To solve the problem, we will follow these steps:
### Step 1: Apply Conservation of Momentum
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
Let:
- Mass of the bullet, \( m_b = 0.25 \, \text{kg} \)
- Velocity of the bullet, \( v_b = 302 \, \text{m/s} \)
- Mass of the block, \( m_1 = 37.5 \, \text{kg} \)
- Mass of the second block, \( m_2 = 1.25 \, \text{kg} \)
Before the collision, the momentum of the system is:
\[
\text{Initial momentum} = m_b \cdot v_b + m_1 \cdot 0 = 0.25 \cdot 302 + 0 = 75.5 \, \text{kg m/s}
\]
After the bullet embeds into the block, the total mass is \( m_b + m_1 \) and let \( V \) be the common velocity after the collision:
\[
\text{Final momentum} = (m_b + m_1) \cdot V
\]
Setting initial momentum equal to final momentum:
\[
75.5 = (0.25 + 37.5) \cdot V
\]
\[
75.5 = 37.75 \cdot V
\]
\[
V = \frac{75.5}{37.75} \approx 2.0 \, \text{m/s}
\]
### Step 2: Calculate the Force of Friction
The force of friction \( F_f \) acting on block \( m_1 \) is given by:
\[
F_f = \mu \cdot N
\]
where \( \mu = 0.5 \) (coefficient of friction) and \( N \) is the normal force. The normal force \( N \) is equal to the weight of the block \( m_1 \) plus the bullet:
\[
N = (m_1 + m_b) \cdot g = (37.5 + 0.25) \cdot 9.81 \approx 37.75 \cdot 9.81 \approx 370.5 \, \text{N}
\]
Thus, the force of friction is:
\[
F_f = 0.5 \cdot 370.5 \approx 185.25 \, \text{N}
\]
### Step 3: Calculate the Displacement
The kinetic energy \( KE \) of the block \( m_1 \) after the collision is:
\[
KE = \frac{1}{2} (m_1 + m_b) V^2 = \frac{1}{2} (37.75) (2.0)^2 = \frac{1}{2} \cdot 37.75 \cdot 4 = 75.5 \, \text{J}
\]
The work done by the friction force to stop the block is equal to the kinetic energy:
\[
F_f \cdot y = KE
\]
\[
185.25 \cdot y = 75.5
\]
\[
y = \frac{75.5}{185.25} \approx 0.407 \, \text{m}
\]
### Final Results
- Common velocity \( V \approx 2.0 \, \text{m/s} \)
- Displacement \( y \approx 0.407 \, \text{m} \)
