Question

A ball of mass $0.2kg$ is thrown from a height of $1m$ and with an initial velocity of $\sqrt{10}m/s$ at an angle of $45^{\circ }$ with the horizontal. Assuming, acceleration due to gravity $g=10m/s^2$, then modulus of momentum increment during the total time of motion in $kgm/s$ is

• A. $\frac{2+\sqrt{10}}{\sqrt{10}}$
• B. $\frac{1+\sqrt{10}}{\sqrt{5}}$
• C. $\frac{1+\sqrt{5}}{\sqrt{5}}$
• D. $\frac{\sqrt{5}-1}{\sqrt{5}}$

Answer 1

Answer: (C)

There is no change in velocity in $x$ -direction, so momentum change occurs only in $y$ -direction.

Now initial momentum in $y$ direction is
$p_y($ initial $)=m{v}_y=0.2\left(\sqrt{10}\sin 45^{\circ }\hat{j}\right)$
$=\frac{1}{\sqrt{5}}\cdot \hat{j}k{g}^{-}m{s}^{-1}$
Final momentum in $y$ -direction is
$p{y}_{\text{final }}=m{v}_y=m\left(\sqrt{u_y^2+2gh}\right)$
$=\left(0.2\sqrt{\left({\sqrt{5}}^2\right)+2(-10)(-1)}\right)\hat{j}$
$=-1\hat{j}kg-m{s}^{-1}$ downwards
So, change in momentum
$\Delta p=p{y}_{(\text{final })}-p{x}_{\text{(initial) }}=-1\hat{j}-\frac{1}{\sqrt{5}}\hat{j}$
$=-\left(1+\frac{1}{\sqrt{5}}\right)\hat{j}k{g}^{-}m{s}^{-1}$
$\Rightarrow =|\Delta p|=1+\frac{1}{\sqrt{5}}=\frac{1+\sqrt{5}}{\sqrt{5}}$