Question

Two cars $A$ and $B$ each of mass $10^3$ kg are moving on parallel tracks separated by a distance of $10$ m, in same direction with speeds $72$ km/h and $36$ km/h. The magnitude of angular momentum of car $A$ with respect to car $B$ is _________ J.s.

• A. $3 \times 10^5$
• B. $10^5$
• C. $3.6 \times 10^5$
• D. $2 \times 10^5$

Hint:

In relative motion problems involving parallel paths, angular momentum simplifies to $m \cdot v_{rel} \cdot r_{\perp}$. Always ensure units are converted to SI before calculation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Angular momentum of a particle relative to an observer is given by the product of its mass, its relative velocity, and the perpendicular distance to the line of motion of the relative velocity.

Step 2: Key Formula or Approach:
\[ L = m \cdot v_{rel} \cdot d \]
Where \( v_{rel} \) is the relative speed and \( d \) is the perpendicular separation between the tracks.

Step 3: Detailed Explanation:
First, convert speeds to m/s:
\[ v_A = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s} \]
\[ v_B = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]
The relative speed of A with respect to B is:
\[ v_{rel} = v_A - v_B = 20 - 10 = 10 \text{ m/s} \]
The perpendicular distance is \( d = 10 \text{ m} \).
The mass of car A is \( m = 10^3 \text{ kg} \).
Calculating magnitude of angular momentum:
\[ L = m \cdot v_{rel} \cdot d = 10^3 \times 10 \times 10 = 10^5 \text{ kg}\cdot\text{m}^2/\text{s} \]

Step 4: Final Answer:
The angular momentum is $10^5$ J.s.