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Question: A cricket ball of mass 150 g has an initial velocity \(\overrightarrow{u} = (3\widehat{i} + 4\wideha...

A cricket ball of mass 150 g has an initial velocity u→=(3i^+4j^)ms−1\overrightarrow{u} = (3\widehat{i} + 4\widehat{j})ms^{- 1} and a final velocity v→=−(3i^+4j^)ms−1\overrightarrow{v} = - (3\widehat{i} + 4\widehat{j})ms^{- 1} after being hit. The change in momentum (final momentum – initial momentum) is (in kg ms−1ms^{- 1})

A

Zero

B

−(0.45i^+0.6j^)- (0.45\widehat{i} + 0.6\widehat{j})

C

−(0.9i^+1.2j^)- (0.9\widehat{i} + 1.2\widehat{j})

D

−5(i^+j^)- 5(\widehat{i} + \widehat{j})

Answer

−(0.9i^+1.2j^)- (0.9\widehat{i} + 1.2\widehat{j})

Explanation

Solution

Here m = 150g =0.5kg= 0.5kg

u→=(3i∧+4j∧)ms−1\overset{\rightarrow}{u} = (3\overset{\land}{i} + 4\overset{\land}{j})ms^{- 1}

v→=−(3i∧+4j∧)ms−1\overset{\rightarrow}{v} = - (3\overset{\land}{i} + 4\overset{\land}{j})ms^{- 1}

Initial momentum pi→=mu→\overset{\rightarrow}{p_{i}} = m\overset{\rightarrow}{u}

pi→=(0.15kg)(3i∧+4j∧)ms−1\overset{\rightarrow}{p_{i}} = (0.15kg)(3\overset{\land}{i} + 4\overset{\land}{j})ms^{- 1} =

(0.45i∧+0.6j∧)kgms−1(0.45\overset{\land}{i} + 0.6\overset{\land}{j})kgms^{- 1}

Final momentum pf→=mu→\overset{\rightarrow}{p_{f}} = m\overset{\rightarrow}{u}

}{= ( - 0.45\overset{\land}{i} - 0.6\overset{\land}{j})kgms^{- 1}}$$ Change in momentum $\Delta\overset{\rightarrow}{p} = \overset{\rightarrow}{p_{f}} - \overset{\rightarrow}{p_{i}}$ $$= ( - 0.45\overset{\land}{i} - 0.6\overset{\land}{j})kgms^{- 1} - (0.45\overset{\land}{i} + 0.6\overset{\land}{j})kgms^{- 1}$$ $$= - (0.9\overset{\land}{i} + 1.2\overset{\land}{j})kgms^{- 1}$$