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Circular Motion

Section: Physics

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Question : 7 of 76

Marks: +1, -0

Two cars

A

B

10^3 \text{ kg}

are moving on parallel tracks separated by a distance of 10 m , in same direction with speeds

72 \text{ km/h}

36 \text{ km/h}

. The magnitude of angular momentum of car

A

with respect to car

B

\_\_\_\_

\mathrm{J}\cdot\mathrm{s}

[21 Jan 2026 Shift 2]

$3.6 \times 10^5$
$10^5$
$3 \times 10^5$
$2 \times 10^5$

Solution:

Concept:

Angular momentum of a mass about a point is

\vec{L} = \vec{r} \times m\vec{v}

. For parallel motion,

L = m \cdot v_{\text{rel}} \cdot d

, where

d

is the perpendicular separation.

Explanation:

Convert speeds:

72\ \text{km/h} = 20\ \text{m/s}

,

36\ \text{km/h} = 10\ \text{m/s}

.

Relative speed of A w.r.t. B (same direction):

v_{\text{rel}} = 20 - 10 = 10\ \text{m/s}

.

Perpendicular distance between tracks:

d = 10\ \text{m}

.

Mass:

m = 10^3\ \text{kg}

.

Angular momentum magnitude:

L = m \cdot v_{\text{rel}} \cdot d = 10^3 \times 10 \times 10 = 10^5 \, \mathrm{kg \cdot m^2/s}

.

Thus

L = 1.0 \times 10^5\ \mathrm{J\cdot s}

.

Answer:

10^5

(Option B).

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