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  • A cricket ball of mass 150 g has an initial velocity u = (3 ˆ i + 4 ˆ j ) m s^ − 1 and a final velocity v = − (3 ˆ i + 4 ˆ j ) m s^ − 1 after being hit.

A cricket ball of mass 150 g has an initial velocity u = (3 \widehat{i} + 4 \widehat{j} ) m s ^{-1}
and a final velocity v= -(3 \widehat{i} + 4 \widehat{j} ) m s ^{-1} after being hit. The change
in momentum (final momentum-initial momentum) is (in kg m s1 )
(a) zero
(b) -(0.45 \widehat{i} + 0.6 \widehat{j} )
(c)-(0.9 \widehat{i} + 1.2 \widehat{j} )
(d) -5( \widehat{i} + \widehat{j} )

Answers (1)

The answer is the option (c) -(0.9 \widehat{i} + 1.2 \widehat{j} )

Explanation: Here, m = 150g = 0.15 kg

u = (3 \widehat{i} + 4 \widehat{j} ) m s ^{-1} &

v= -(3 \widehat{i} + 4 \widehat{j} ) m s ^{-1}

Now, we know that

Change in momentum \Delta (\overrightarrow{p})= Final momentum – initial momentum

                                                               =m\overrightarrow{v}-m\overrightarrow{u}

                                                               =m(\overrightarrow{v}-\overrightarrow{u})

                                                                = 0.15 [ -(3\widehat{i} + 4\widehat{j}) -(3\widehat{i} + 4\widehat{j})]

                                                                = 0.15 [ -6\widehat{i} - 8\widehat{j}) ]

                                Therefore, \Delta \overrightarrow{p} = 0.9\widehat{i} -1.2\widehat{j} = - [0.9\widehat{i} + 1.2\widehat{j}]

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