Question

A particle moving with a speed $v$ changes direction by an angle $\theta$, without change in speed, then which of the following are correct.

• A. The change in the magnitude of its velocity is zero.
• B. The change in the magnitude of its velocity is $2v$ $sin$ ($\theta$/2).
• C. The magnitude of the change in its velocity is $2v$ $sin$ ($\theta$/2).
• D. The magnitude of the change in its velocity is $v$(1 - cos $\theta$).

Answer 1

Answer: (A), (C)

The change in the magnitude of its velocity is zero.
The magnitude of the change in its velocity is $2v$ $sin$ ($\theta$/2).

$\text{Step 1: Change in the magnitude of velocity}$

The magnitude of velocity, i.e speed, is constant

$|{\overrightarrow{v}}_f|=|{\overrightarrow{v}}_i|$

Therefore, $|{\overrightarrow{v}}_f|-|{\overrightarrow{v}}_i|=0$

Hence no change in magnitude of velocity.

$\text{Step 2: The magnitude of change in velocity}$

Angle between ${\overrightarrow{v}}_f$ and ${\overrightarrow{v}}_i$ is $\theta$

$|{\overrightarrow{v}}_f|=|{\overrightarrow{v}}_i|=v$

Therefore, magnitude of change in velocity is given as:

$|\overrightarrow{v_f}-\overrightarrow{v_i}|=\sqrt{v^2+v^2-2.v.v.\cos \theta }=v\sqrt{2(1-\cos \theta )}$ $[\because 1-\cos \theta =2{\sin }^2\theta /2]$

Thus, magnitude of change in velocity $=v\sqrt{4{\sin }^2\theta /2}=2v\sin \left(\theta /2\right)$

Hence, Option A and C are correct