# Homework Help: A projectile exploding at the top of its trajectory

1. Nov 7, 2012

### anban

1. The problem statement, all variables and given/known data

An object is launched from the origin. At the peak of it's trajectory, it explodes into two fragments, one of mass m and the other of mass 3m. The smaller fragment of mass m lands back at the trajectory.

How far from the origin does the second fragment land?

2. Relevant equations

Conservation of Momentum: P = mv
Conservation of Energy: KE = .5mv^2 only because explosion happens at singular height, so GPE irrelevant
Center of Mass: 1/M Ʃ mx
Kinematic equations

3. The attempt at a solution

The small mass m is at the origin, and the large mass 3m is some distance away. I will call this distance x.

Using the Center of Mass formula, (m(0)+3m(x))/(3m+m) = 3/4 x. So we know that the center of mass is 3/4 x. We also know that if the projectile did not explode, it would land at this spot.

Now, there are no times or velocities given. Using the kinematic equations and considering the even of the original projectile launching from the ground to the top of it's trajectory, we can say that:

x0=0
xf= 3/8 x
ax=0
ay=g
y0=0

This info combined with different kinematic equations gives these expressions:

0 = voy^2 +2gh
voy = gt
h = gt^2 + .5gt^2 = 3/2gt^2
0=g^2 t^ + 2gh
-g^2 t^4 / 2g = (3gt^2)/2
t^2 = 3
t = sqrt (3) This is the time it takes for our projectile to reach the top of it's path.

Now, I am ultimately out to get the x distance where the large mass is laying.
I think that I should continue manipulating the kinematic equations, but I am not sure that I have enough information.

I could fiddle around and get velocities for each of the pieces, I think. I could plug them into the conservation of momentum equation: but this wouldn't do me any good with finding a distance! Likewise with the conservation of energy.

Another useful piece of info: if m1 = 3m2, then according to cons. of momentum, I'm pretty sure that 3v1= v2.

Basically, I am not sure which direction to go in.... I've been up and down this problem for a long time already, and I'm hoping that I am just forgetting some obvious point! Thanks in advance!

Last edited: Nov 7, 2012
2. Nov 7, 2012

### PeterO

Like all good projectiles, the shell [before exploding] will have had a constant horizontal component to its velocity.
Also, the projectile got to maximum height in time T, and the bits will take an additional time T to get back to the ground.
In order to land at the origin, that means the small fragment has been blown back at exactly the same speed as it was travelling before the explosion - giving you the change in velocity of that fragment.
Equal impulses on the two fragments will show you by how much the velocity of the 3m mass changes, and thus how far - relative to the distance travelled so far - the 3m mass will go.