Question

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 \(𝑡=2 s\) is found to be \(\sqrt{K}\  kgm^2/s.\) The value of \(K\) is ________. (Take \(g=10\) \(ms^{−2}\))

Answer 1

Correct Answer: 800

The angular momentum about the starting point is: \[ L = m \cdot v_x \cdot h, \] where: \(m = 0.1 \, \text{kg}\), \(v_x = v \cos 45^\circ = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} \, \text{ms}^{-1}\), \(h = \text{vertical displacement at } t = 2 \, \text{s}\). The vertical displacement is: \[ h = v_y t - \frac{1}{2} g t^2 = 20 \cdot \frac{\sqrt{2}}{2} \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2 = 20\sqrt{2} - 20 = 20(\sqrt{2} - 1) \, \text{m}. \] Substitute values into \(L\): \[ L = (0.1) \cdot (10\sqrt{2}) \cdot [20(\sqrt{2} - 1)] = 20\sqrt{2} \cdot (\sqrt{2} - 1) = \sqrt{800} \, \text{kgm}^2/\text{s}. \]