Oblique Collisions and Momentum Conservation in Identical Mass Objects

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In oblique collisions involving identical mass objects, momentum conservation principles dictate that the two objects move at right angles to each other. The equations presented indicate that momentum is conserved only when the collision occurs at these right angles, as they are vector quantities. The discussion highlights a confusion regarding the terms in the equations, particularly the absence of p2.p2 in the final line, suggesting a need for clarification in the derivation. The correct substitution shows that p.p should equal p1.p1 plus p2.p2, reinforcing the vector nature of momentum. Understanding these relationships is crucial for accurately analyzing oblique collisions in physics.
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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?
 
Last edited:
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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
 
haruspex said:
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?
 
Last edited:
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.
 

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