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A particle of mass $m$ is projected with velocity $y$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be
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Verified Answer
The correct answer is:
$m v \sqrt{2}$
Key Idea: Required momentum is the difference of final and initial momentum. The situation is shown in figure.
Change in momentum,
$\begin{aligned}
\Delta \overrightarrow{\mathbf{p}} & =\overrightarrow{\mathbf{p}}_f-\overrightarrow{\mathbf{p}}_i \\
& =m\left(\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right) \\
& =m\left[\left(v \cos 45^{\circ} \hat{\mathbf{i}}-v \sin 45^{\circ} \hat{\mathbf{j}}\right)\right. \\
& \left. -\left(v \cos 45^{\circ} \hat{\mathbf{i}}+v \sin 45^{\circ} \hat{\mathbf{j}}\right)\right] \\
& =m\left[\left(\frac{v}{\sqrt{2}} \hat{\mathbf{i}}-\frac{v}{\sqrt{2}} \hat{\mathbf{j}}\right)-\left(\frac{v}{\sqrt{2}} \hat{\mathbf{i}}+\frac{v}{\sqrt{2}} \hat{\mathbf{j}}\right)\right] \\
& =-\sqrt{2} m v \hat{\mathbf{j}} \\
\therefore & {[\Delta \overrightarrow{\mathbf{p}}]=\sqrt{2} m v }
\end{aligned}$
Alternative :
The horizontal momentum does not change. The change in vertical momentum is $m v \sin \theta-(-m v \sin \theta)=2 m v \frac{1}{\sqrt{2}}=\sqrt{2} m v$
Change in momentum,
$\begin{aligned}
\Delta \overrightarrow{\mathbf{p}} & =\overrightarrow{\mathbf{p}}_f-\overrightarrow{\mathbf{p}}_i \\
& =m\left(\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right) \\
& =m\left[\left(v \cos 45^{\circ} \hat{\mathbf{i}}-v \sin 45^{\circ} \hat{\mathbf{j}}\right)\right. \\
& \left. -\left(v \cos 45^{\circ} \hat{\mathbf{i}}+v \sin 45^{\circ} \hat{\mathbf{j}}\right)\right] \\
& =m\left[\left(\frac{v}{\sqrt{2}} \hat{\mathbf{i}}-\frac{v}{\sqrt{2}} \hat{\mathbf{j}}\right)-\left(\frac{v}{\sqrt{2}} \hat{\mathbf{i}}+\frac{v}{\sqrt{2}} \hat{\mathbf{j}}\right)\right] \\
& =-\sqrt{2} m v \hat{\mathbf{j}} \\
\therefore & {[\Delta \overrightarrow{\mathbf{p}}]=\sqrt{2} m v }
\end{aligned}$
Alternative :
The horizontal momentum does not change. The change in vertical momentum is $m v \sin \theta-(-m v \sin \theta)=2 m v \frac{1}{\sqrt{2}}=\sqrt{2} m v$
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