Step 1: Understanding the Concept:
This problem deals with the application of impulsive forces and the conditions for pure rolling.
When an impulsive force \(F\) acts for a very short time \(\Delta t\), it produces a linear impulse \(J = \int F dt\) and an angular impulse \(\tau \Delta t\).
For "rolling without slipping immediately after impact," the sphere must possess a linear velocity \(v\) and an angular velocity \(\omega\) such that the point of contact has zero relative velocity with the ground.
On a horizontal surface, this condition is \(v = R\omega\).
Step 2: Key Formula or Approach:
1. Linear Impulse: \(J = \Delta P = Mv - 0 = Mv\).
2. Angular Impulse about the center of mass: \(J \cdot h = \Delta L = I\omega - 0 = I\omega\).
3. Moment of Inertia for a solid sphere about its center: \(I = \frac{2}{5}MR^2\).
4. Pure rolling condition: \(v = R\omega \implies \omega = \frac{v}{R}\).
Step 3: Detailed Explanation:
Let the horizontal cue apply an impulse \(J\) at a height \(h\) above the center of the sphere.
The linear velocity \(v\) acquired by the center of mass is given by the relation:
\[ J = Mv \quad \text{--- (Equation 1)} \]
The torque produced by this impulse about the center of mass is \(\tau = F \times h\).
The total angular impulse is the integral of torque over time, which equals the change in angular momentum:
\[ \int (F \cdot h) dt = I\omega \]
\[ h \int F dt = I\omega \]
\[ J \cdot h = I\omega \quad \text{--- (Equation 2)} \]
We substitute the value of the moment of inertia for a solid sphere, \(I = \frac{2}{5}MR^2\), into Equation 2:
\[ J \cdot h = \left( \frac{2}{5}MR^2 \right) \omega \]
To satisfy the condition of rolling without slipping immediately after the impact, we require \(v = R\omega\).
Substituting \(\omega = v/R\) into the equation above:
\[ J \cdot h = \frac{2}{5}MR^2 \left( \frac{v}{R} \right) \]
\[ J \cdot h = \frac{2}{5}MRv \]
From Equation 1, we know that \(J = Mv\). We substitute this into the LHS:
\[ (Mv) \cdot h = \frac{2}{5}MRv \]
By canceling \(M\) and \(v\) from both sides (since they are non-zero):
\[ h = \frac{2}{5}R \]
This implies that striking the sphere at precisely \(0.4R\) above its center will result in instantaneous pure rolling without any initial sliding phase.
Step 4: Final Answer:
The required height \(h\) is \(2R/5\).
Hence, the correct option is (B).