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

### Related Formula L_textrel = m cdot V_textrel cdot r_perp$$L_{\text{rel}} = m \cdot V_{\text{rel}} \cdot r_{\perp}$$ ### Core Logic The relative angular momentum of a particle translating uniformly with respect to another moving reference frame can be found using their relative velocity and the perpendicular distance between their lines of motion. Relative velocity of A with respect to B: V_textrel = V_A - V_B = 72 - 36 = 36 text km/h$$V_{\text{rel}} = V_A - V_B = 72 - 36 = 36 \text{ km/h}$$ ### Step 1: Conversion to SI units V_textrel = 36 times frac518 text m/s = 10 text m/s$$V_{\text{rel}} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$$ ### Step 2: Final Conclusion Using the perpendicular distance (r_perp = 10 text m$r_{\perp} = 10 \text{ m}$) and mass (m = 10^3 text kg$m = 10^3 \text{ kg}$): L = m cdot V_textrel cdot r_perp$$L = m \cdot V_{\text{rel}} \cdot r_{\perp}$$ L = 1000 times 10 times 10 = 10^5 text kg m^2text/s text (or Jcdottexts)$$L = 1000 \times 10 \times 10 = 10^{5} \text{ kg m}^2\text{/s} \text{ (or J}\cdot\text{s)}$$ ### Pattern Recognition For parallel tracks, the angular momentum of one translating body relative to another is simply m times v_textrelative times texttrack separation$m \times v_{\text{relative}} \times \text{track separation}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Kinematics