What special condition is necessary for the law of conservation of momentum to be valid?

Question

Help me please!

Q#2: In the collision of objects, is the total momentum of the system always conserved? How or why?

q#2 emphasis on ALWAYS

thanks

Answer 1

The real answer is that it is ALWAYS conserved. Even when another force, like friction, acts on the system (you must then consider the force as part of the system and on the atomic level you will still have momentum conservation).

The necessary condition for momentum conservation is that the Universe must be homogenous, or that the laws of physics are invariant under linear translation, which they are (this condition leads mathemtically to conservation of momentum).

Similarly for angular momentum (the laws of physics must be invariant under spherical translation).

Answer 2

Ans 1.The special condition according me is that there should not be any external forces acting on the system in the direction in which you are applying the conservation of momentum.

Ans 2. Not the total momentum but yes the momentum along the direction along which there is no effect of external forces.

Answer 3

Yes; the total momentum of the system after the collision is equal to the momentum of the system before the collision. That's what "conservation of momentum" means.

Answer 4

The condition is that there are no other forces acting on the system. A force is equal to some change in momentum per unit time. So if a force is exerted, momentum is added to (or subtracted from) the system.

Answer 5

The regulations of angular momentum didn't exist within the few planck instances after the gigantic bang. So the occasion might now not violate legislation that weren't lifestyles. Inflation is a latest concept that has promise as to how the universe improved so speedily. B

Answer 6

yes, momentum of the system is always conserved provided that there are no external forces acting on the objects (e.g. friction is absent.)
This is written mathematically as follows:
m1 u1 + m2 u2 = m1 v1 + m2 v2
where
m1 = mass of object 1
m2 = mass of object 2
u1 = initial velocity 0f object 1
u2 = initial velocity of object 2
v1 = final velocity of object 1
v2 = final velocity of object 2
Answer to ques #2 The proof is as follows:
since there is a change of momentum involved,
change of momentum of object 1 = m1v1 - m1u1
Rate of change of momentum in time t= (m1v1 - m1u1) / t
which is equal to the force acting on the object, by Newton's Second Law or else
F = (m1v1 - m1 u1) / t --------- (1)
Similarly change of momentum of object 2 = (m2v2 - m2u2)
And rate of change of momentum in time t = (m2v2 - m2u2)/t
which is also equal to the force on this object,
F = (m2v2 - m2u2) / t ----------- (2)
Now since the force acting on obj 1= force acting on obj 2 ( by third law of newton)
therfore
(m1v1 - m1u1) /t = - (m2v2 - m2u2) / t ( the negative sign indicates that the forces act in opposite directions.
multiplying both sides by t,
m1v1 - m1u1 = -(m2v2 - m2u2)
m1v1 - m1u1 = m2u2 - m2v2
bringing the initial and final velocities together,
m1v1 + m2v2 = m1u1 +m2u2
hence proved.

Answer 7

well.
it is exactly.for that u do it in absenting other external forces.