Basic Conservation of Energy Problem

[studentteacher]

So, I was tutoring a student and we got to a question of a type I've seen before. A block attached to a spring is resting on a frictionless horizontal surface. A bullet is fired into it. You're told the amplitude of the oscillation and the spring constant. So, you find the total energy. You get the right answer.

 

Another part of this question is to find the speed of the bullet before it hit the block. So, you set 1/2 mv^2 equal to the total energy found above, and solve for v. But somehow, I kept getting the wrong answer, and this was very embarrassing in front of my student. Is there a trick here?

[timo]

It is a bit hard to give a definite answer without knowing which parameters are given an which are not (your second paragraph suggests the mass of the bullet is given even though your first paragraph suggests it is not, for example). My guess would be the following: Collisions not only need to conserve energy, but also momentum. Additionally, the initial kinetic energy does not need to equal the final kinetic energy, since some of the original kinetic energy can be transformed into deformation energy. Hence, the kinetic energy of the bullet was larger than the energy of the resulting oscillation. Taking this into account (which becomes easy if, for example, you are given the mass of the block and the bullet) should give a correct answer, as far as I can tell.

[studiot]

Yes momentum balance is the key to this one, we had a long discussion a while back about bullets and blocks, where I posted links to a good Ytube video on the subject.

 

Perhaps someone can find it, the search function is poor here.

[swansont]

Additionally, the initial kinetic energy does not need to equal the final kinetic energy, since some of the original kinetic energy can be transformed into deformation energy.

I'll go a step further and say that for the situation where a bullet is embedded in a block, we know that there is a (maximum) loss of kinetic energy (it's a totally inelastic collision), so one can't apply conservation of mechanical energy to the collision. The only time you can is if the collision is elastic.

[studentteacher]

We're given the mass of the block, the mass of the bullet, the spring constant, and the amplitude. So, I used the spring constant and amplitude to find total energy. And then set the kinetic energy of the bullet before the collision equal to the total energy, and solved for v. I don't understand why that was wrong, or what is right.

[John Cuthber]

A bullet that has just been shot into a block of wood will be warm. Work will have been done forcing the metal through the wood and some of that work will have been converted into heat.

Since you do not (directly) know how much heat is produced you can not know how much energy is left as kinetic energy.

 

However the momentum of the system remains constant and you can do a calculation with it that is similar to the equation you used for conservation of energy.

[studiot]

I'm stuck out in the north of Scotland right now, so resources are limited but look here.

 

https://www.google.co.uk/#q=sienceforums.net+momentum+bullet+and+block&*

https://www.google.co.uk/#q=youtube+momentum+bullet+and+block&*

[J.C.MacSwell]

Assume an ideal spring.

You can assume energy is conserved without dissipation from the point after the collision where the masses are then at the same velocity.

You have to assume the collision was instantaneous or you will not have enough information.

From the spring constant and amplitude you can for that energy.

Knowing the total mass you can solve for what the velocity was at that point, just after the instantaneous collision.

From that velocity and the masses you can solve for the initial velocity of the bullet, and you can also solve for the initial kinetic energy and how much energy was lost in the inelastic collision.

 

You can also solve it by making other assumptions about the collision being not instantaneous, but it gets more complicated because the spring will displace during the duration of the collision.

[Bender]

Good video about it

[swansont]

We're given the mass of the block, the mass of the bullet, the spring constant, and the amplitude. So, I used the spring constant and amplitude to find total energy. And then set the kinetic energy of the bullet before the collision equal to the total energy, and solved for v. I don't understand why that was wrong, or what is right.

 

 

In the collision, KE is not conserved, but momentum is. So you have to apply conservation of momentum from the bullet to the bullet+block system, From that you can determine the KE of the bullet+block system, and subsequent motion. Or you can do all of that in reverse, as is required here.

[Sriman Dutta]

How do we find the total energy of the spring+block? Is it 0.5kx^2 ?

[KipIngram]

J.C.McSwell seemed to be on the right track to me. You know the energy of the system "block with embedded bullet" from the spring constant and the amplitude. If you neglect the "collision time" (basically that means the bullet has come to a halt in the block before the block has moved very far), you note that the spring is uncompressed at that point, and all the energy has to be kinetic energy of "block + bullet." That gives you the velocity of "block + bullet," and that lets you calculate the momentum.

 

That momentum must equal the momentum of the bullet prior to collision, from which you can ascertain the bullet speed.

 

I was given a similar question in an interview once, only in an electrical system. There were two equal capacitors, one charged to a voltage V and one with no voltage, with a switch between them. I was asked what the voltage would be after the switch closed. I used charge conservation to come up with V/2. But energy conservation gives a different answer, because it presumes no energy was lost. Even with no resistor in the circuit, energy is still lost. A thorough analysis would show that it was radiated away, since there's no loss component in the (ideal) circuit to consume it. But there's no way for charge to escape, which is why it's appropriate to use charge conservation.

 

Energy is always conserved, but energy can change form on you in tricky ways. Momentum is always conserved, and it's easier to keep track of it - none of it "sneaks away."

[mistermack]

I'm not sure that you can use momentum in this case.

You aren't told what the spring is attached to. So how do you calculate the momentum of an unknown object?

 

I think the question is an idealised one, about kinetic energy, without the losses that you know will actually happen.

