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

The problem requires us to find the angular momentum of car A with respect to car B. To solve this, we need to understand the basic concepts of angular momentum in physics.

Angular momentum L of a particle with respect to a point is given by:

L = m \cdot v \cdot r

where:

• m is the mass of the particle,
• v is the linear velocity of the particle,
• r is the perpendicular distance from the point to the line of motion of the particle.

Given:

• Mass of car A (and B): m = 10^3 kg,
• Speed of car A: 72 km/h,
• Speed of car B: 36 km/h,
• Distance between the tracks (perpendicular distance r): 10 m.

First, convert the velocities from km/h to m/s:

• Speed of car A: 72 km/h = 72 \times \frac{1000}{3600} = 20 m/s
• Speed of car B: 36 km/h = 36 \times \frac{1000}{3600} = 10 m/s

The relative velocity of car A with respect to car B is:

v_{\text{relative}} = v_A - v_B = 20 \, \text{m/s} - 10 \, \text{m/s} = 10 \, \text{m/s}

Using the formula for angular momentum, we have:

L = m \cdot v_{\text{relative}} \cdot r = 10^3 \, \text{kg} \cdot 10 \, \text{m/s} \cdot 10 \, \text{m} = 10^5 \, \text{kg} \cdot \text{m}^2/\text{s}

Therefore, the magnitude of angular momentum of car A with respect to car B is 10^5 J.s.

Hence, the correct answer is: 10^5.