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A cricket ball of mass 150 g has an intial velocity u = `(3 hati + 4 hatj)ms^(-1)` and a final velocity `v = -(3hati + 4 hatj) ms^(-1)` , after being hit. The change in momentum (final momentum - initial momentum ) is (in `Kg ms^(1)`)A. zeroB. `-(0.45 hati + 0.6 hatj)`C. `-(0.9 hatj + 1.2 hatj)`D. `-5(hati + hatj ) hati` |
Answer» Correct Answer - C Given , `" " v = (3 hati + 4 hatj)` m/s and `" " v = - (3 hati + 4 hatj)` m/s Mass of the ball = 150 g = `0.15` kg `Delta`p = Change in momentum = Final momentum - initial momentum = mv - mu = `m(v - u) = (0.15) [-(3hati + 4 hatj ) - (3hati + 4 hatj)]` = `(0.15)[-6 hati - 8 hatj]` = `-[0.15 xx 6 hati + 0.15 xx 8 hatj]` = `-[0.9 hati + 1.20 hatj ]` Hence , `" " Delta p = -[0.9 hat i + 1.2 hatj]` |
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