The projectile motion of a particle of mass

Question:

The projectile motion of a particle of mass $5 \mathrm{~g}$ is shown in the figure.

The initial velocity of the particle is

$5 \sqrt{2} \mathrm{~ms}^{-1}$ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points $A$ and $B$ is

$x \times 10^{-2} \mathrm{kgms}^{-1}$. The value of $\mathrm{x}$, to the nearest integer, is_______.

Solution:

$|\overrightarrow{\mathrm{u}}|=|\overrightarrow{\mathrm{v}}|$    $\ldots(1)$

$\overrightarrow{\mathrm{u}}=u \cos 45 \hat{\mathrm{i}}+\mathrm{u} \sin 45 \hat{\mathrm{j}} \quad \ldots(2)$

$\vec{v}=v \cos 45 \hat{i}-v \sin 45 \hat{j}$ ......(3)

$|\overrightarrow{\Delta \mathrm{P}}|=|\mathrm{m}(\overrightarrow{\mathrm{v}}-\overrightarrow{\mathrm{u}})|$ .......(4)

$\Delta \mathrm{P}=2 \mathrm{mu} \sin 45^{\circ}$

$=2 \times 5 \times 10^{-3} \times 5 \sqrt{2} \times \frac{1}{\sqrt{2}}$

$=50 \times 10^{-3}$

$=5 \times 10^{-2}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now