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

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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