Dunno why he gets the wrong answer though.

[KipIngram]

I think you're meant to assume that the other end of the spring is rigidly fixed. Otherwise there's no way even to start. Losses are inevitable, once you assume the bullet comes to a stop in the block. You must conserve momentum in the instant of collision (while you're neglecting the spring compression). Energy is conserved only in the case of a completely elastic collision (when the bullet rebounds). If the bullet becomes embedded in the block you now only have one free velocity to play with so you can't conserve both momentum and energy. With an elastic rebound you are assuming no loss and you have two free velocities to work with so you can conserve both momentum and energy.

[mistermack]

I don't think you can assume the spring is rigidly fixed, for momentum purposes.

If you do, momentum can never be conserved.

[KipIngram]

Of course you can. Only one end of the spring is rigidly fixed - the other end is attached to the block. J.C. MacSwell had the key insight into this problem - what you neglect is the time spent by the bullet "halting" in the block. Just as the bullet makes initial contact with the block, the spring is uncompressed. We assume that the compression that occurs between then and the time the bullet comes to a halt is negligible, so we neglect force exerted on the block during that tiny initial period of compression. At that point in time momentum conservation requires

 

(bullet mass)*(initial bullet velocity) = (bullet mass + block mass)*(velocity of block/bullet system)

 

That still has unknowns in it. But the kinetic energy of the block/bullet system right then is what compresses the spring, and that has to match the energy obtained from the given amplitude later when the spring reaches maximum compression. So

 

0.5 * (bullet mass + block mass) * (velocity of block/bullet system)^2 = (spring energy at max compression)

 

That lets us solve for (velocity of block/bullet system), and then we can use the momentum conservation equation I listed first above to solve for (initial bullet velocity).

 

The point of my last reply is that you are not allowed to neglect losses if you treat the bullet and block as one object with one velocity after the collision. It just doesn't work.

 

Obviously momentum is not going to be match the initial momentum total once the spring has significant compression; some of the momentum has been absorbed by force applied by the spring on the block/bullet system. That's why you have to make that assumption of negligible collision duration.

[mistermack]

Obviously momentum is not going to be match the initial momentum total once the spring has significant compression; some of the momentum has been absorbed by force applied by the spring on the block/bullet system. That's why you have to make that assumption of negligible collision duration.

That's what I said. You can't have conservation of momentum if the xx (end of the spring) is rigidly fixed.

So if momentum is not conserved, you can't use the conservation of momentum in your calculation.

 

! edited, "block" changed to end of the spring.

I didn't mean that the block was fixed.

Edited March 27, 2017 by mistermack

[KipIngram]

Ok, you are correct, but the block is not rigidly fixed; we're told how far it's moved when the spring reaches maximum compression. It moves. The other end of the spring is assumed to be rigidly fixed. And in the strictest possible sense momentum is not conserved once the block has moved at all, because the spring is now slightly compressed and has produced a force. But if you follow J.C. MacSwell and neglect that, which truly will be reasonably negligible in a real situation of this type, then you can conserve momentum over that time range.

 

It's really the only way you can get an answer to this problem without understanding the deep and gory details of the ballistic impact physics. To remove that assumption you'd need a model for precisely how the relative velocity between the bullet and the block dropped to zero during the impact event. With such a model in hand you could solve the problem exactly.

[mistermack]

I don't think you can use momentum at all, unless you know the mass of the object to which the spring is attached, and it's final velocity.

It's like shooting a bullet into the ground. The bullet stops, (nearly) and the Earth accelerates (tiny amount) and momentum is conserved.

Even in that case, nothing is fixed, otherwise conservation of momentum becomes useless.

[KipIngram]

You're incorrect. J.C. MacSwell gave precisely the condition for applying momentum conservation to this problem, and we've stated it several times above. However, in a completely strict sense you are right - the answer we obtain using J.C.'s approach is an approximation. A very very good one, in my opinion, but an approximation nonetheless.

 

The block is not fixed. It is free to move. Like I pointed out, we absolutely know that it moves, because the original problem statement tells us exactly how far it moves - the "amplitude." That provides a spring energy which can be used to calculate block/bullet velocity immediately following the collision. But you cannot get the initial bullet velocity from energy conservation, because the total energy to which the pre-impact bullet's motion contributes has not been conserved. The only way to get initial bullet velocity is with a momentum conservation, and that requires you to assume an instantaneous impact event, per J.C.

 

We even have confirmation that trying to use energy conservation alone to get the bullet velocity doesn't work - the OP tried and he couldn't get the right answer. That's why he posted here in the first place.

 

I'm letting this one go now - we have all the salient points the OP should need to help him along.

[swansont]

The assumption usually applied it that the collision is essentially instantaneous. The displacement of the block in the time of the collision is negligible, so one can ignore external forces.

[J.C.MacSwell]

The assumption usually applied it that the collision is essentially instantaneous. The displacement of the block in the time of the collision is negligible, so one can ignore external forces.

Right. Also the spring is assumed massless, but fixed at one end or attached to a massive object (can be assumed infinite) that can absorb (in case of an idealized damper) or can reverse (in this case of an idealized spring) any momentum in the remainder of the system.

It is always good to look at it and ask if the assumptions are reasonable for the purpose. If, say, the bullet takes any significant time to imbed, the results will be off.