Center of mass problem involving shell

In summary: Therefore, the direction of the initial speed of the 2nd fragment must be taken into account when applying the conservation of momentum equation.In summary, the problem involves a shell being shot at an angle of 54° with an initial velocity of 23 m/s. At the top of its trajectory, the shell explodes into two fragments of equal mass. One fragment falls vertically while the other lands at a certain distance from the gun. Using the conservation of momentum equation, it is found that the speed of the second fragment immediately after the explosion is 27.04 m/s, which is twice the initial speed. The direction of this speed is needed to accurately solve the problem.
  • #1
mmattson07
31
0

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.
 
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  • #2
just realized I posted this twice. My apologies.
 
  • #3
Apparently the new speed is 2(13.52m) ?
 
  • #4
mmattson07 said:

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?

mmattson07 said:
Apparently the new speed is 2(13.52m) ?
Please explain why and indicate its direction
 
  • #5
Nobody could point this out I guess but
m*v=0.5m*u
2v=u

So the new speed is twice the initial
 
  • #6
mmattson07 said:
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?
 
  • #7
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
mmattson07 said:
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.
 

1. What is the center of mass?

The center of mass is a point in an object or system that represents the average position of all the mass in that object or system. It is the point where the object or system can be balanced.

2. How is the center of mass determined for a shell?

The center of mass for a shell is determined by finding the average position of all the mass in the shell. This is done by dividing the total mass of the shell by the total volume of the shell.

3. What is the significance of the center of mass in a shell?

The center of mass is significant in a shell because it is the point where the shell can be balanced and it is also the point around which the shell will rotate if an external force is applied.

4. How does the shape of a shell affect its center of mass?

The shape of a shell can affect its center of mass because it determines the distribution of mass within the shell. A more compact and symmetrical shape will have a more centralized center of mass, while a more irregular shape will have a less centralized center of mass.

5. What is an example of a practical application of center of mass in a shell?

A practical application of center of mass in a shell is in the design and stability of boats and ships. By understanding the center of mass of a vessel, engineers can design the hull and weight distribution to keep the center of mass as low as possible, making the vessel more stable and less likely to capsize.

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