Question

At high altitude, a body explodes which is at rest into two equal fragments with one fragment receiving horizontal velocity of $10\, m \cdot s ^{-1}$. Time taken by the two radius vectors connecting point of explosion to fragments to make $90^{\circ}$ is

• A. 10 s
• B. 4 s
• C. 2 s
• D. 1 s

Answer 1

Answer: (C)

The given situation is shown in the following figure.

According to question, angle between two radius vector is $90^{\circ}$.
i.e., $\theta+\theta=90^{\circ}$
$\Rightarrow \theta=45^{\circ}$
$\therefore $ Time taken by the two radius vectors connecting of explosion to fragments is $t$ second, then
According to conservation of linear momentum,
$m_{1} v_{1}=m_{2} v_{2}$
$\Rightarrow m v_{1}=m v_{2}$
$\Rightarrow v_{1}=v_{2}$
Since, $v_{1}$ and $v_{2}$ are perpendicular to each other, hence
$v_{1} \cdot v_{2}=0$
$(-10 \hat{i}-g \hat{j}) \cdot 10 \hat{i}-g \hat{j}=0$
$\Rightarrow -100+g^{2} t^{2}=0$
$\Rightarrow x^{2}=\frac{100}{g^{2}}$
$=\frac{100}{10^{2}}=1\, s$