How Far Does a Pellet Fly After Midair Explosion?

[bodensee9]

Hello:

I am wondering if anyone can help with the following problem:

A pellet of mass M is fired at an angle of 60 to the horizontal. In midair (when the vertical velocity is zero) the pellet explodes into 2 pieces. A piece of mass 1/2*M has a speed of 0 and falls immediately to the ground. The second piece of mass 1/2*M continues flying. How far does the second piece fly?

I am wondering if someone can explain the significance of the explosion? Wouldn't the vertical acceleration be the same (-g)? And I thought that there is no horizontal acceleration?

This is in the center of mass chapter, but I don't see how this has to do with center of mass?
Sorry for not understanding.

 

[bodensee9]

sorry, initial velocity of the pellet is 20m/s.

 

[Hootenanny]

Hello:

I am wondering if anyone can help with the following problem:

A pellet of mass M is fired at an angle of 60 to the horizontal. In midair (when the vertical velocity is zero) the pellet explodes into 2 pieces. A piece of mass 1/2*M has a speed of 0 and falls immediately to the ground. The second piece of mass 1/2*M continues flying. How far does the second piece fly?

I am wondering if someone can explain the significance of the explosion? Wouldn't the vertical acceleration be the same (-g)? And I thought that there is no horizontal acceleration?

This is in the center of mass chapter, but I don't see how this has to do with center of mass?
Sorry for not understanding.

sorry, initial velocity of the pellet is 20m/s.

HINT: Consider the kinetic energy of the centre of mass immediately before and after the explosion.

 

[tiny-tim]

Bang!

I am wondering if someone can explain the significance of the explosion? Wouldn't the vertical acceleration be the same (-g)? And I thought that there is no horizontal acceleration?

Hi bodensee9!

An explosion is just a collision … in reverse!

And forget acceleration for an instant … the explosion only lasts an instant!

 

[bodensee9]

Sorry, so if we ignore drag, this means that kinetic energy is conserved? So if before the explosion, energy = 1/2*M*(20)^2. After the explosion, KE = 1/2*1/2*M*(v)^2 (because the other half pellet has speed 0)? And then you can find the v. Then you can use v^2 = v(initial)^2 -1/2*g*d to find the height and calculate the time it would take for the half pellet to fall to the ground and then multiply by v to find the distance traveled? Thanks. Seems like I'm missing something.

 

[tiny-tim]

inelastic collision

Sorry, so if we ignore drag, this means that kinetic energy is conserved? So if before the explosion, energy = 1/2*M*(20)^2. After the explosion, KE = 1/2*1/2*M*(v)^2 (because the other half pellet has speed 0)

Hi bodensee9!

No … it's an inelastic collision, so energy is not conserved during the collision (of course, it is still conserved until just before, and from just after).

Momentum is conserved in all collisions.

In other words: the centre of mass obeys good ol' Newton's first law in all collisions!

 

[Hootenanny]

Sorry, so if we ignore drag, this means that kinetic energy is conserved? So if before the explosion, energy = 1/2*M*(20)^2. After the explosion, KE = 1/2*1/2*M*(v)^2 (because the other half pellet has speed 0)? And then you can find the v. Then you can use v^2 = v(initial)^2 -1/2*g*d to find the height and calculate the time it would take for the half pellet to fall to the ground and then multiply by v to find the distance traveled? Thanks. Seems like I'm missing something.

Oops my bad! An explosion is an example of an inelastic collision and therefore kinetic energy is not conserved. My previous hint should have said "consider the momentum of the centre of mass immediatly before and after the collision". Sorry !

Edit: It seems tiny-tim has already caught my slip, thanks Tim!

 

[bodensee9]

Hello

So does this mean then before the explosion, p = M*20cos60, but then after the collision, p = 1/2*M*v (because the other half pellet has speed of 0)? And then you have the v of the second pellet and then you can find out how tall it is at the time and how long it takes to drop to the ground to find the distance traveled? Thank!

 

[tiny-tim]

Hello

So does this mean then before the explosion, p = M*20cos60, but then after the collision, p = 1/2*M*v (because the other half pellet has speed of 0)?

erm … nooo …

20 is the initial velocity …

use conservation of energy to find the velocity just before the explosion.

 

[bodensee9]

Hello,

Sorry again, but I thought that the horizontal component doesn't change just before the explosion? And so I don't understand why I need to use the conservation of energy to find the velocity of the pellet just before the explosion?

But I think this should be:

1/2*M*20^2 = 1/2*M*v^2 + m*g*h where h is the height reached by the pellet. And then I can solve for v.

Thanks!

 

[tiny-tim]

… oops!

Sorry again, but I thought that the horizontal component doesn't change just before the explosion? And so I don't understand why I need to use the conservation of energy to find the velocity of the pellet just before the explosion?

Hi bodensee9!

oh dear … sorry … I misread the question … I didn't notice that the vertical velocity was zero during the collision!

Yes … the horizontal component of the centre of mass does stay the same (well, until the bits hit the ground! ).

So you're right … you don't need energy to find v (though you do need it later, to find h).