A uniform solid sphere of mass M and radius R is placed on a smooth horizontal surface. It is given a horizontal impulse J at a height h above the centre of mass and sphere starts rolling. Then the value of h and speed of ccentre of mass are

Options

A. `h=(2)/(5) R and v=(J)/(M)`
B. `h=(2)/(5) R and v= (2)/(5)(J)/(M)`
C. `h=(7)/(5) R and v= (7)/(5)(J)/(M)`
D. `h=(7)/(5) R and v=(J)/(M)`

Explanation

To solve the problem, we need to analyze the motion of the uniform solid sphere after it receives a horizontal impulse \( J \) at a height \( h \) above its center of mass. We will determine the value of \( h \) and the speed of the center of mass \( v_{cm} \) after the impulse is applied.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform solid sphere of mass \( M \) and radius \( R \).
- The sphere is placed on a smooth horizontal surface, meaning there is no friction.
- An impulse \( J \) is applied at a height \( h \) above the center of mass.

2. **Torque and Angular Acceleration**:
- The impulse \( J \) creates a torque about the center of mass. The torque \( \tau \) due to the impulse is given by:
\[ \tau = J \cdot h \]
- The moment of inertia \( I \) of a solid sphere about its center of mass is:
\[ I = \frac{2}{5} M R^2 \]
- The angular acceleration \( \alpha \) can be related to the torque by:
\[ \tau = I \alpha \implies J \cdot h = \frac{2}{5} M R^2 \alpha \]

3. **Linear Acceleration**:
- The linear acceleration \( a \) of the center of mass is related to the angular acceleration \( \alpha \) by the equation for rolling without slipping:
\[ a = R \alpha \]
- Substituting \( \alpha \) from the torque equation, we get:
\[ a = R \cdot \frac{J \cdot h}{\frac{2}{5} M R^2} = \frac{5J \cdot h}{2M R} \]

4. **Impulse-Momentum Theorem**:
- The impulse \( J \) also relates to the change in linear momentum of the sphere. The change in momentum is:
\[ J = \Delta p = M v_{cm} - 0 \implies M v_{cm} = J \implies v_{cm} = \frac{J}{M} \]

5. **Finding the Height \( h \)**:
- For pure rolling to occur, the center of mass must rise to a height where the impulse is effectively applied. The height \( h \) can be determined from the condition that the angular momentum about the center of mass must be consistent with the linear motion:
\[ h = \frac{2}{5} R \]
- This height corresponds to the point where the impulse causes the sphere to start rolling without slipping.

6. **Final Results**:
- The value of \( h \) is:
\[ h = \frac{2}{5} R \]
- The speed of the center of mass \( v_{cm} \) is:
\[ v_{cm} = \frac{J}{M} \]

### Summary of Results:
- Height \( h = \frac{2}{5} R \)
- Speed of center of mass \( v_{cm} = \frac{J}{M} \)