Two particles of equal masses moving with same speed collide perfectly inelastically. After the collision the combined mass moves with half of the speed of the individual masses. The angle between the initial momenta of individual particle is
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a
60∘
b
90∘
c
120∘
d
45∘
answer is C.
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Detailed Solution
As P→1=P→2=|P→|=P0 (say), P→ is final momentum P=2mv2=P1=P2From conservation of linear momentum P→1+P→2=P→ ∴ P=P12+P22+2P1P2cosθ or P=P2+P2+2P2cosθ∴ cosθ=−12⇒θ=120∘