Heat energy in an inelastic collision

[vcsharp2003]

Homework Statement

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m/s. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?

The problem I am facing is that the answer in book is given as 8.82 J and it mentions that it will be the same even when lift is stationary. The answer that I get is 25.5 J as explained below. May be I am missing something here.

Homework Equations

I am assuming that the observer is on ground and not in elevator i.e. frame of reference is ground.
Also, g is assumed to be 10 m/s².

KE1 + PE1 = KE2 + PE2, where 1 refers to when bolt starts its fall and 2 refers to just before it hits the floor.

Also for the bolt fall,
v = u + at and v² = u² + 2 x a x s

The Attempt at a Solution

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The bolt falls from a height with initial velocity of 7 m/s since bolt will have same velocity as the elevator when it starts it's fall. When it hits the floor it has fallen a distance of (3 + 7t) since floor of elevator descends by 7t

So, if the floor of the elevator when bolt hits the floor is taken as datum level for potential energy, then
KE1 + PE1 = KE2 + PE2, where 1 refers to when bolt starts its fall and 2 refers to just before it hits the floor.

.KE2 + PE2 = .5 x .3 x v² + 0

Also, using kinematics, we have
v = 7 + 10 x t and v² = 7² + 2 x 10 x (3 + 7t)

Solving above equations, t = .78 seconds, v= 14.8 m/s

Since bolt does not rebound after hitting the floor, so KE2 + PE2 - (KE of bolt after hitting floor) = Heat generated. = .5 x .3 x 14.8² + 0 - .5 x .3 x 7² = 25.5 J.

But answer at back of book is given as 8.82 J. I am not getting where I am making a mistake.

 

[PeroK]

But answer at back of book is given as 8.82 J. I am not getting where I am making a mistake.

What's the speed of the bolt after hitting the floor?

 

[vcsharp2003]

What's the speed of the bolt after hitting the floor?

The question says that the bolt does not rebound, so it means that it moves with the same velocity as the elevator i.e. 7 m/s.

 

[PeroK]

The question says that the bolt does not rebound, so it means that it moves with the same velocity as the elevator i.e. 7 m/s.

Yes, I see now you got that right. This is tricky. The bolt definitely loses more energy if you analyse the problem in the ground frame, than if you analyse it in the moving elevator frame. But, the increase in temperature can't depend on the frame! So, logically, not all the lost KE can go to heat in the moving frame.

Any ideas why not?

What is heat energy anyway?

 

[vcsharp2003]

Yes, I see now you got that right. This is tricky. The bolt definitely loses more energy if you analyse the problem in the ground frame, than if you analyse it in the moving elevator frame. But, the increase in temperature can't depend on the frame! So, logically, not all the lost KE can go to heat in the moving frame.

Any ideas why not?

What is heat energy anyway?

What is funny is if I use the elevator as frame of reference then I get 8.82 J using the same principles I used when observer was on ground. But the law of conservation of energy is true for all inertial frames of reference, which means that the answers should be the same whether frame of reference is ground or elevator. I am sure that there is something missing in my approach when using ground as reference.

 

[PeroK]

What is funny is if I use the elevator as frame of reference then I get 8.82 J using the same principles I used when observer was on ground. But the conservation of energy is true for all inertial frames of reference, so it also means that the answers should be same whether frame of reference is ground or elevator. I am sure that there is some missing point in my approach when using ground as reference.

No, I don't think your calculations are necessarily wrong. Let me try to explain. It's really more about what heat is.

The kinetic energy definitely went into the kinetic energy of particles in the floor and the bolt. More energy in the ground frame. But, all the particles in the bolt and the floor have an added $7m/s$ in the ground frame. So, in order to increase the speed of one of those particles requires more energy in the ground frame than in the elevator frame - for exactly the same reason more KE was lost by the bolt.

The moral: analyse heat in the frame where the object being heated up is at rest!

 

[TSny]

Interesting problem.

Here is one way to make some sense of it. You are assuming that the elevator maintains a constant speed during the collision of the bolt with the floor of the elevator. So, there must be a net external force that acts on the elevator during the time interval of the collision.

In the inertial frame of the elevator, this force doesn't do any work (why?).
But for the ground observer, this force does do work (why?).
See if you can show that this accounts for the "extra" loss of KE of the system in the ground frame compared to the elevator frame.

 

[vcsharp2003]

Interesting problem.

Here is one way to make some sense of it. You are assuming that the elevator maintains a constant speed during the collision of the bolt with the floor of the elevator. So, there must be a net external force that acts on the elevator during the time interval of the collision.

In the inertial frame of the elevator, this force doesn't do any work (why?).
But for the ground observer, this force does do work (why?).
See if you can show that this accounts for the "extra" loss of KE of the system in the ground frame compared to the elevator frame.

From what you said the conservation of energy equations for the bolt would be as below in any inertial frame of reference.

KE1 + PE1 = KE2 + PE2 and KE2 + PE2 - (Work done against collision force) = (KE of bolt after impact) + (Heat energy generated by collision)

So, the bolt does work against the force of collision for a small period of time when viewed from ground, but when viewed from elevator the bolt does no work against this collision force since displacement relative to elevator is zero ( ball does not rebound). When viewing from ground, it is impossible with given data to determine the KE i.e.velocity of bolt after impact, and also it's impossible to determine the force of collision on the bolt. But if inertial frame of reference is the elevator, then KE of bolt after impact is zero since bolt does not move relative to elevator and also work done against collision force is zero due to no displacement. Therefore, it's best to take elevator as frame of reference.

The collision force is what decreases the velocity of the bolt so it moves with the elevator rather than rebounding.

With elevator as frame of reference the equations become: KE1 + PE1 = KE2 + PE2 and KE2 + PE2 - (0) = (0) + (Heat energy generated by collision). From first equation we have 0 + .3 x 10 x 3 = KE2 + 0. Then this KE2 is converted to heat energy after collision since after collision the bolt becomes stationary with respect to elevator i..e KE of bolt after impact) = 0 . This gives an answer of 9 J for heat energy generated, but if I take g= 9.8 m/s² then I get the exact answer of 8.82 J.

 

[TSny]

OK. I was thinking in terms of the system as a whole (bolt + elevator) and work done by external forces on the system. But I like your idea of just considering the collision force between the bolt and the floor of the elevator. I believe you have the right idea.

In the elevator frame, the bolt moves a small distance d during the collision while coming to a stop. If F is the magnitude of the average force that the bolt exerts on the floor during the collision, then the work Fd is responsible for transforming the KE of the bolt just before the collision into heat, Q.

In the ground frame, F acts over an extra distance D = v_e Δt, where v_e is the constant speed of the elevator and Δt is the time of the collision. So, in the ground frame, F does extra work FD = (FΔt)v_e = Δp v_e, where Δp is the change in momentum of the bolt. (Δp is the same in the two frames of reference). You can show that this extra work, Δp v_e, accounts for the extra loss of KE in the ground frame compared to the elevator frame.