An object is project with a speed `10ms^(-1)` at an angle of `30^@` with the horizontal. It breaks into n equal fragments during its motion . One fragment strikes the ground at a distance of `sqrt(3)m` from the point of projection. The centre of mass of the remaining fragments strikes the ground at a distance of `7sqrt(3)m` from the point of projection . If all fragments strike the ground at the same time , Find the value of n.

Options

A. 2
B. 3
C. 4
D. 525

Answer

To solve the problem step by step, we will analyze the motion of the object and the behavior of the fragments after it breaks apart.

### Step 1: Determine the initial parameters
The object is projected with a speed \( u = 10 \, \text{m/s} \) at an angle \( \theta = 30^\circ \) with the horizontal.

### Step 2: Calculate the horizontal range of the projectile
The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Where \( g \approx 9.81 \, \text{m/s}^2 \).

Calculating \( \sin(60^\circ) \):
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]

Now substituting the values:
\[ R = \frac{(10)^2 \cdot \frac{\sqrt{3}}{2}}{9.81} = \frac{100 \cdot \frac{\sqrt{3}}{2}}{9.81} \approx \frac{50\sqrt{3}}{9.81} \]

### Step 3: Analyze the fragments
Let the object break into \( n \) equal fragments. The mass of each fragment is \( \frac{m}{n} \).

One fragment strikes the ground at a distance of \( \sqrt{3} \, \text{m} \) from the point of projection.

### Step 4: Center of mass of the remaining fragments
The center of mass of the remaining \( n-1 \) fragments strikes the ground at a distance of \( 7\sqrt{3} \, \text{m} \).

### Step 5: Set up the equation for the center of mass
The center of mass of the system can be calculated using the formula:
\[ \text{Center of mass} = \frac{m_1 x_1 + m_2 x_2 + \ldots + m_n x_n}{m_1 + m_2 + \ldots + m_n} \]
For our case:
- \( m_1 = \frac{m}{n} \) (the fragment that lands at \( \sqrt{3} \))
- \( m_2 = \frac{(n-1)m}{n} \) (the remaining fragments that land at \( 7\sqrt{3} \))

Setting up the equation:
\[ \frac{\frac{m}{n} \cdot \sqrt{3} + \frac{(n-1)m}{n} \cdot 7\sqrt{3}}{m} = \text{Center of mass position} \]

### Step 6: Simplify the equation
Since the total mass \( m \) cancels out:
\[ \frac{\sqrt{3}}{n} + \frac{(n-1) \cdot 7\sqrt{3}}{n} = 5\sqrt{3} \]

### Step 7: Multiply through by \( n \) to eliminate the denominator
\[ \sqrt{3} + (n-1) \cdot 7\sqrt{3} = 5n\sqrt{3} \]

### Step 8: Factor out \( \sqrt{3} \)
\[ \sqrt{3} + 7(n-1)\sqrt{3} = 5n\sqrt{3} \]
Dividing through by \( \sqrt{3} \):
\[ 1 + 7(n-1) = 5n \]

### Step 9: Solve for \( n \)
Expanding and rearranging gives:
\[ 1 + 7n - 7 = 5n \implies 2n = 6 \implies n = 3 \]

### Conclusion
The value of \( n \) is \( 3 \). Therefore, the object breaks into 3 equal fragments.

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