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Asked by Akhileshwar Last Modified
Answer 
Aarti Chowbey
(a) For head on collission:
Conservation of momentum ⇒ 2mv0 = mv1 + mv2
Since e <1 ⇒ v1 has the same sign as v0, therefore the ball moves on after collission.
(b) Conservation of momentum ⇒ p = p1+ p2
⇒ p2/2m > p22/2m + p22
Thus p, p1 and p2 are related as shown in the figure.
θ is acute (less than 90°) (p2 = p12 + p22 would give θ = 90°)
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