To solve the problem, we need to follow these steps:
### Step 1: Understand the scenario
A bullet of mass \( m_1 = 10 \, \text{g} = 0.01 \, \text{kg} \) is moving with a velocity \( u_1 = 400 \, \text{m/s} \) and strikes a wooden block of mass \( m_2 = 2 \, \text{kg} \). The block is suspended and after the collision, it rises by a vertical distance of \( h = 10 \, \text{cm} = 0.1 \, \text{m} \).
### Step 2: Calculate the velocity of the block after the collision
When the block rises, it converts its kinetic energy into potential energy. The potential energy gained by the block is given by:
\[
PE = m_2 g h
\]
where \( g = 9.8 \, \text{m/s}^2 \).
Substituting the values:
\[
PE = 2 \times 9.8 \times 0.1 = 1.96 \, \text{J}
\]
This potential energy is equal to the kinetic energy of the block just after the collision:
\[
KE = \frac{1}{2} m_2 v_2^2
\]
Setting these equal gives:
\[
1.96 = \frac{1}{2} \times 2 \times v_2^2
\]
\[
1.96 = v_2^2
\]
\[
v_2 = \sqrt{1.96} \approx 1.4 \, \text{m/s}
\]
### Step 3: Apply the conservation of momentum
Before the collision, the momentum of the system is given by the bullet alone since the block is at rest:
\[
\text{Initial momentum} = m_1 u_1
\]
After the collision, the momentum of the system is the sum of the momenta of the bullet and the block:
\[
\text{Final momentum} = m_2 v_2 + m_1 v_1
\]
By conservation of momentum:
\[
m_1 u_1 = m_2 v_2 + m_1 v_1
\]
Rearranging to find \( v_1 \):
\[
v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}
\]
### Step 4: Substitute the known values
Substituting \( m_1 = 0.01 \, \text{kg} \), \( m_2 = 2 \, \text{kg} \), \( u_1 = 400 \, \text{m/s} \), and \( v_2 = 1.4 \, \text{m/s} \):
\[
v_1 = \frac{0.01 \times 400 - 2 \times 1.4}{0.01}
\]
Calculating the numerator:
\[
0.01 \times 400 = 4
\]
\[
2 \times 1.4 = 2.8
\]
Thus,
\[
v_1 = \frac{4 - 2.8}{0.01} = \frac{1.2}{0.01} = 120 \, \text{m/s}
\]
### Final Answer
The speed of the bullet after it emerges out horizontally from the block is \( v_1 = 120 \, \text{m/s} \).
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