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A cricket ball of mass 150 g has an initial velocity `bar(u)=(3hati-4hatj)ms^(-1)` and a final velocity `bar(v)=-(3hati-4hatj)ms^(-1)` after being hit. The change in momentum (final momentum -initial momentum) is `("in kg ms"^(-1))`A. zeroB. `-(0.45hati+0.6hatj)`C. `-(0.9hati+1.2hatj)`D. `-5(hati+hatj)` |
Answer» Correct Answer - C Here, m=150g =0.15kg `bar(u)=(3 hati+4 hatj)ms^(-1)` `bar(v)=-(3 hati+4 hatj)ms^(-1)` Initial momentum, `bar(p)_(i)=mbar(u)` `bar(p)_(f)=(0.15kg) (3 hati+4 hatj)ms^(-1)=(0.45hati+0.6hatj)kg ms^(-1)` Final momentum, `bar(p)_(i)=mbar(u)` `bar(p)_(f)=(0.15kg)(-3hati-4hatj)ms^(-1)` `=-(-0.45hati-0.6 hatj)kg ms^(-1)` Change in momentum, `Deltabar(p)=bar(p)_(f)-bar(p)_(i)` `=(-0.45hati-0.6hatj)kg ms^(-1) -(0.45 hati+0.6 hatj) kg ms^(-1)` `=(-0.9hati+1.2hatj)kg ms^(-1) ` |
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