Question

The speed of a particle changes from $\sqrt{5}m{s}^{-1}$ to $2\sqrt{5}m{s}^{-1}$ in a time If the magnitude of change in its velocity is $5m{s}^{-1}$, the angle between the initial and final velocities of the particle is

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. $60^{\circ }$
• D. $90^{\circ }$

Answer 1

Answer: (D)

Given, $v_i=\sqrt{5}m{s}^{-1},v_f=2\sqrt{5}m{s}^{-1}$ and
$\Delta v=5m{s}^{-1}$
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
$R^2=A^2+B^2+2AB\cos \theta$
Hence, the angle between the velocities,
$\cos \theta =\frac{\Delta v^2-v_i^2-v_f^2}{2v_i{v}_f}$
Putting the given values, we get
$\cos \theta =\frac{(5{)}^2-(\sqrt{5}{)}^2-(2\sqrt{5}{)}^2}{2(\sqrt{5})(\sqrt{5})}$
$\Rightarrow \cos \theta =\frac{25-5-20}{10}$
$=\frac{0}{10}=0$
$\Rightarrow \theta =90^{\circ }$