Q117 / 125 · BITSAT 2015 · Physics
A particle of mass $2 \mathrm{~m}$ is projected at an angle of $45^{\circ}$ with the horizontal with a velocity of $20 \sqrt{2} \mathrm{~m} / \mathrm{s}$. After $1 \mathrm{~s}$, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is
- A$50 \mathrm{~m}$
- B$25 \mathrm{~m}$
- C$40 \mathrm{~m}$
- Correct answerD$35 \mathrm{~m}$
Solution
Given : Initial velocity $u_{0}=20 \sqrt{2} \mathrm{~m} / \mathrm{s} ;$ angle of projection $\theta=45^{\circ}$
Therefore horizontal and vertical components of initial velocity are $\mathrm{u}_{\mathrm{x}}=20 \sqrt{2} \cos 45^{\circ}=20 \mathrm{~m} / \mathrm{s}$
and $\mathrm{u}_{\mathrm{y}}=20 \sqrt{2} \sin 45^{\circ}=20 \mathrm{~m} / \mathrm{s}$
After $1 \mathrm{~s}$, horizontal component remains unchanged while the vertical component becomes $v_{y}=u_{y}-g t$
Due to explosion, one part comes to rest. Hence, from the conservation of linear momentum, vertical component of second
part will become $v_{y}^{\prime}=20 m / s$ Therefore, maximum height attained by the second part will be $\mathrm{H}=\mathrm{h}_{1}+\mathrm{h}_{2}$
Using, $\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$
$\Rightarrow \mathrm{h}_{1}=(20 \times 1)-\frac{1}{2} \times 10 \times(1)^{2}=15 \mathrm{~m}$
$\mathrm{a}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$
$\mathrm{h}_{2}=\frac{\mathrm{v}_{\mathrm{y}}^{\prime 2}}{2 \mathrm{~g}}=\frac{(20)^{2}}{2 \times 10}=20 \mathrm{~m}$
$\mathrm{H}=20+15=35 \mathrm{~m}$
Hence, the correct answer is option: 'D'.