A bullet of mass 20 kg travelling horizontally at `100 ms^(-1)` gets embedded at the centre of a block of wood of mass 1kg, suspended by a light vertical string of 1m in length. Calculate the maximum inclination of the string to the vertical.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a bullet of mass \( m = 20 \, \text{g} = 0.02 \, \text{kg} \) traveling horizontally at a speed of \( v = 100 \, \text{m/s} \) that embeds itself into a block of wood of mass \( M = 1 \, \text{kg} \). The block is suspended by a string of length \( L = 1 \, \text{m} \). We need to find the maximum angle \( \theta \) that the string makes with the vertical after the bullet embeds into the block.

### Step 2: Apply Conservation of Momentum
Since there are no external horizontal forces acting on the system, we can use the conservation of momentum. The initial momentum of the bullet before it hits the block is equal to the momentum of the bullet-block system after the collision.

\[ \text{Initial momentum} = \text{Final momentum} \]
\[ m \cdot v = (M + m) \cdot V \]
Where \( V \) is the velocity of the combined mass after the collision.

Substituting the values:
\[ 0.02 \cdot 100 = (1 + 0.02) \cdot V \]
\[ 2 = 1.02 \cdot V \]
\[ V = \frac{2}{1.02} \approx 1.96 \, \text{m/s} \]

### Step 3: Calculate the Kinetic Energy
The kinetic energy of the bullet-block system just after the collision is given by:
\[ KE = \frac{1}{2} (M + m) V^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot 1.02 \cdot (1.96)^2 \]
Calculating \( (1.96)^2 \):
\[ (1.96)^2 \approx 3.8416 \]
Thus,
\[ KE \approx \frac{1}{2} \cdot 1.02 \cdot 3.8416 \approx 1.96 \, \text{J} \]

### Step 4: Convert Kinetic Energy to Potential Energy
As the block swings up, the kinetic energy will convert into potential energy at the maximum height \( h \):
\[ PE = (M + m) g h \]
Setting \( KE = PE \):
\[ 1.96 = (1.02) \cdot 9.81 \cdot h \]
Solving for \( h \):
\[ h = \frac{1.96}{1.02 \cdot 9.81} \approx \frac{1.96}{10} \approx 0.196 \, \text{m} \]

### Step 5: Determine the Maximum Inclination Angle
The height \( h \) is the vertical rise of the block. The length of the string is \( L = 1 \, \text{m} \). The horizontal distance \( x \) from the lowest point to the point where the string makes an angle \( \theta \) with the vertical can be found using the Pythagorean theorem:
\[ L^2 = h^2 + x^2 \]
Substituting \( h = 0.196 \):
\[ 1^2 = (0.196)^2 + x^2 \]
Calculating \( (0.196)^2 \):
\[ (0.196)^2 \approx 0.038416 \]
Thus,
\[ 1 = 0.038416 + x^2 \]
\[ x^2 = 1 - 0.038416 \approx 0.961584 \]
\[ x \approx \sqrt{0.961584} \approx 0.98 \, \text{m} \]

### Step 6: Calculate the Angle \( \theta \)
Using the cosine of the angle:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{0.8}{1} \]
Where \( 0.8 = L - h = 1 - 0.196 \).

Thus,
\[ \cos \theta = 0.8 \]
\[ \theta = \cos^{-1}(0.8) \approx 36.87^\circ \approx 37^\circ \]

### Final Answer
The maximum inclination of the string to the vertical is approximately \( \theta \approx 37^\circ \).

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