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A cricket ball of mass `150g` has an initial velocity `(3 hati + 4hatj)ms^(-1)` and a final velocity `upsilon = -(3hati + 4hatj)ms^(-1)` after beigh hit The change in momentum (final momentum initial momentum) is (in kg `ms^(-1)`)A. zeroB. `-(0.45hati +0.6 hatj)`C. `-(0.9 hatj +1.2 hatj)`D. `-5(hati +hatj)hati` |
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Answer» Correct Answer - C Given `u = (3 hati + 4hatj)m//s` and `v = - (3hati +4hatj) m//s` mass of the ball `=150 g = 0.15 kg` `DeltaP = mv -"mu"` `DeltaP =m(v -u) = - (0.15) [(3hati + 4hatj)-(3hati + 4hatj)]` `= (0.15) [-6hati -8hatj]` Hence `Deltap = - [0.9hati + 1.2hatj]` . |
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