Q122 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Physics

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

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

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).

Hence, the correct answer is option: 'D'.

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