Physics mcq help

[bloooooo]

Plz.....anyone help me with OCT/NOV 2008 p1 question no.10. i just cant do it. i get option a but marking scheme says option C?????????????
PS i tried to upload the pic but cudnt but i'll copy the txt though it duznt giv a clear diagram
Two spheres approach each other along the same straight line. Their speeds are u1 and u2
before collision, and v1 and v2 after collision, in the directions shown below.
before collision:
u1----------> <----------------u2


after collision:
v1-----------> ---------------->v2
Which equation is correct if the collision is perfectly elastic?
A u1 – u2 = v2 + v1
B u1 – u2 = v2 – v1
C u1 + u2 = v2 + v1
D u1 + u2 = v2 – v1

 

[hassam]

C

 

[bloooooo]

C

No dude....its D. nd btw how did u got it????

 

[jumana94]

look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..

and good luck in ur paper 1

 

[bloooooo]

look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..

and good luck in ur paper 1

Thanks DUDE!!!! u rock!!

 

[papajohn]

look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..

and good luck in ur paper 1

Three year later, Papa john get benefit from it.May it continue to help other people

 

[mjbjhb]

look ... this question deals with relative speeds rather than just momentum.... in a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation.. but what is meant by relative speed of the 2 objects.... imagine u are moving in a car at a speed of 100 km/h beside another car moving at the same speed that is 100 km/h in the same direction.. u will feel that the other car is not moving.....also if u move at a speed of 100 km/h and u look at another car moving in opposite direction at a speed of 100 km/h u feel that the other car is moving very fast .. that is at a speed of 200 km/h......... so to find relative speed of 2 objects ...add their speeds if they are moving in opposite direction and subtract their speeds when they are moving in the same direction... but only in a perfectly elastic collision.
In this question, the relative speed of approach is equal to u1 + u2 since they are moving in opposite directions.. this is equal to the relative speed of separation which is equal to v2 - v1 since they are moving in the same direction after the collision..

and good luck in ur paper 1

I have no idea how to thank you enough, this was confusing me for a while now, THANK YOU!! 8)

 

[Hassan Altaf Cheema]

Use newtons law of restitution
u1+u2 = -(v1-v2)