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

Concept:

• Total momentum is the vector sum of the individual momenta.

• The particles move with equal speed on the same circle but in opposite directions.

• Due to symmetry, horizontal components cancel.

Step 1: Write velocity vectors
Let the speed of each particle be \(v\). At time \(t\), \[ \theta=\omega t \] Velocity of first particle, \[ \vec v_1 = v(-\sin\theta\,\hat i+\cos\theta\,\hat j) \] Velocity of second particle, \[ \vec v_2 = v(\sin\theta\,\hat i+\cos\theta\,\hat j) \]

Step 2: Find resultant momentum
Since masses are unity, \[ \vec P = \vec v_1+\vec v_2 \] \[ \vec P = 2v\cos\theta\,\hat j \] Therefore, \[ P = 2v|\cos\theta| \] \[ P = 2v|\cos(\omega t)| \]

Step 3: Study the variation
At \[ t=0 \] \[ P=0 \] Then \(P\) increases to a maximum value. At the midpoint, \[ P=0 \] again. Finally it increases and decreases symmetrically. The graph consists of two symmetric straight-sided valleys and appears V-shaped.

Step 4: Choose the correct graph
Hence the correct graph is \[ \boxed{\text{Option (C)}} \]