InterviewSolution
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InterviewSolution
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A ball strickes a horizontal floor at `45^(@). 25%` of its kinetic energy is lost in collision. Find the coefficient of restitution.A. `(1)/(2)`B. `(1)/(sqrt2)`C. `(1)/(2 sqrt2)`D. `(1)/(4)` |
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Answer» Correct Answer - C Fraction of `KE` lost in collision `Delta K % = Delta K % = ((1)/(2) mu^(2) - (1)/(2) mv^(2))/((1)/(2) m u^(2)) = 1 - ((v)/(u))^(2) = (1)/(4)` (given) `v = y sqrt((3)/(4)` The ball strikes at `45^(@)` will hot change while component of velocity normal to wall will change. `v_(x) = u cos 45^(@) = (u)/(sqrt(2)` `v_(y) = eu cos 45^(@) = (eu)/(sqrt(2)` `v = sqrt(v_(x)^(2) + v_(y)^(2)) = [((u)/(sqrt2))^(2) + ((eu)/(sqrt2))^(2)])^(1)/(2)` `implies v = u [(1)/(2) + (e^(2))/(2)]^(1)/(2)` Solving (i) and (ii), we get `e = (1)/(sqrt2)`. |
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