0
$\begingroup$

First of all, I tell you that I know there is same question available which is already answered, but no answer cleared my doubt, that's why I am asking the question in more detail.

Let there be a collision between two bodies with velocity in same direction before and after the collision. During the collision of two bodies, heat energy is produced, so particles of the bodies start vibrating rapidly about there positions, due to transfer of kinetic energy of the bodies to constituent particles. This must lead to reduce in velocity of the two bodies, and consequently reduce in total momentum.

Thus, I am saying there may be loss of kinetic energy, but there should be loss of momentum too.

Please tell me where I am wrong.

Also please, don't state the law of conservation of momentum or its modified form and try to explain in practical terms.

$\endgroup$

marked as duplicate by Rob Jeffries, Qmechanic Nov 14 '17 at 23:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Your last statement - "in practical terms". There is always errors with experiments. In experiments you refine, refine refine, and you do enough of them to convince yourself that your though experiments, your hypothesis indeed true. $\endgroup$ – docscience Nov 14 '17 at 21:14
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/92051/2451 , physics.stackexchange.com/q/133759/2451 and links therein. $\endgroup$ – Qmechanic Nov 14 '17 at 21:29
  • 2
    $\begingroup$ Possible duplicate of How can momentum but not energy be conserved in an inelastic collision? $\endgroup$ – Chris Nov 14 '17 at 22:23
  • $\begingroup$ "Please tell me where I am wrong." Here: "This must lead to reduce in velocity of the two bodies, and consequently reduce in total momentum." Consider two equal balls $m$. One moves with $v$ and hits the stationary ball. They now move together $2m$ with $\dfrac{v}{2}$. The total momentum remains $mv$. The kinetic energy was $\dfrac{m}{2}v^2$, but now is $\dfrac{2m}{2}\left(\dfrac{v}{2}\right)^2=\dfrac{m}{4}v^2$ or a half of the original. The first ball lost a half the velocity, but the second ball gained just as much. Still a half of kinetic energy was lost, due to the velocity squared. $\endgroup$ – safesphere Nov 15 '17 at 0:56
  • $\begingroup$ The conceptual answer to your question of why energy is lost, but the momentum is not, is that there are different types of energy, so energy can convert from one type to another (e.g. kinetic to heat) in a non elastic collision. However, there is only one type of momentum in this case. There is nothing else for the momentum to convert to, so it must remain unchanged. $\endgroup$ – safesphere Nov 15 '17 at 1:02
1
$\begingroup$

Perhaps it would help to consider that momentum, unlike energy has direction. Think of two identical massive balls that impact one another at the same (but one has exactly the negative direction of the other) velocities, then stick together. Initially the system of the two balls had zero momentum, and also the final two ball mass has zero velocity thus zero momentum.

$\endgroup$
  • $\begingroup$ Yes, but when there is some final velocity of both the objects? and the 2 objects don't have initially equal velocity (when the objects have equal velocity, then only the objects in your answer will come to rest)? $\endgroup$ – TontyTon Nov 14 '17 at 21:06
  • $\begingroup$ That's a different thought experiment. And in that experiment if you have a small velocity afterwards you have to conclude that (1) the balls were not exactly the same mass or (2) their speeds were not exactly the same or (3) their velocities were not exactly colinear or (4) any combination of the prior three possibilities. $\endgroup$ – docscience Nov 14 '17 at 21:10
  • $\begingroup$ In the real world you will indeed see a final velocity from the original experiment since you cannot make everything perfect. But still in the real world momentum is conserved, just as in the world of thought experiments! $\endgroup$ – docscience Nov 14 '17 at 21:12
  • $\begingroup$ Sir, I am not contradicting momentum conservation theory, I accept it. But I am unable to understand how it is taking hold here? How even after reduce in speed of bodies momentum is conserved? $\endgroup$ – TontyTon Nov 14 '17 at 21:16
  • $\begingroup$ Please think more about my example and how you start with zero momentum, then wind up with zero momentum, masses being equal. $mv + (-mv) = 0$ v is a vector, therefor momentum is a vector. $\endgroup$ – docscience Nov 14 '17 at 21:23
1
$\begingroup$

Possibly one way to help to make the result more intuitive is to look at docscience's answer, but in one dimension only. In this case, these are the two things to notice:

  1. kinetic energy is always positive,

  2. momentum can be negative or positive.

That's how you can have loss of energy (and therefore speeds) without having loss of total momentum. Because, considering for simplicity the case where the absolute value of both speeds decrease with the collision, a sum of their squares will necessarily decrease, while a sum of their signed values can remain constant (since, e.g., $7-4=5-2$).

As a numerical example, consider (SI units) unit masses at speeds $-1$ and $2$: kinetic energy is $K_i=(-1)^2+(2)^2=5$, and $P_i=-1+2=1$ before the collision; after the collision, the common speed is $(-1+2)/2=1/2 \to$ $K_f=(1/2)^2+(1/2)^2=1/2<K_i$, and $P_f=1/2+1/2=1=P_i$.

Yet another example is the one you propose, let's make it numeric: say we have unit masses at speeds $1$ and $3$ (same direction): the kinetic energy is $K_i=(1)^2+(3)^2=10$, and $P_i=1+3=4$ before the collision; after the collision, the common speed is $(1+3)/2=2 \to$ $K_f=(2)^2+(2)^2=8<K_i$, and $P_f=2+2=4=P_i$. Once again energy is lost, a speed is reduced, but momentum is conserved.

there should be loss of momentum too. Please tell me where I am wrong. [...] in practical terms.

In very practical terms, the mistake is to ignore experimental results, which show momentum is conserved. Plus, your argumentation is theoretical, so the burden's on you to prove that "there should be loss of momentum".

$\endgroup$
  • $\begingroup$ I had already said in comments of docscience's, I am not contradicting the Law of conservation of momentum. So I am trying to prove myself how it is holding in my example. $\endgroup$ – TontyTon Nov 14 '17 at 21:58
  • $\begingroup$ I am talking about Inelastic collision, so velocity of objects (objects with equal masses) with exchanging will also get reduced. Am I right? $\endgroup$ – TontyTon Nov 14 '17 at 22:03
  • $\begingroup$ A perfectly inelastic collision is precisely my numerical example. Yes, both speeds in that example have their absolute values reduced (from 1 and 2 to 1/2), but the momentum stays at a constant 1. $\endgroup$ – stafusa Nov 14 '17 at 22:25
  • $\begingroup$ @TontyTon, I now added a numerical example for exactly the situation you mention in your question. $\endgroup$ – stafusa Nov 14 '17 at 23:29
  • 1
    $\begingroup$ @stafusa - no just for providing additional detailed example and explanation. Hopefully it's stuck! $\endgroup$ – docscience Nov 15 '17 at 17:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.