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)

$90^{\circ}$

Answer 2

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 $\cos\, \theta=\frac{(5)^{2}-(\sqrt{5})^{2}-(2 \sqrt{5})^{2}}{2(\sqrt{5})(\sqrt{5})}$ $\Rightarrow \cos\, \theta=\frac{25-5-20}{10}$ $=\frac{0}{10}=0$ $\Rightarrow \theta=90^{\circ}$