Question

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

The initial velocity of the particle is $5\sqrt{2}\mathrm{m}/\mathrm{s}$ 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 $x$, to the nearest integer, is ……….

Answer 1

Step 1: Given data:

Initial velocity $u=5\sqrt{2}\mathrm{m}/\mathrm{s}$

Mass of particle $m=5\mathrm{g}=5\times {10}^{-3}\mathrm{kg}$

Air resistance is negligible

Change in momentum $P=x\times {10}^{-2}{\mathrm{kgms}}^{-1}$

Step 2: Draw the diagram:

Step 3: Formula used:

From the formula of momentum

$P=mv$

where, $m,v$ are mass and velocity of particle

Step 4: Calculating the change in momentum

First find x and y component of initial velocity-

$u=u\cos 45\overbrace{i}+u\sin 45\overbrace{j}$

Similarly, for final velocity-

$v=v\cos 45\overbrace{i}-v\sin 45\overbrace{j}$

Change in momentum-

$\begin{array}{rcl}∆\mathrm{P}& =& \mathrm{m}(\mathrm{v}-\mathrm{u})\\ & =& \mathrm{m}\left[\left(\mathrm{vcos}45\hat{\mathrm{i}}-\mathrm{vsin}45\hat{\mathrm{j}}\right)-\left(\mathrm{ucos}45\hat{\mathrm{i}}+\mathrm{usin}45\hat{\mathrm{j}}\right)\right]\\ & =& \mathrm{m}\left[2\mathrm{u}\left(\frac{1}{\sqrt{2}}\right)\right]\\ \mathrm{x}\times {10}^{-2}& =& \left(5\times {10}^{-3}\right)\times \left(2\mathrm{u}\right)\times \left(\frac{1}{\sqrt{2}}\right)\\ \mathrm{x}\times {10}^{-2}& =& \left(5\times {10}^{-3}\right)\times \left(2\times 5\sqrt{2}\right)\times \left(\frac{1}{\sqrt{2}}\right)\\ \mathrm{x}\times {10}^{-2}& =& 5\times {10}^{-2}\\ \mathrm{x}& =& 5\end{array}$

Hence, The magnitude of the change in momentum between the points A and B is $x\times {10}^{-2}{\mathrm{kgms}}^{-1}$Where, the value of $x$, to the nearest integer, is $5$.