A ball of radius 11 cm and mass 8 kg rolls from rest down a ramp of length 2m. The ramp is inclined at `35^(@)` to the horizontal. When the ball reaches the bottom, its velocity is `(sin35=0.57)`

Options

A. 2 m/s
B. 5 m/s
C. 4 m/s
D. 6 m/s

Answer

To solve the problem step by step, we will follow the principles of rotational motion and energy conservation.

### Step 1: Identify the given values
- Radius of the ball, \( r = 11 \, \text{cm} = 0.11 \, \text{m} \)
- Mass of the ball, \( m = 8 \, \text{kg} \)
- Length of the ramp, \( s = 2 \, \text{m} \)
- Angle of inclination, \( \theta = 35^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- \( \sin(35^\circ) = 0.57 \)

### Step 2: Calculate the acceleration of the ball
The forces acting on the ball can be analyzed. The gravitational force can be resolved into two components: one parallel to the ramp and one perpendicular to the ramp.

1. The component of gravitational force acting down the ramp is:
\[ F_{\text{parallel}} = mg \sin \theta \]

2. The frictional force \( F \) opposes the motion. For a solid sphere rolling without slipping, we have:
\[ F = \frac{2}{5} m a \]

3. According to Newton's second law, the net force acting on the ball is:
\[ mg \sin \theta - F = ma \]
Substituting \( F \):
\[ mg \sin \theta - \frac{2}{5} ma = ma \]

4. Rearranging gives:
\[ mg \sin \theta = ma + \frac{2}{5} ma = \frac{7}{5} ma \]

5. Solving for \( a \):
\[ a = \frac{5}{7} g \sin \theta \]

### Step 3: Substitute values to find acceleration
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( \sin(35^\circ) = 0.57 \):
\[ a = \frac{5}{7} \cdot 9.8 \cdot 0.57 \]
Calculating:
\[ a = \frac{5 \cdot 9.8 \cdot 0.57}{7} = \frac{28.05}{7} \approx 4.007 \, \text{m/s}^2 \]

### Step 4: Use kinematic equation to find final velocity
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where \( u = 0 \) (initial velocity), \( a \) is the acceleration we just calculated, and \( s = 2 \, \text{m} \):
\[ v^2 = 0 + 2 \cdot 4.007 \cdot 2 \]
Calculating:
\[ v^2 = 8.014 \cdot 2 = 16.028 \]
Taking the square root:
\[ v = \sqrt{16.028} \approx 4.0 \, \text{m/s} \]

### Final Answer
The velocity of the ball when it reaches the bottom of the ramp is approximately \( 4.0 \, \text{m/s} \).

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