A bullet of mass `0.01 kg` and travelling at a speed of `500 ms^(-1)` strikes a block of mass `2 kg` which is suspended by a string of length `5 m`. The centre of gravity of the block is found to raise a vertical distance of `0.2 m`. What is the speed of the bullet after it emerges from the block?

Options

A. `15ms^(-1)`
B. `20ms^(-1)`
C. `100ms^(-1)`
D. `50ms^(-1)`

Explanation

To solve the problem, we will use the principles of conservation of momentum and conservation of energy.

### Step 1: Understand the problem
We have a bullet of mass \( m_1 = 0.01 \, \text{kg} \) traveling at a speed \( u_1 = 500 \, \text{m/s} \) that strikes a block of mass \( m_2 = 2 \, \text{kg} \). The block is suspended and rises to a height \( h = 0.2 \, \text{m} \) after the collision. We need to find the speed of the bullet after it emerges from the block, denoted as \( v_1 \).

### Step 2: Apply conservation of momentum
Before the collision, the total momentum is given by:
\[ p_{\text{initial}} = m_1 u_1 + m_2 \cdot 0 = 0.01 \cdot 500 + 0 = 5 \, \text{kg m/s} \]
After the collision, let \( v_1 \) be the speed of the bullet and \( v_2 \) be the speed of the block. The total momentum after the collision is:
\[ p_{\text{final}} = m_1 v_1 + m_2 v_2 \]
Setting the initial momentum equal to the final momentum, we have:
\[ 5 = 0.01 v_1 + 2 v_2 \tag{1} \]

### Step 3: Apply conservation of energy
The kinetic energy before the collision is:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \cdot 0.01 \cdot (500)^2 = 1250 \, \text{J} \]
After the collision, the kinetic energy of the bullet and block plus the potential energy gained by the block is:
\[ KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + mgh \]
where \( h = 0.2 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \). The potential energy gained by the block is:
\[ PE = m_2 g h = 2 \cdot 10 \cdot 0.2 = 4 \, \text{J} \]
Setting the initial kinetic energy equal to the final kinetic energy plus potential energy, we have:
\[ 1250 = \frac{1}{2} \cdot 0.01 v_1^2 + \frac{1}{2} \cdot 2 v_2^2 + 4 \tag{2} \]

### Step 4: Solve the equations
From equation (1):
\[ 2 v_2 = 5 - 0.01 v_1 \implies v_2 = \frac{5 - 0.01 v_1}{2} \]
Substituting \( v_2 \) into equation (2):
\[ 1250 = \frac{1}{2} \cdot 0.01 v_1^2 + \frac{1}{2} \cdot 2 \left(\frac{5 - 0.01 v_1}{2}\right)^2 + 4 \]
Simplifying:
\[ 1250 - 4 = \frac{0.01}{2} v_1^2 + \frac{1}{2} \cdot 2 \cdot \frac{(5 - 0.01 v_1)^2}{4} \]
\[ 1246 = 0.005 v_1^2 + \frac{(5 - 0.01 v_1)^2}{4} \]
Expanding the square:
\[ 1246 = 0.005 v_1^2 + \frac{25 - 0.1 v_1 + 0.0001 v_1^2}{4} \]
Multiplying through by 4 to eliminate the fraction:
\[ 4984 = 0.02 v_1^2 + 25 - 0.1 v_1 \]
Rearranging gives:
\[ 0.02 v_1^2 + 0.1 v_1 - 4960 = 0 \]
This is a quadratic equation in \( v_1 \).

### Step 5: Solve the quadratic equation
Using the quadratic formula \( v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 0.02 \), \( b = 0.1 \), and \( c = -4960 \).
Calculating the discriminant:
\[ D = b^2 - 4ac = (0.1)^2 - 4 \cdot 0.02 \cdot (-4960) = 0.01 + 396.8 = 396.81 \]
Calculating \( v_1 \):
\[ v_1 = \frac{-0.1 \pm \sqrt{396.81}}{2 \cdot 0.02} \]
Calculating \( \sqrt{396.81} \approx 19.93 \):
\[ v_1 = \frac{-0.1 \pm 19.93}{0.04} \]
Taking the positive root:
\[ v_1 = \frac{19.83}{0.04} \approx 495.75 \, \text{m/s} \]

### Step 6: Find \( v_2 \)
Using \( v_2 \) from equation (1):
\[ v_2 = \frac{5 - 0.01 \cdot 495.75}{2} \approx 1.21 \, \text{m/s} \]

### Final Result
The speed of the bullet after it emerges from the block is approximately \( 495.75 \, \text{m/s} \).