Question

A frictionless circular wire of unit radius is fixed on a horizontal plane. Two point particles of unit mass start moving simultaneously from point \(A\) \((\theta=\pi/2)\) with identical uniform angular speeds in opposite directions and meet again at point \(B\). During this time, which graph correctly represents the magnitude of total linear momentum \(P\) of the system as a function of time?

• A. Sine shaped graph
• B. Cosine shaped graph
• C. V-shaped graph
• D. Linear graph

Hint:

For particles moving symmetrically on a circle, always resolve velocity vectors first and then add momenta vectorially. The modulus sign in \[ P=2v|\cos\omega t| \] creates the V-shaped behaviour.

Answer 1

The Correct Option is C

Step 1: Set up the motion.
Two equal masses (unit mass) start at $A$ and run around a unit-radius circle with the same angular speed $\omega$ but in opposite directions. Let each move with speed $v$.
Step 2: Write the velocity of each particle.
With angle swept $\theta = \omega t$, one particle has $\vec v_1 = v(-\sin\theta\,\hat i + \cos\theta\,\hat j)$ and the other $\vec v_2 = v(\sin\theta\,\hat i + \cos\theta\,\hat j)$.
Step 3: Add the momenta.
Since masses are unity, $\vec P = \vec v_1 + \vec v_2$. The $\hat i$ parts cancel and the $\hat j$ parts add: $\vec P = 2v\cos\theta\,\hat j$.
Step 4: Take the magnitude.
\[ P = 2v\,|\cos(\omega t)| \]
Step 5: Track how P changes.
At $t=0$, $\cos 0 = 1$ so $P$ is maximum; as the particles approach the meeting point, $\cos(\omega t)$ passes through zero, so $P$ dips to zero and rises again. The absolute value turns the smooth cosine dip into a sharp valley.
Step 6: Identify the shape.
Near the zero, $|\cos|$ looks like straight lines meeting at a point, giving a V-shaped graph, which is option C.
\[ \boxed{\text{V-shaped graph}} \]