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Asked by Rayyan Last Modified
Answer 
Let u = initial velocity of the first particle
V1, V2 ard the velocities of first and second particle respectively
θ1 & θ2 are tbe angles made by the first & second particle respectively with the original direction of U
Let u = initial velocity of the first particle
V1, V2 ard the velocities of first and second particle respectively
Equting conservation of momentum in the direction of initial direction of U, & perpendicular to it, one get,
mU + 0 = m V1 cos θ1+ m cosθ2............. i
0 + 0 = m V1 sinθ1 + m V2 sinθ2.............ii
For elastic collision kinetic energy is also conserved.
Equating i initial & final kinetic energener,
(1/2)mu2 + 0 = (1/2) mV12 + (1/2) m V22 .. Iii
& adding eqn. i, ii,
we get
U2 = V12 + V22 + 2 V1 V2 cos ( θ1 + θ2 )
This equation will be identical to eqn. iii, only when ( θ1 + θ2 ) = 90 degree.
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