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Step-by-Step Solution
Step 1: Identify the known quantities
• Mass of the particle, m = 100\,\text{g} = 0.1\,\text{kg} .
• Initial speed, v = 20\,\text{m/s} .
• Angle of projection, \theta = 45^\circ .
• Time at which angular momentum is desired, t = 2\,\text{s} .
• Acceleration due to gravity, g = 10\,\text{m/s}^2.
Step 2: Express the horizontal displacement at time t
The horizontal velocity component is v \cos\theta . For \theta = 45^\circ , \cos 45^\circ = \frac{\sqrt{2}}{2} . Thus,
v \cos \theta = 20 \times \frac{\sqrt{2}}{2} = 10\sqrt{2}.
The horizontal displacement after time t is then
x = (10\sqrt{2})\,t.
At t = 2\,\text{s} ,
x = 10\sqrt{2} \times 2 = 20\sqrt{2}\,\text{m}.
Step 3: Determine torque due to weight
The angular momentum of the particle about the point of projection changes because of the torque due to the gravitational force. The torque from weight at any time t with horizontal displacement x is:
\tau = m g \, x \quad(\text{direction into the plane, i.e., } -\hat{k}).
Since torque \tau is related to the rate of change of angular momentum L by
\frac{d\vec{L}}{dt} = \vec{\tau},
we can write:
\frac{d\vec{L}}{dt} = m g x \,(-\hat{k}).
Step 4: Integrate to find the angular momentum at t = 2 s
The horizontal displacement x increases linearly with time, x = 10\sqrt{2}\,t . Therefore,
\frac{d\vec{L}}{dt} = m g \, (10\sqrt{2}\,t)\, (-\hat{k}).
Integrating from 0 to t = 2\,\text{s} :
\vec{L} = \int_0^2 m g \, (10\sqrt{2}\,t)\, (-\hat{k})\, dt.
Substitute the known values m = 0.1\,\text{kg} and g = 10\,\text{m/s}^2 :
\vec{L}
= \int_0^2 0.1 \times 10 \times \bigl(10\sqrt{2}\,t\bigr)\, (-\hat{k})\, dt
= \int_0^2 10\sqrt{2}\, t \, (-\hat{k})\, dt.
Compute the integral:
10\sqrt{2} \int_0^2 t \,dt = 10\sqrt{2}\, \left[\frac{t^2}{2}\right]_0^2
= 10\sqrt{2}\, \left(\frac{2^2}{2}\right)
= 10\sqrt{2} \times 2
= 20\sqrt{2}.
Therefore,
\vec{L} = -\,20\sqrt{2}\,\hat{k}\quad(\text{the negative sign indicates the direction along }-\hat{k}).
Step 5: Find the magnitude and determine K
The magnitude of the angular momentum vector at t=2\,\text{s} is
|\vec{L}| = 20\sqrt{2}\,\text{kg}\,\text{m}^2/\text{s}.
According to the question, |\vec{L}| = \sqrt{K}\,\text{kg}\,\text{m}^2/\text{s} . Hence,
\sqrt{K} = 20\sqrt{2} \quad \Longrightarrow \quad K = (20\sqrt{2})^2 = 20^2 \times 2 = 800.
Final Answer
K = 800 .
