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Momentum question

Question 3 of this paper:

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FAQA%2F04-Mechanics-and-Materials%2FSet-M%2FMomentum%2520QP.pdf

quick question:
how do we that both the particles do not move at the same velocity together? if they are both moving together at 2.8ms^-1 then kinetic energy is still conserved, yet the answer is A? why is this? is there something I am misunderstanding about elastic collisions?
Original post by Glooc
Question 3 of this paper:

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FAQA%2F04-Mechanics-and-Materials%2FSet-M%2FMomentum%2520QP.pdf

quick question:
how do we that both the particles do not move at the same velocity together? if they are both moving together at 2.8ms^-1 then kinetic energy is still conserved, yet the answer is A? why is this? is there something I am misunderstanding about elastic collisions?

Momentum would not be conserved for c)
Original post by Glooc
Question 3 of this paper:

https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FPhysics%2FA-level%2FTopic-Qs%2FAQA%2F04-Mechanics-and-Materials%2FSet-M%2FMomentum%2520QP.pdf

quick question:
how do we that both the particles do not move at the same velocity together? if they are both moving together at 2.8ms^-1 then kinetic energy is still conserved, yet the answer is A? why is this? is there something I am misunderstanding about elastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved.

Initial momentum = 8kg ms^-1.
By looking at option C, final momentum = 2.8x2x2=11.2kg ms^-1.
So momentum is not conserved.

Let Va and Vb be velocities after collision.
In order to solve the problem we can see that:
To conserve momentum, after collision: (Va + Vb)2 = 8
So Va + Vb = 4

KE before = 1/2 x 2.0 x 16 = 16J
Thus 1/2 x 2 (Va^2 + Vb^2) = 16
So Va^2 + Vb^2 = 16

We can solve these equations to get Va = 4 and Vb = 0 (or vice versa).

As momentum is a vector, B is impossible as it would be negative so not conserved. Thus A is the answer.
If a collision is perfectly elastic, you should know that there is a rule about velocity of approach and velocity of separation.
In this case, the velocity with which the 2 objects separate must be the same as when they approach. 4m/s in this case.
The only way for the 2 objects to separate at 4 m/s (and conserve momentum) is for T1 to stop and T2 to move away from it at 4m/s.

Your suggestion that they both move at 2.8 m/s is not an elastic collision. (The same goes for the other option of 1.4m/s)
If they both moved at that speed in the same direction they would have merged and that is a perfectly inelastic collision. (No separation after impact.)
If one moved one way and the other moved the other way both at 2.8 m/s then the velocity of separation would be 2.8+2.8 = 5.6 m/s which is impossible, as it cannot be greater that 4m/s.
(edited 1 year ago)
For elastic collision, a short cut can be to consider the 3 special cases: this is one of them, where the 2 masses are equal. The particles undergoing collision exchange velocities.

So particle 1 will inherit the velocity of particle 2 and vice versa.

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