A particle of mass 100 g is projected at time t = 0 with a speed of 20 ms^-1 at an angle 45^ to the horizontal. The magnitude of the angular momentum of the particle about the starting point at time…

Given: m = 0.1 kg, initial speed v = 20 ms^-1, angle of projection 45^, and time t = 2 s. Find: The value of K when angular momentum magnitude is K kgm^2/ s. The angular momentum about the starting point is: L = m vx h where vx = v 45^ = 20 22 = 10 2 ms^-1 and h is the vertical displacement at t = 2 s. The vertical displacement is: h = vy t - 12 g t^2 = 20 22 2 - 12 10 2^2 = 20 2 - 20 = 20( 2 - 1) m Substitute these values into L: L = (0.1) (10 2) [20( 2 - 1)] = 20 2 ( 2 - 1) = 800 kgm^2/ s Therefore, K = 800.

A particle of mass 100 g is projected at time t = 0 with a speed… — Solution | JEE Physics

NVAMediumJEE 2023Projectile Motion

A particle of mass $100 \, \text{g}$ is projected at time $t = 0$ with a speed of $20 \, \text{ms}^{-1}$ at an angle $45^\circ$ to the horizontal. The magnitude of the angular momentum of the particle about the starting point at time $t = 2 \, \text{s}$ is found to be $\sqrt{K} \, \text{kgm}^2/\text{s}$ . The value of $K$ is:

Answer

Correct answer:800

Step-by-step solution

Standard Method

Given: $m = 0.1 \, \text{kg}$ , initial speed $v = 20 \, \text{ms}^{-1}$ , angle of projection $45^\circ$ , and time $t = 2 \, \text{s}$ .

Find: The value of $K$ when angular momentum magnitude is $\sqrt{K} \, \text{kgm}^2/\text{s}$ .

The angular momentum about the starting point is:

L = m \cdot v_x \cdot h

where $v_x = v \cos 45^\circ = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} \, \text{ms}^{-1}$ and $h$ is the 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 these 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}

Therefore, $K = 800$ .

Common mistakes

Using the total speed instead of the horizontal component in angular momentum. Here the perpendicular distance method gives $L = m v_x h$ , so $v_x = v\cos 45^\circ$ must be used.
Calculating vertical displacement incorrectly by ignoring gravity. The height at $t = 2 \, \text{s}$ is not $v_y t$ alone; use $h = v_y t - \frac{1}{2}gt^2$ .
Keeping mass as $100$ instead of converting $100 \, \text{g}$ to $0.1 \, \text{kg}$ . SI units are required for angular momentum in $\text{kgm}^2/\text{s}$ .

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