Question

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

• A. $Zero$
• B. $-(0.45\hat{i}+0.6\hat{j})$
• C. $-(0.9\hat{i}+1.2\hat{j})$
• D. $-5(\hat{i}+\hat{j})$

Answer 1

The correct option is C $-(0.9\hat{i}+1.2\hat{j})$

Here, m$=150g=0.15kg$

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

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

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

$=-(0.9\hat{i}+1.2\hat{j})kgm{s}^{-1}$