0

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.

stafusa
  • 13,064
TontyTon
  • 95

2 Answers2

1

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.

docscience
  • 11,779
1

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".

stafusa
  • 13,064