When a particle suffers an oblique elastic collision with another particle of equal mass and initially at rest, the two particles would move in mutually perpendicular directions after collision, explain.

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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...
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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 V1V2 cos (θ1 + θ2 ) This equation will be identical to eqn. iii, only when ( θ1 +θ2 ) = 90 degree. + read less
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