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Question
A ball of mass m, moving with a speed 2v0, collides inelastically (e > 0) with an identical ball at rest. Show that for a general collision, the angle between the two velocities of scattered balls is less than 90°.
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Solution
Consider the diagram below for a general collision.
By the principle of conservation of linear momentum,
P = P1 + P2
For inelastic collision, some KE is lost, hence `p^2/(2m) > p_1^2/(2m) + p_2^2/(2m)`
∴ `p^2 > p_1^2 + p_2^2`
Thus, p, p1 and p2 are related as shown in the figure.
θ is acute (less than 90) `(p^2 = p_1^2 + p_2^2 "would given" θ = 90^circ)`
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