Question

A ball of mass $0.2\, kg$ is thrown from a height of $1\, m$ and with an initial velocity of $\sqrt{10}\, m / s$ at an angle of $45^{\circ}$ with the horizontal. Assuming, acceleration due to gravity $g=10\, m /s ^{2}$, then modulus of momentum increment during the total time of motion in $kg\, m / s$ 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 $)=m v_{y}=0.2\left(\sqrt{10} \sin 45^{\circ} \hat{ j }\right)$
$=\frac{1}{\sqrt{5}} \cdot \hat{ j } kg ^{-} ms ^{-1}$
Final momentum in $y$ -direction is
$p y_{\text {final }} =m v_{y}=m\left(\sqrt{u^{2}_{y}+2 g h}\right)$
$=\left(0.2 \sqrt{\left(\sqrt{5}^{2}\right)+2(-10)(-1)}\right) \hat{ j }$
$=-1 \hat{ j } kg - ms ^{-1}$ downwards
So, change in momentum
$\Delta p=p y_{(\text {final })}-p x_{\text {(initial) }}=-1 \hat{ j }-\frac{1}{\sqrt{5}} \hat{ j }$
$=-\left(1+ \frac{1}{\sqrt{5}}\right) \hat{ j } kg ^{-} ms ^{-1}$
$\Rightarrow = |\Delta p|=1+\frac{1}{\sqrt{5}}=\frac{1+\sqrt{5}}{\sqrt{5}}$