Quote by mmattson07 View Post
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 Quote by mmattson07 View Post
Apparently the new speed is 2(13.52m) ?
Please explain why and indicate its direction
mmattson07
#5
Nov22-10, 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
PhanthomJay
#6
Nov22-10, 10:09 PM
Sci Advisor
HW Helper
PF Gold
PhanthomJay's Avatar
P: 6,043
Quote Quote by mmattson07 View Post
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
#7
Nov22-10, 10:22 PM
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) ;)
PhanthomJay
#8
Nov22-10, 10:32 PM
Sci Advisor
HW Helper
PF Gold
PhanthomJay's Avatar
P: 6,043
Quote Quote by mmattson07 View Post
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.


Register to reply

Related Discussions
Center of mass problem involving shell Introductory Physics Homework 0
Center of Mass involving three cubes Introductory Physics Homework 1
Center of mass of semicirculardisk calculated from center of mass of semicircular arc Introductory Physics Homework 5
Center of mass involving changing positions Introductory Physics Homework 8
Help with problem of Center of mass, linear mass density and total mass Introductory Physics Homework 1