Question

The speed of a particle changes from $\sqrt{5} \,ms^{- 1}$ to $2 \sqrt{5} \; ms^{-1}$ in a time If the magnitude of change in its velocity is $5 \; ms^{-1}$, 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, $v_{i}=\sqrt{5} ms ^{-1}, v_{f}=2 \sqrt{5} ms ^{-1}$ and
$\Delta v=5 \,ms ^{-1}$
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
$ \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 $
$\Rightarrow \theta=90^{\circ}$