Projectile Motion/ Explosion Mid air

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1. Mar 1, 2017

RED119

1. The problem statement, all variables and given/known data
So in the problem there is a projectile that is in the air and has no forces acting up, only the -9.8 acceleration down, the projectile is at the maximum height of say 10m and at this height (with a horizontal acceleration of around 40 m/s) explodes into two even 20g pieces (original mass is 40g, not sure where the energy for the explosion is coming from since all the mass is conserved and non is converted in a chemical reaction to release heat or energy, instructor said to ignore it), one of the pieces now has 0 acceleration horizontally, what is the distance the second piece traveled from the explosion?

2. Relevant equations

W =FxCos KE = 12mv2 KE = KEf-KEi

WTotal=KE

PEg=mgh PEg=mgy

PEs=12kx2 PEs=PEsf-PEsi

ME = KE+PEg+PEs

KEi+PEgi+PEsi=KEf+PEgf+PEsf

ME = Wfriction

p = mv p = pf-pi

I = p I = F*t

3. The attempt at a solution
So... my basic ideas of physics (which could be misguided here) are telling me that if the mass is halved and the same force is imparted on that new lighter object that got a heavier object to accelerate to X (40m/s) then the new acceleration of the lighter object will need to be greater than X. My issue is I am not sure how to figure out that initial force or whether I am even on the right track here...

The last part I am pretty sure is just you figure out how long it would be in the air for (free fall conditions, so pretty easy, then take that and plug it in to what you found about its acceleration and figure out how far the delta X is in that time-frame to get that answer

Last edited: Mar 1, 2017
2. Mar 1, 2017

kuruman

You mean horizontal velocity, of course.
Is anything conserved just before and just after this explosion?

3. Mar 1, 2017

RED119

This is all the information that is given, unless you're getting at that the momentum is conserved?

4. Mar 1, 2017

kuruman

That's what I am getting at. Explosions are like inelastic collisions with time running backwards. If inelastic collisions conserve momentum, so do explosions. Can you write a momentum conservation equation (or two) for this case?

5. Mar 1, 2017

RED119

So if I did this right the momentum is 1600 for the first interaction, so it needs to be equal to the momentum after the explosion, so the stationary block has a momentum of zero, so I got 20X = 1600. Divide by 20 and get 80m/s for the velocity?

6. Mar 1, 2017

kuruman

That is correct. You can see that immediately because all of the initial horizontal momentum is transferred to half the mass. Therefore the velocity of the fragment will be twice the initial velocity. Can you finish the problem now?

7. Mar 1, 2017

RED119

I can, thanks so much for the time and help