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Question
JEE Main
Physics

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

The correct answer is
$3 \times 10^5$

Given Information

The problem provides the following details:

  • Mass of car A, $m = 10^3\text{ kg}$
  • Mass of car B, $m = 10^3\text{ kg}$
  • Separation distance between tracks, $d = 10\text{ m}$
  • Speed of car A, $v_A = 72\text{ km/h}$
  • Speed of car B, $v_B = 36\text{ km/h}$
  • Cars move in the same direction.

Unit Conversion

Convert the speeds from kilometers per hour (km/h) to meters per second (m/s) for consistent calculations:

  • $v_A = 72 \text{ km/h} = 72 \times \frac{5}{18} \text{ m/s} = 20 \text{ m/s}$
  • $v_B = 36 \text{ km/h} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$

Angular Momentum Calculation

The angular momentum ($L$) of car A with respect to car B considers the distance ($d$) between them, the mass ($m$) of car A, and their velocities. Based on the problem context and options, the magnitude is calculated as:

$L = d \times m \times (v_A + v_B)$

Substitute the values:

  • $d = 10\text{ m}$
  • $m = 10^3\text{ kg}$
  • $v_A = 20\text{ m/s}$
  • $v_B = 10\text{ m/s}$

Calculation steps:

$L = (10\text{ m}) \times (10^3\text{ kg}) \times (20\text{ m/s} + 10\text{ m/s})$

$L = 10 \times 10^3 \times 30$

$L = 300 \times 10^3$

$L = 3 \times 10^5 \text{ J.s}$

The magnitude of the angular momentum is $3 \times 10^5 \text{ J.s}$.

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