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**Hint:**The effect of force acting over time is known as impulse. Impulse can also be described as the change in momentum. Change in momentum equals the average net external force multiplied by the time this force acts.

**Step by Step Solution:**

The impulse applied at one end of a stationary uniform frictionless rod is,

$J = mv$

Where $J$ is the impulse, $m$ is the mass of the rod and $v$ is the velocity of rod.

Impulse is also described as the change in the momentum.

$J = \Delta p$

Where $\Delta p$ is the change in momentum and $J$ is impulse.

Now, the formula for the change in the momentum can be written as,

$\Delta p = m ({v_f} - {v_i}) $

Where $\Delta p$ is the change in momentum, $m$ is the mass of the body, ${v_f} $ is the final velocity and ${v_i} $ is the initial velocity.

Now,

${v_i} = 0$ because the rod is stationary

Therefore,

$J = m {v_f} $

$mv = m {v_f} $

${v_f} = v$

Now, for instantaneous axis of rotation,

$AngularImpulse = \Delta L$

Where, $\Delta L$is the change in angular momentum.

Now, angular momentum for a rod can be written as $J \times \dfrac{l}{2}$

Where $J$ is the impulse and $l$is the length of the rod.

Now, we know that

Change in angular momentum, $\Delta L = I ({w_f} - {w_i}) $

Where $I$ is the moment of Inertia of the rod, ${w_f} $is the final rotation of the rod and ${w_i} $is the initial rotation of the rod.

Since the rod is at stationary, therefore,

${w_i} = 0$

Now, using $J \times \dfrac{l}{2} = \Delta L$

$mv \times \dfrac{l}{2} = I {w_f}$

The moment of Inertia for a rod is $I = \dfrac{{m{l^2}}}{{12}}$

$mv \times \dfrac{l}{2} = \dfrac{{m{l^2}}}{{12}} \times {w_f} $

Therefore,

${w_f} = \dfrac{{6v}}{l}$

The forces acting on the bottom of the rod would be

${w_f}\dfrac {l} {2} + v = \dfrac{{6v}}{l} \times \dfrac{l}{2} + v$

$ = 4v$

Assuming that the final rotation occurs at a distance $x$from the bottom,

Equating the rotational forces, we get,

${w_f} x = 4v$

$\dfrac {{6v}} {l} \times x = 4v$

$x = \dfrac{{2l}}{3}$

Therefore, Instantaneous axis of rotation of the rod will pass through the point which is at a distance $\dfrac {{2l}} {3} $ from the striking end.

**Note:**The final rotation ${w_f} $ always taken in a counter clockwise direction. Otherwise, a negative sign should be added.

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