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Question

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


A
Zero
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B
(0.45^i+0.6^j)
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C
(0.9^i+1.2^j)
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D
5(^i+^j)
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Solution

The correct option is C $$-(0.9\hat{i}+1.2\hat{j})$$
Here, m$$= 150g=0.15 kg$$

$$\overrightarrow{u}=(3\hat{i}+4\hat{j})ms^{-1}$$
$$\overrightarrow {v}=-(3\hat{i}+4\hat{j})ms^{-1}$$ 

Initial momentum, $$\overline{p}_i=m\overline{u}$$
$$\overline{p}_i=(0.15 kg)(3\hat{i}+4\hat{j})ms^{-1}=(0.45\hat{i}+0.6\hat{j})kg ms^{-1}$$
Final momentum, $$\overline{p}_i=m\overline{u}$$
$$\overline{p}_f=(0.15kg)(-3\hat{i}-4{j})ms^{-1}$$
$$=(-0.45\hat{i}-0.6\hat{j})kg ms^{-1}$$

Change in momentum, $$\triangle \overline {p}=\overline {p}_f-\overline{p}_i$$
$$=(-0.45\hat{i}-0.6\hat{j})kg ms^{-1} - (0.45\hat{i} + 0.6\hat{j})kg ms^{-1}$$
$$=(-0.9\hat{i}-1.2\hat{j})kg ms^{-1}$$
$$= - (0.9\hat{i} + 1.2\hat{j})kg ms^{-1}$$

Physics

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