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

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Aicharya · Accepted Answer

The given situation is shown in the following figure.

According to question, angle between two radius vector is $90^{\circ}$. $\text {i.e., } \theta+\theta=90^{\circ} \Rightarrow \theta=45^{\circ}$ $\therefore$ 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, $\mathbf{v}_1$ and $\mathbf{v}_2$ are perpendicular to each other, hence $\begin{aligned} & \mathbf{v}_1 \cdot \mathbf{v}_2=0 \\ & (-10 \hat{\mathbf{i}}-g \hat{\mathbf{j}}) \cdot 10 \hat{\mathbf{i}}-g t \hat{\mathbf{j}}=0 \\ \Rightarrow & -100+g^2 t^2 =0 \\ \Rightarrow & x^2=\frac{100}{g^2} =\frac{100}{10^2}=\mathrm{ls} \end{aligned}$

At high altitude, a body explodes which is at rest into two equal fragments with one fragment receiving…

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