A particle of mass 100 g is projected at time t = 0 with a speed 20 ms⁻¹ at an angle 45° to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time t = 2s is found to be √K kg m²/s. The value of K is ___________.
(Take g = 10 ms⁻²)

Solution

Physics sir · 4 steps

Gravity is the only force, so its torque about O builds up the angular momentum from zero.

1. Start from zero. At t = 0 the particle sits at O, so r⃗ = 0 and L = 0. After that, gravity is the only force, so dL/dt equals the torque of mg about O.
2. Torque of gravity. The weight mg acts straight down, so its lever arm about O is the horizontal distance x = u cos 45° t = 10√2 t. The torque is τ = mg x, and it always points the same way (into the page).

   L = ∫₀ᵗ mg u cos θ t dt = 1/2 mg u cos θ t²

3. Put in numbers. With m = 0.1 kg, g = 10 m s⁻², u cos θ = 10√2 m s⁻¹ and t = 2 s: L = 1/2 × 0.1 × 10 × 10√2 × 4 = 20√2 kg m²/s. As a check, m(xv_y - yvₓ) with x = 20√2, y = 20√2 - 20, vₓ = 10√2, v_y = 10√2 - 20 gives the same magnitude.
4. So. L = 20√2 = √800 kg m²/s, so K = 800.

The common slip: 8000000. Keeping the mass as 100 instead of 0.1 kg makes L = 2000√2 and K = 8 000 000. Convert grams to kilograms first.

The idea to keep. For a projectile, angular momentum about the launch point is 1/2mg u cos θ t². Gravity's torque builds it from zero, because the weight's lever arm is just the horizontal distance x.

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