Explanation
Solution:
Hint:
· Find the magnitude of velocity by applying the Kinetic energy formula.
· Find the distance for two revolutions.
· Apply third equation of motion to find the acceleration.
Step 1: Find the velocity of the particle from the given kinetic energy.
$$\dfrac{1}{2}m{v^2} = 8 \times {10^{ - 4}}$$
Here, m is the mass of the particle and v is the velocity.
$$\dfrac{1}{2} \times 10 \times {10^{ - 3}} \times {v^2} = 8 \times {10^{ - 4}}$$
$$v = 0.4\;m/s$$
Step 2: Write the total distance covered.
$$s = 2(2\pi r)$$ ( particle covered two revolutions)
Here, s is the total distance covered.
Step 3: Apply the third equation of motion.
$${v^2} -{u^2} = 2as$$ (Where v is final velocity, u is initial velocity (given u=0), a is acceleration and s is distance)
$$ \Rightarrow {(0.4)^2} = 2 \times a \times (4\pi \times 6.4 \times {10^{ - 2}})$$
$$\therefore a = 0.1\;m/{s^2}$$
Hence, option (a) is the correct answer.
The velocities are given as,
$$\dfrac{{dx}}{{dt}} = 6$$
$${v_x} = 6$$
$$\dfrac{{dy}}{{dt}} = 6$$
$${v_y} = 8 - 10t$$
The velocity at projection is given as,
$$v\left( t \right) = {v_x}\hat i + {v_y}\hat j$$
$$v\left( t \right) = 6\hat i + \left( {8 - 10t} \right)\hat j$$
$$v\left( 0 \right) = 6\hat i + 8\hat j$$
$$\left| {v\left( 0 \right)} \right| = \sqrt {{6^2} + {8^2}} $$
$$\left| {v\left( 0 \right)} \right| = 10\;{\rm{m/s}}$$
The velocity of projection is $$10\;{\rm{m/s}}$$.
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