How to Calculate the Distance of a Projectile Fragment Landing After Explosion

[lemonpie]

Homework Statement

A shell is shot with initial velocity v0 of 20 m/s at an angle 60 degrees. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment has a speed of zero immediately after the explosion and falls vertically. How far from the gun does the other fragment land, assuming terrain is level and there is no air drag?

Homework Equations

i don't know!

The Attempt at a Solution

i have no idea. i want to use v1f = (m1 – m2) / (m1+ m2) * v1i, but that leads to v1f = 0 which is clearly not correct in this situation -- why not? and if i use and v2f = 2m1 / (m1+ m2) * v1i, then i just get v2f = v1i, which is clearly not correct either. or maybe these are correct, like v1f is the fragment that falls vertically, and v2f just equals v1i somehow. but i still can't figure out how to get any distances out of this.

i really hate this problem, having looked at it for way too long, and i think my teacher will use something like this on the exam because he covered it again yesterday in class. this is what he wrote on the board:

vf = (m1+m2)/m2 * v0cos60
r - r0 = vft - 0.5gt^2
= Di - Hj
D = vft
-H = -0.5gt^2
D = vf sqrt(2H/g)

i'm sorry but this makes absolutely no sense to me. i vaguely recognize some of the projectile stuff here, but there's no way i could have done this on my own. in fact, i can't even do it with these notes in front of me. please help me. any advice on how to break this down. thanks.

 

[LowlyPion]

Without wanting to unravel your notes, I would note that the center of mass of the 2 parts will still land at the same destination as it would if it hadn't broken apart. Since you know the masses of the parts, and you know where 1 part lands, then you should be able to figure the point of impact of the other shouldn't you by simple means?

 

[ImAnEngineer]

I like this question :) .

Let's first find out the height the projectile reaches:
At the maximum heigt all inital kinetic energy is converted into potential energy, so
.5mv² = mgh
h = v² / (2g)

Beware though, the v we're talking about is the the velocity in the y direction, which equals v₀sin60 = 20sqrt(3).

Plug it into the formula, and you'll find a height. When you know the height, you can calculate how long it will take an object of any mass to come down (of course neglecting air drag), using h = (1/2)gt². Solve for t.

When you know t, you know how long particle 2 can maintain it's journey in the air. But we must first find its velocity!

This is where conservation of momentum kicks in.

I don't want to spoil the fun. Could you finish the calculations? :)

 

[lemonpie]

1. do you mean just find the center of mass as a regular projectile problem and then figure out the location of the two pieces around that?

2. i'll definitely have to look at this some more this evening. (sorry i have a test this evening and can't do it now.)

 

[ImAnEngineer]

1. do you mean just find the center of mass as a regular projectile problem and then figure out the location of the two pieces around that?

I think what he says is a different way of saying that the time it takes for an object to come down is mass independent. Center of mass isn't important in this problem if I'm not mistaken.

 

[LowlyPion]

1. do you mean just find the center of mass as a regular projectile problem and then figure out the location of the two pieces around that?

2. i'll definitely have to look at this some more this evening. (sorry i have a test this evening and can't do it now.)

It's worth a look.

 

[lemonpie]

you know, i redid the problem last night and came at something like 52.3, which was close to the actual answer, but I'm pretty sure i did it wrong and any similarity was just a fluke. anyway, my test was this morning, and there was a problem quite like this, so your input definitely helped me out. thanks!

 

[ImAnEngineer]

I get 53.0m. Would that be the correct answer?

Glad to help.

 

[lemonpie]

yes, 53 m was the correct answer, as i recall. i got an A- on that exam. thanks!

 

[ImAnEngineer]

yes, 53 m was the correct answer, as i recall. i got an A- on that exam. thanks!

That's great ! I usually get lower marks