Shell fragments exploding - NSL for system of particles

[netrunnr]

A shell is shot with initial velocity $$\vec{}v_{0}$$ is 20ms$$^{-1}$$ at an angle $$\vartheta$$ = 60 degrees. At the top of the trajectory, the shell explodes into two fragments of equal mass. Fragment1 has a speed of 0 immediately after the explosion and falls vertically. How far from the gun does fragment2 land assuming terrain is level and there is no air drag?

Equations used:
F = ma
v$$^{2}$$ = v(of origin)$$^{2}$$ a + 2a(y-y$$_{o}$$)

p = mv

Initial middle
p=20ms$$-{1}$$m p = $$\frac{1}{2}$$mv + $$\frac{1}{2}$$m x 0
=> 20m => $$\frac{1}{2}$$mv + $$\frac{1}{2}$$mx0

=> 20m = $$\frac{1}{2}$$mv
=> 20 = $$\frac{0.5mv}{m}$$ = $$\frac{1}{2}$$v = 40ms$$-{1}$$

so the middle position is p = 40ms$$-{1}$$

breaking this down to x y components:
for x:
cos$$\vartheta$$ = $$\frac{x}{20}$$
cos $$\vartheta$$ x 20 = x
$$\frac{1}{2}$$ x 20 = x
x = 10

for y:
sin$$\vartheta$$ = $$\frac{y}{20}$$
sin$$\vartheta$$ x 20 = y
$$\sqrt{\frac{3}{2}}$$ x 20 = y
[tex]\frac{20 x $$\sqrt{3}$$}{2}[/tex] = 10 x $$\sqrt{3}$$
y = 10 x $$\sqrt{3}$$


using v$$^{2}$$ = v$$_{o}$$$$^{2}$$ a + 2a(y-y$$_{o}$$)
0 = (10 x $$\sqrt{3}$$)$$^{2}$$ + 2 x 9.8ms$$-{1}$$(y - 0)
0 = 100 x 3 + 2 x 9.8 x y
y = $$\frac{300}{19.6}$$ = 15.3

fragment1 dropped from a height of 15.3 meters

I am lost as to what equation to use to calculate the distance fragment 2 traveled from here. I know its simple but somehow I am not able to think of what to do next... I need the distance in the x direction that fragement2 traveled.

hints??

thanx!

 

[Office_Shredder]

It looks like you calculated the momentum for the fragment based on the initial velocity. This is the wrong approach. What you need to do is:

1.) Calculate the velocity of the shell at the top of its arc
2.) Calculate the momentum of the shell at the top of its arc
3.) Calculate the momentum of the second fragment based on conservation of momentum
4.) Calculate fragment 2's velocity based on its momentum
5.) Calculate how far it will go before it hits the ground

 

[thomate1]

I would suggest using the equation for projectile motion in the x-direction, which is x = v_{0}cos\vartheta t. In this case, v_{0} is 40ms^{-1} and \vartheta is the angle at which the shell was shot, which is 60 degrees. We can also assume that the time taken for fragment2 to reach the ground is the same as the time taken for fragment1 to reach the ground, since they were both released at the same time. Therefore, we can set the equation for fragment2 equal to the equation for fragment1, and solve for t. Once we have the value for t, we can plug it back into the equation for fragment2 to find the distance traveled in the x-direction. This will give us the final answer for the distance that fragment2 traveled from the gun. I hope this helps!