Oblique Collisions and Momentum Conservation in Identical Mass Objects

[jsmith613]

In an oblique collision (when one object hits another object of IDENTICAL mass at a slight angle the two objects move at right angles to each other.

my book says

p² = p₁² + p₂²

For momenum conservtation:

p= p₁+ p₂

apparently these equations can only be consistent of the collision occurs at right angles. why?

 

[haruspex]

These are vectors, right?
p.p = (p1+p2).(p1+p2) [2nd equn]
= p1.p1 + 2.p1.p2 + p2.p2
= p.p + 2.p1.p2 [1st equn]
Hence p1.p2 = 0

 

[jsmith613]

These are vectors, right?
p.p = (p1+p2).(p1+p2) [2nd equn]
= p1.p1 + 2.p1.p2 + p2.p2
= p.p + 2.p1.p2 [1st equn]
Hence p1.p2 = 0

where has p2.p2 gone in the final line
surely the final line should be
= p1.p2 + p2.p2, no?

 

[haruspex]

1st equn says p.p = p1.p1 + p2.p2
To get from my second line to my third line, use this to substitute p.p for p1.p1 + p2.p2.

 

[PJC01]

This is because momentum conservation is based on the principle that the total momentum of a system remains constant before and after a collision. In an oblique collision, the two objects have different velocities and therefore different momenta before the collision. If the collision occurs at an angle, the direction of the velocities will change and the resulting momenta of the objects will also be in different directions. This means that the total momentum of the system will not be conserved in this case. However, if the collision occurs at right angles, the direction of the velocities will change but the magnitudes of the momenta will remain the same, allowing for momentum conservation to hold true.