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I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question.
The problem is as follows:

A bullet of mass $25\,g$ is fired horizontally into a ballistic pendulum of mass $5.0\,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10\,cm$, find the speed of the bullet.
(H.C. Verma, Centre of Mass, Q47)


Let the mass of the bullet be $m=25\,g$.
Let the mass of the pendulum be $M=5\,kg$.
Let the peak height be $h=0.1\,m$.
Let the initial and final velocities be $u$ and $v$ respectively.

Method 1: (Momentum Conservation)

$$mu=(M+m)v \implies v=\frac{mu}{M+m}\\ \text{Also, }\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2\biggl(\frac{M+m}{m}\biggr)^2gh\\ \implies u=201\sqrt2 \text{ m/s}$$

Method 2: (Energy Conservation)

Since both the masses move together after the collision, and because the velocity at highest point is null, therefore: $$\frac{1}{2}mu^2=\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2gh\biggl(\frac{M+m}{m}\biggr) \implies u=\sqrt{402}$$


To summarize myself, I am curious about the following:
Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?

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closed as off-topic by AccidentalFourierTransform, John Rennie, Jon Custer, user191954, Cosmas Zachos Nov 3 '18 at 15:43

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – Ben Crowell Nov 2 '18 at 3:30
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Shouldn't the energy be conserved as well as the momentum?

But this is a perfectly inelastic collision, i.e., kinetic energy is not conserved (maximally not conserved in fact).

From the Wikipedia article Inelastic collision:

An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction.

...

A perfectly inelastic collision occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together.

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    $\begingroup$ Oh, so I should have concluded from the fact that the masses stick, therefore inelastic collision. Thanks. $\endgroup$ – Utkarsh Verma Nov 2 '18 at 3:07
  • $\begingroup$ @UtkarshVerma think of the heat produced and absorbed in "embedding" $\endgroup$ – anna v Nov 2 '18 at 4:38
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    $\begingroup$ @UtkarshVerma A way to convince yourself that objects sticking is inelastic is consider that you can change reference frame to one in which the combined object has zero speed (this is the zero momentum frame). In this frame the system has kinetic energy before the collision and no kinetic energy after the collision. This frame is not useful for solving the rest of the problem though. $\endgroup$ – user668074 Nov 2 '18 at 9:08
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You can use energy conservation, but you have to take into account the heat produced by friction as the bullet entered the block, the rotational energy imparted into the mass itself in case the bullet hits it off-center, not to mention the energy used to deform both bullet and mass (does that automatically transfer into heat?) I'm sure i've missed a few as well.

It's theoretically possible, but not recommended.

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  • $\begingroup$ Either object may, in principle, end with some elastic energy, and that will not go to heat. In practice I think that there will be essentially no stored elastic energy in the metal (lead being ductile and likely being hot at the time to boot), and that the amount stored in the wood is likely to be small by comparison to the that converted to thermal energy. $\endgroup$ – dmckee Nov 2 '18 at 21:33
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As in another answer, it's just that some of the kinetic energy is spent on heating the bob of the pendulum.

Energy conservation figures in the calculations when either there is effectively zero energy converted into heat, as in, say, the gravitational interaction of celestial bodies; or the heat figures in the dynamics, such as in the propagation of a shock, or flow of gas along a duct.

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Energy is conserved only in elastic collisions, which is not this case. You can think it as energy is dissipated as heat when the bullet hits the pendulum, thus it's conserved too, though that's not seems so clear.

The point of using momentum is precisely that we don't need worry about the heat transference of the system and get along using just the given data. Also, momentum gives us very usefull information over the direction of the resulting velocity, which isn't present at energy calculations.

Be aware that some problems must be solved using both momentum and energy conservation.

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The energy IS conserved .... And I urge you to ignore any answer that says it isn't since it can lead to confusion.

It just happens to be in heat/sound rather than kinetic; and since these aren't usually measured or included in the question details, there is not sufficient information to calculate using conservation of energy.

Were you told in the question that a soundwave of (say) 30J and heating of 50J was done to the system; then you could subtract that 80J from your speeding bullet energy and get the same answer.

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  • $\begingroup$ This is, of course, perfectly correct. But to be really useful you have to get the student to notice (explicitly!) that they were trying to conserve only one channel (bulk kinetic energy) and to recognize that there is no general rule suggesting that such a thing is reasonable. $\endgroup$ – dmckee Nov 2 '18 at 21:30
  • $\begingroup$ @dmckee agreed - and the simplest way to describe it would be "how loud do you think it will be when this bullet hits this metal block?" $\endgroup$ – UKMonkey Nov 2 '18 at 23:28

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