Define the initial and final velocity vectors.

Question

Define the initial and final velocity vectors.

Asked on March 29, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-29T12:18:12.718Z

Step 1: Define the initial and final velocity vectors. Let the particle be at point $P$ initially and point $Q$ finally. The speed of the particle is uniform, $v$. The velocity vector is always tangential to the circular path. Let $\vec{v_P}$ be the velocity vector at point $P$ and $\vec{v_Q}$ be the velocity vector at point $Q$. The magnitude of both velocity vectors is $v$, i.e., $|\vec{v_P}| = v$ and $|\vec{v_Q}| = v$. Step 2: Determine the angle between the velocity vectors. The angle between the position vectors $\vec{OP}$ and $\vec{OQ}$ is given as $\theta$. Since the velocity vector at any point on a circular path is perpendicular to the position vector at that point, the angle between the velocity vectors $\vec{v_P}$ and $\vec{v_Q}$ will also be $\theta$. Step 3: Calculate the magnitude of the change in velocity. The change in velocity is $\Delta \vec{v} = \vec{v_Q} - \vec{v_P}$. The magnitude of the change in velocity can be found using the law of cosines for vector subtraction: $$ |\Delta \vec{v}|^2 = |\vec{v_Q}|^2 + |\vec{v_P}|^2 - 2|\vec{v_Q}||\vec{v_P}| \cos \theta $$ Substitute $|\vec{v_P}| = v$ and $|\vec{v_Q}| = v$: $$ |\Delta \vec{v}|^2 = v^2 + v^2 - 2(v)(v) \cos \theta $$ $$ |\Delta \vec{v}|^2 = 2v^2 - 2v^2 \cos \theta $$ Factor out $2v^2$: $$ |\Delta \vec{v}|^2 = 2v^2 (1 - \cos \theta) $$ Step 4: Use a trigonometric identity to simplify. Recall the trigonometric identity: $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$. Substitute this into the equation for $|\Delta \vec{v}|^2$: $$ |\Delta \vec{v}|^2 = 2v^2 \left(2 \sin^2 \left(\frac{\theta}{2}\right)\right) $$ $$ |\Delta \vec{v}|^2 = 4v^2 \sin^2 \left(\frac{\theta}{2}\right) $$ Step 5: Take the square root to find the magnitude. $$ |\Delta \vec{v}| = \sqrt{4v^2 \sin^2 \left(\frac{\theta}{2}\right)} $$ $$ |\Delta \vec{v}| = 2v \left|\sin \left(\frac{\theta}{2}\right)\right| $$ For typical angles $\theta$ swept in a circular motion ($0 \le \theta \le 2\pi$), $\frac{\theta}{2}$ will be in the range $0 \le \frac{\theta}{2} \le \pi$. In this range, $\sin\left(\frac{\theta}{2}\right) \ge 0$. Therefore, we can remove the absolute value sign: $$ |\Delta \vec{v}| = 2v \sin \left(\frac{\theta}{2}\right) $$ Comparing this result with the given options: (a) $2v \sin (2\theta)$ (b) शून्य (c) $2v \sin \left(\frac{\theta}{2}\right)$ (d) $2v \cos \left(\frac{\theta}{2}\right)$ The calculated magnitude of the change in velocity matches option (c). The final answer is $\boxed{\text{(c) } 2v \sin \left(\frac{\theta}{2}\right)}$.

Accepted Answer · 2026-03-29T12:18:12.718Z

Step 1: Define the initial and final velocity vectors. Let the particle be at point $P$ initially and point $Q$ finally. The speed of the particle is uniform, $v$. The velocity vector is always tangential to the circular path. Let $\vec{v_P}$ be the velocity vector at point $P$ and $\vec{v_Q}$ be the velocity vector at point $Q$. The magnitude of both velocity vectors is $v$, i.e., $|\vec{v_P}| = v$ and $|\vec{v_Q}| = v$. Step 2: Determine the angle between the velocity vectors. The angle between the position vectors $\vec{OP}$ and $\vec{OQ}$ is given as $\theta$. Since the velocity vector at any point on a circular path is perpendicular to the position vector at that point, the angle between the velocity vectors $\vec{v_P}$ and $\vec{v_Q}$ will also be $\theta$. Step 3: Calculate the magnitude of the change in velocity. The change in velocity is $\Delta \vec{v} = \vec{v_Q} - \vec{v_P}$. The magnitude of the change in velocity can be found using the law of cosines for vector subtraction: $$ |\Delta \vec{v}|^2 = |\vec{v_Q}|^2 + |\vec{v_P}|^2 - 2|\vec{v_Q}||\vec{v_P}| \cos \theta $$ Substitute $|\vec{v_P}| = v$ and $|\vec{v_Q}| = v$: $$ |\Delta \vec{v}|^2 = v^2 + v^2 - 2(v)(v) \cos \theta $$ $$ |\Delta \vec{v}|^2 = 2v^2 - 2v^2 \cos \theta $$ Factor out $2v^2$: $$ |\Delta \vec{v}|^2 = 2v^2 (1 - \cos \theta) $$ Step 4: Use a trigonometric identity to simplify. Recall the trigonometric identity: $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$. Substitute this into the equation for $|\Delta \vec{v}|^2$: $$ |\Delta \vec{v}|^2 = 2v^2 \left(2 \sin^2 \left(\frac{\theta}{2}\right)\right) $$ $$ |\Delta \vec{v}|^2 = 4v^2 \sin^2 \left(\frac{\theta}{2}\right) $$ Step 5: Take the square root to find the magnitude. $$ |\Delta \vec{v}| = \sqrt{4v^2 \sin^2 \left(\frac{\theta}{2}\right)} $$ $$ |\Delta \vec{v}| = 2v \left|\sin \left(\frac{\theta}{2}\right)\right| $$ For typical angles $\theta$ swept in a circular motion ($0 \le \theta \le 2\pi$), $\frac{\theta}{2}$ will be in the range $0 \le \frac{\theta}{2} \le \pi$. In this range, $\sin\left(\frac{\theta}{2}\right) \ge 0$. Therefore, we can remove the absolute value sign: $$ |\Delta \vec{v}| = 2v \sin \left(\frac{\theta}{2}\right) $$ Comparing this result with the given options: (a) $2v \sin (2\theta)$ (b) शून्य (c) $2v \sin \left(\frac{\theta}{2}\right)$ (d) $2v \cos \left(\frac{\theta}{2}\right)$ The calculated magnitude of the change in velocity matches option (c). The final answer is $\boxed{\text{(c) } 2v \sin \left(\frac{\theta}{2}\right)}$.