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

• A

  $2 m v$
• B

  $m v / \sqrt{2}$
• Correct answer
  C

  $m v \sqrt{2}$
• D

  zero

Solution

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 $m v \sin \theta-(-m v \sin \theta)=2 m v \frac{1}{\sqrt{2}}=\sqrt{2} m v$

Hence, the correct answer is option: 'C'.