The speed of a particle changes from $\sqrt{5} \mathrm{~ms}^{-1}$ to $2 \sqrt{5} \mathrm{~ms}^{-1}$ in a…
The speed of a particle changes from $\sqrt{5} \mathrm{~ms}^{-1}$ to $2 \sqrt{5} \mathrm{~ms}^{-1}$ in a time $t$. If the magnitude of change in its velocity is $5 \mathrm{~ms}^{-1}$, the angle between the initial and final velocities of the particle is
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
Solution
Given, $v_i=\sqrt{5} \mathrm{~ms}^{-1}, v_f=2 \sqrt{5} \mathrm{~ms}^{-1}$ and
$
\Delta v=5 \mathrm{~ms}^{-1}
$
Since, both $v_i$ and $v_f$ are extreme speeds, i.e. at $t=0$ and $t=t$.
So, they can be considered as magnitude of the velocities at time, $t=0$ and $t=t$.
As we know that
$
R^2=A^2+B^2+2 A B \cos \theta
$
Hence, the angle between the velocities,
$
\cos \theta=\frac{\Delta v^2-v_i^2-v_f^2}{2 v_i v_f}
$
Putting the given values, we get
$
\begin{aligned}
\cos \theta & =\frac{(5)^2-(\sqrt{5})^2-(2 \sqrt{5})^2}{2(\sqrt{5})(\sqrt{5})} \\
\Rightarrow \quad \cos \theta & =\frac{25-5-20}{10}=\frac{0}{10}=0 \Rightarrow \theta=90^{\circ}
\end{aligned}
$
Hence, the correct option is (d)