A body of mass 5 kg moving with a uniform speed 3 √2 ms⁻¹ in X-Y plane along the line y=x+4. The angular momentum of the particle about the origin will be _________ kg m² s⁻¹.

Solution

Physics sir · 4 steps

For straight-line motion, L = mv d, where d is the perpendicular distance of the line from the origin.

1. Moment of momentum. Angular momentum about a point is momentum times the perpendicular distance of its line of action from that point, L = mv d. The speed is uniform and the line is fixed, so L stays constant along the whole path.
2. Perpendicular distance. Write the line as x - y + 4 = 0. Its distance from the origin is d = |4|/√(1² + 1²) = 4/√2 = 2√2 m.

   L = mv d = 5 × 3√2 × 2√2 = 60 kg m² s⁻¹

3. Vector check. The velocity along the line is 3î + 3ĵ, and (0, 4) is a point on it. Then L_z = m(xv_y - yvₓ) = 5(0 × 3 - 4 × 3) = -60, so the magnitude is again 60.
4. So. The angular momentum about the origin is 60 kg m² s⁻¹.

The common slip: 84.85. Taking the y-intercept 4 as the distance instead of the perpendicular distance 2√2 gives 60√2 ≈ 84.85, not 60.

The idea to keep. A particle moving in a straight line has constant angular momentum about any point. Find the perpendicular distance from that point to the line, then use L = mv d.

Answer from the NTA final answer key. Solution by PrepWiser.