To solve the problem step by step, we will use the principles of conservation of momentum and conservation of energy.
### Step 1: Understanding the Problem
We have a bullet of mass \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) moving with a velocity \( u = 400 \, \text{m/s} \) that strikes a wooden block of mass \( M = 2 \, \text{kg} \). After the collision, the center of gravity of the block rises by \( h = 10 \, \text{cm} = 0.1 \, \text{m} \).
### Step 2: Apply Conservation of Energy
The initial kinetic energy of the bullet is converted into potential energy of the block and the kinetic energy of the bullet after it exits the block.
The potential energy gained by the block when it rises to height \( h \) is given by:
\[
PE = Mgh
\]
Substituting the values:
\[
PE = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 0.1 \, \text{m} = 2 \, \text{kg} \times 10 \times 0.1 = 2 \, \text{J}
\]
### Step 3: Write the Energy Conservation Equation
The initial kinetic energy of the bullet is:
\[
KE_{initial} = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.01 \, \text{kg} \times (400 \, \text{m/s})^2
\]
Calculating this:
\[
KE_{initial} = \frac{1}{2} \times 0.01 \times 160000 = 800 \, \text{J}
\]
The final kinetic energy of the bullet after passing through the block is:
\[
KE_{final} = \frac{1}{2} m v_1^2
\]
where \( v_1 \) is the speed of the bullet after it exits the block.
### Step 4: Set Up the Energy Conservation Equation
According to the conservation of energy:
\[
KE_{initial} = PE + KE_{final}
\]
Substituting the known values:
\[
800 \, \text{J} = 2 \, \text{J} + \frac{1}{2} \times 0.01 \, \text{kg} \times v_1^2
\]
This simplifies to:
\[
800 - 2 = \frac{1}{2} \times 0.01 \times v_1^2
\]
\[
798 = \frac{1}{2} \times 0.01 \times v_1^2
\]
### Step 5: Solve for \( v_1 \)
Multiply both sides by 2:
\[
1596 = 0.01 \times v_1^2
\]
Dividing by 0.01:
\[
v_1^2 = 159600
\]
Taking the square root:
\[
v_1 = \sqrt{159600} \approx 399.5 \, \text{m/s}
\]
### Step 6: Apply Conservation of Momentum
Now, we can use conservation of momentum to find the final speed of the bullet after it exits the block. The initial momentum is:
\[
P_{initial} = m u = 0.01 \times 400 = 4 \, \text{kg m/s}
\]
The final momentum is:
\[
P_{final} = m v_1 + M v_2
\]
where \( v_2 \) is the velocity of the block after the collision.
Using the height gained, we can find \( v_2 \) using energy conservation:
\[
\frac{1}{2} M v_2^2 = Mgh
\]
Substituting:
\[
\frac{1}{2} \times 2 v_2^2 = 2 \times 10 \times 0.1
\]
\[
v_2^2 = 2 \times 2 = 4 \implies v_2 = 2 \, \text{m/s}
\]
### Step 7: Substitute into Momentum Equation
Now substituting back into the momentum equation:
\[
4 = 0.01 v_1 + 2
\]
Solving for \( v_1 \):
\[
4 - 2 = 0.01 v_1
\]
\[
2 = 0.01 v_1 \implies v_1 = \frac{2}{0.01} = 200 \, \text{m/s}
\]
### Conclusion
The speed of the bullet after it emerges horizontally from the block is approximately \( 200 \, \text{m/s} \).
