Internal and center of mass energies

[negation]

Homework Statement

An object with kinetic energy K explodes into two pieces, each of which moves with twice the speed of the original object.


Compare the internal and center-of-mass energies after the explosion.

The Attempt at a Solution

Let K₁ be the kinetic energy of the main object before the explosion. k₂ and k₃ be the kinetic energies after explosion.

K₁ = k₂ + k₃

Let K₁ = 0.5m₁v₁² = K

Then

0.5m₁v₁² = 0.5m₂v₂² + 0.5m₃v₃²

and since each pieces has twice the velocity of the original piece befor explosion

0.5m₁v² = 0.5m₂(2v)² + 0.5m₃(2v)²

∴ 0.5m₁v² = 0.5(4v²)(m₂+m₃)

 

[Andrew Mason]

Homework Statement

An object with kinetic energy K explodes into two pieces, each of which moves with twice the speed of the original object.


Compare the internal and center-of-mass energies after the explosion.

The Attempt at a Solution

Let K₁ be the kinetic energy of the main object before the explosion. k₂ and k₃ be the kinetic energies after explosion.

K₁ = k₂ + k₃

Let K₁ = 0.5m₁v₁² = K

Then

0.5m₁v₁² = 0.5m₂v₂² + 0.5m₃v₃²

and since each pieces has twice the velocity of the original piece befor explosion

0.5m₁v² = 0.5m₂(2v)² + 0.5m₃(2v)²

∴ 0.5m₁v² = 0.5(4v²)(m₂+m₃)

I am not following what you are doing. You seem to be equating the original kinetic energy to the sum of the final kinetic energies of the two parts? Why would k₁ = k₂ + k₃?

Think of the problem this way: In the frame of reference of the centre of mass does the body have any energy before the explosion? How about after?

AM

 

[negation]

I am not following what you are doing. You seem to be equating the original kinetic energy to the sum of the final kinetic energies of the two parts? Why would k₁ = k₂ + k₃?

Think of the problem this way: In the frame of reference of the centre of mass does the body have any energy before the explosion? How about after?

AM

I equate them because the total energy is the sum of the KE of each particle.

However, I may be wrong.

I think it's good to ask what internal kinetic energy is and what center of mass kinetic enegry is and why k_int + k_cm = K_total

 

[negation]

I am not following what you are doing. You seem to be equating the original kinetic energy to the sum of the final kinetic energies of the two parts? Why would k₁ = k₂ + k₃?

Think of the problem this way: In the frame of reference of the centre of mass does the body have any energy before the explosion? How about after?

AM

What's the definition of the frame of reference of the center of mass?

 

[ehild]

The problem asks the center-of-mass energy and the internal energy after the explosion. I do not see any of them in your post.

I see, you assumed conservation of energy. It is not true in an explosion.

ehild

 

[negation]

The problem asks the center-of-mass energy and the internal energy after the explosion. I do not see any of them in your post.

I see, you assumed conservation of energy. It is not true in an explosion.

ehild

Where can we go from here?

K_total = [itex]\sum[/itex]0.5m_iv_cm² + [itex]\sum[/itex]0.5m_iv_irel² + [itex]\sum[/itex]m_iv_irelv_cm

Edit: why is a particle relative velocity to COM zero?

 

[negation]

k_int = 0.5m₁v_cm = m₁v_cm²
there are 2 pieces so 2m1vcm^2
k_int/k_cm = 4 but the answer is 3.

I spent 4 hrs on this. It's not going anywhere and very agonizing/

 

[ehild]

The CoM will travel with the same velocity as before the explosion, as the forces are internal. So the center-of mass KE is the same as the initial KE:

KE_CoM=1/2 m₁v₁².

If m2 and m3 are the masses after the explosion, and m2+m3=m1, the total KE is

KE(total)=1/2 (m2v₁²+m₃v₃²=2v₁²m₁.


The internal energy is the difference between the total energy and the energy of the translation of the CoM:

KE(internal)= KE(total)-KE(CoM)

So what is the ratio KEi/KE(CoM)?

ehild

 

[negation]

The CoM will travel with the same velocity as before the explosion, as the forces are internal. So the center-of mass KE is the same as the initial KE:

KE_CoM=1/2 m₁v₁².

If m2 and m3 are the masses after the explosion, and m2+m3=m1, the total KE is

KE(total)=1/2 (m2v₁²+m₃v₃²=2v₁²m₁.


The internal energy is the difference between the total energy and the energy of the translation of the CoM:

KE(internal)= KE(total)-KE(CoM)

So what is the ratio KEi/KE(CoM)?

ehild

I have managed to solve it by drawing it out and using geometry. However, one question persist when I attempt to derive K= Kint + Kcm.

Why is [itex]\sum[/itex] m_iv_irel² = 0?

 

[ehild]

Are vi rel the relative velocities with respect to the CoM? How did you get that ∑ m_iv_irel² = 0? It is certainly wrong. What are the relative velocities? The pieces move in opposite directions, and have different masses. The velocity of one piece is v2=2v1, its relative velocity with respect to the CoM is v2rel=v1.
The other piece has velocity v3=-2v1. Its relative velocity is -2v1-v1=-3v1.

ehild

 

[negation]

Are vi rel the relative velocities with respect to the CoM? How did you get that ∑ m_iv_irel² = 0? What are the relative velocities? ehild

Yes

My initial premise is

K = [itex]\sum[/itex]0.5 m_iv_i²

but v_i = (v_cm + v_irel) (it would be great if someone can give me an intuitive sense of why v_i = (v_cm + v_irel)

then we have K = [itex]\sum[/itex]0.5m_i(v_irel+v_cm).(v_cm+v_irel)

= [itex]\sum[/itex]0.5m_iv_cm² + [itex]\sum[/itex]m_iv_cm . v_irel² + [itex]\sum[/itex] m_iv_irel²

and then the book states [itex]\sum[/itex]m_iv_cm.v_irel = 0 because v_irel = 0 since it is the particle's velocity relative to the COM.
Unsure what this means.

 

[ehild]

Yes

My initial premise is

K = [itex]\sum[/itex]0.5 m_iv_i²

but v_i = (v_cm + v_irel) (it would be great if someone can give me an intuitive sense of why v_i = (v_cm + v_irel)

then we have K = [itex]\sum[/itex]0.5m_i(v_irel+v_cm).(v_cm+v_irel)= [itex]\sum[/itex]0.5m_iv_cm² + [itex]\sum[/itex]m_iv_cm . v_irel² + [itex]\sum[/itex] m_iv_irel²

and then the book states [itex]\sum[/itex]m_iv_cm.v_irel = 0 because v_irel = 0 since it is the particle's velocity relative to the COM.
Unsure what this means.

I edited my previous post, read it.

Your formula is wrong. There is no square in the second term.

Recall how the CoM is defined.
The velocity of the CoM is V_CoM=(∑m_iv_i)/∑m_i.

Virel=Vi-V_CoM. Virel is not zero, but ∑m_iV_irel=0

ehild

 

[negation]

I edited my previous post, read it.

Your formula is wrong. There is no square in the second term.

Recall how the CoM is defined.
The velocity of the CoM is V_CoM=(∑m_iv_i)/∑m_i.

Virel=Vi-V_CoM. Virel is not zero, but ∑m_iV_irel=0

ehild

The square was a typo. It's hard to see what is going on with all the brackets when using tex.