How much energy is released in the explosion?

[Joza]

A particle with mass m is fired with initial velocity v at an angle theta with the horizontal. At maximum y, it explodes into 2 paticles of equal mass m/2. One of these has zero initial velocity.

I want to find the energy released in the explosion.

My equation is:

mgy + (1/2)mv^2 = E + (m/2)gy + (1/2)(m/2)(2v)^2 + (m/2)gy

max y distance is ((vsintheta)^2)/g

Thru conservation of momentum, one particle after the explosion has 2 times the original velocity.

Thru cancelling out, I got: (1/2)m(vcostheta)^2 - (1/4)m(2vcostheta)^2

Is this correct? I think my reasoning is correct:
Potential and kinetic energy of particle before explosion = potential energy of 1 particle, + potential and kinetic of the other, + energy released


Sorry if I left anything out. I need this one tho

 

[Shooting Star]

Your v on the LHS can't be the same v as initial velo.

 

[Joza]

Oh right, I need the x component? I actually have that, I just left it out, my bad.

It is: costheta(v)

 

[Shooting Star]

Your LHS denotes the mechanical energy of the particle in Earth's grav field. Why should that be related in any way to the energy of explosion?

 

[Joza]

Well I was thinking that the total energy before hand would equal the total energy after? But I think I see the flaw now.

The only energies I have are my kinetic and gravitational potential. I need to relate these somehow to the energy lost in the explosion, which would be a change in internal energy??



Is it that my change in kinetic energy plus change potential energy plus the change in internal energy equals zero?

 

[Shooting Star]

The horizontal component of the momentum remains conserved. One of the fragments is at rest initially at the highest point. Then the other piece must have all the momentum, from which you can find out how much energy it has, and how much is extra.

 

[Joza]

The particle that is initially at rest has energy of (m/2)gy. The other has energy of (m/2)gy and (1/2)(m/2)(2v_x)^2...before the explosion, it has energy of mgy and (1/2)m(v_x)^2

Is that correct? Do I just find the difference now?

 

[Shooting Star]

Why can't we forget the mgy? Anyway, they cancel out.

There was horizontal motion and thus initial KE and momentum, then an explosion. One piece stayed at home and the other flew off carrying off all the initial momentum. It also carried off all the energy.

Is it possible now to find the extra energy? Find it in terms of the things given in the problem.

 

[Joza]

The particle that is initially at rest has energy of (m/2)gy. The other has energy of (m/2)gy and (1/2)(m/2)(2v_x)^2...before the explosion, it has energy of mgy and (1/2)m(v_x)^2

Is that correct? Do I just find the difference now?

 

[Shooting Star]

Forget the PEs. They are not contributing in any way.

(1/2)m*vx^2 + E = 0 + (1/2)(m/2)u^2 and m*vx = (m/2)u. You have to find E in terms of vx = v*cos(theta).