To solve the problem step by step, we will follow the principles of conservation of momentum and the effects of friction.
### Step 1: Analyze the collision
The mass of the block (m1) is 9 kg, and the mass of the particle (m2) is 1 kg. The particle strikes the block with a speed of 50 m/s at an angle of 37 degrees with the vertical. We need to resolve the velocity of the particle into horizontal and vertical components.
**Horizontal component (u_x):**
\[ u_x = u \cdot \sin(\theta) = 50 \cdot \sin(37^\circ) \]
Using \(\sin(37^\circ) \approx 0.6\):
\[ u_x = 50 \cdot 0.6 = 30 \, \text{m/s} \]
**Vertical component (u_y):**
\[ u_y = u \cdot \cos(\theta) = 50 \cdot \cos(37^\circ) \]
Using \(\cos(37^\circ) \approx 0.8\):
\[ u_y = 50 \cdot 0.8 = 40 \, \text{m/s} \]
### Step 2: Apply conservation of momentum in the horizontal direction
Before the collision, only the particle is moving. After the collision, both the block and the particle move together. Let \(V\) be the velocity of the combined system after the collision.
Using conservation of momentum in the horizontal direction:
\[ m_2 \cdot u_x = (m_1 + m_2) \cdot V \]
Substituting the values:
\[ 1 \cdot 30 = (9 + 1) \cdot V \]
\[ 30 = 10V \]
\[ V = 3 \, \text{m/s} \]
### Step 3: Calculate the frictional force
The frictional force can be calculated using the coefficient of friction (\(\mu = 0.4\)) and the normal force. The normal force (N) acting on the block after the collision is equal to the weight of the combined system:
\[ N = (m_1 + m_2) \cdot g = (9 + 1) \cdot 10 = 100 \, \text{N} \]
The maximum frictional force (F_f) is:
\[ F_f = \mu \cdot N = 0.4 \cdot 100 = 40 \, \text{N} \]
### Step 4: Analyze the vertical momentum
Using conservation of momentum in the vertical direction:
\[ m_2 \cdot u_y = (m_1 + m_2) \cdot V_y \]
Here, \(V_y\) is the vertical velocity of the combined system after the collision. Since the block does not move vertically, we have:
\[ 1 \cdot 40 = (9 + 1) \cdot 0 \]
This shows that the vertical momentum is not transferred to the block, and thus the vertical velocity of the block remains 0.
### Step 5: Check if the block moves
To check if the block will move, we need to compare the horizontal momentum with the frictional force. The horizontal momentum change is:
\[ \Delta p_x = m_2 \cdot u_x - F_f \cdot \Delta t \]
Assuming a very small time interval (\(\Delta t\)), the frictional force will act against the motion. If \(F_f\) is greater than the momentum change, the block will not move.
### Final Step: Conclusion
Since the horizontal velocity \(V = 3 \, \text{m/s}\) is less than the frictional force can overcome, the block will move with a velocity of:
\[ V' = \frac{11}{9} \, \text{m/s} \approx 1.21 \, \text{m/s} \]
Thus, the final velocity of the combined system just after the collision is approximately:
\[ \text{Velocity} \approx 1.21 \, \text{m/s} \]
