Laws of Motion 4 Question 2
2. A particle is moving along a circular path with a constant speed of $10 ms^{-1}$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60^{\circ}$ around the centre of the circle?
(2019 Main, 11 Jan I)
(a) $10 \sqrt{2} m / s$
(b) $10 m / s$
(c) $10 \sqrt{3} m / s$
(d) Zero
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Answer:
Correct Answer: 2. (b)
Solution:
- Let $v _1$ be the velocity of the particle moving along the circular path initially, $v _1$ and $v _2$ be the velocity when it moves through an angle of $60^{\circ}$ as shown below.
From the figure,
$$ \begin{aligned} \Delta \mathbf{v} & =\mathbf{v} _2-\mathbf{v} _1 \\ \Rightarrow \quad|\Delta \mathbf{v}| & =2 v \sin \frac{\theta}{2}=2 \mathbf{v} \sin 30^{\circ} \quad\left[\because\left|\mathbf{v} _1\right|=\left|\mathbf{v} _2\right|\right] \\ & =2 v \times \frac{1}{2}=v \quad(\text { Given, } v=10 m / s) \\ \Rightarrow \quad|\Delta \mathbf{v}| & =10 m / s \end{aligned} $$
Alternate method
$\because \Delta \mathbf{v}=\mathbf{v} _2-\mathbf{v} _1=\mathbf{v} _2+\left(-\mathbf{v} _1\right)$
$\therefore|\Delta \mathbf{v}|^{2}=v _1^{2}+v _2^{2}+2 v _1 v _2 \cos 120^{\circ}$
$$ =v^{2}+v^{2}+2 v \times v \times-\frac{1}{2} $$
$\Rightarrow|\Delta \mathbf{v}|=v=10 m / s$.
