To solve this problem, we need to analyze the situation using the principles of conservation of momentum and the effects of friction. Let's break it down step by step.
Step 1: Calculate the Common Velocity After Collision
When the bullet embeds itself into the block of wood, we can use the law of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.
The initial momentum of the system can be calculated as follows:
- Momentum of the bullet (m1): p_bullet = mass_bullet × velocity_bullet = 0.25 \, \text{kg} × 302 \, \text{m/s} = 75.5 \, \text{kg m/s}
- Momentum of the block (m2): p_block = mass_block × velocity_block = 37.5 \, \text{kg} × 0 \, \text{m/s} = 0 \, \text{kg m/s}
The total initial momentum (p_initial) is:
p_initial = p_bullet + p_block = 75.5 \, \text{kg m/s} + 0 \, \text{kg m/s} = 75.5 \, \text{kg m/s}
After the bullet embeds itself in the block, the combined mass (m_total) is:
m_total = mass_bullet + mass_block = 0.25 \, \text{kg} + 37.5 \, \text{kg} = 37.75 \, \text{kg}
Let the common velocity after the collision be v. According to the conservation of momentum:
p_initial = m_total × v
Substituting the values:
75.5 \, \text{kg m/s} = 37.75 \, \text{kg} × v
Now, solving for v:
v = 75.5 \, \text{kg m/s} / 37.75 \, \text{kg} ≈ 2 \, \text{m/s}
Step 2: Analyze the Motion of the Blocks
Now that we have the common velocity of the combined mass after the collision, we need to determine how far the blocks will move together on the smooth surface. The block m1 (the wooden block with the bullet embedded) is resting on block m2, which has a coefficient of friction of 0.5.
The force of friction (F_friction) acting on m1 due to m2 can be calculated using:
F_friction = μ × N
Where:
- μ = coefficient of friction = 0.5
- N = normal force = m1 × g (where g = 9.81 m/s² is the acceleration due to gravity)
The normal force acting on m1 is:
N = 37.5 \, \text{kg} × 9.81 \, \text{m/s²} ≈ 367.875 \, \text{N}
Now, substituting this into the friction force equation:
F_friction = 0.5 × 367.875 \, \text{N} ≈ 183.9375 \, \text{N}
Step 3: Calculate the Acceleration of the Blocks
The acceleration (a) of the blocks due to friction can be found using Newton's second law:
F = m × a
Here, the total mass that is being accelerated is the mass of m1 (37.5 kg) and m2 (12.5 kg):
m_total = 37.5 \, \text{kg} + 12.5 \, \text{kg} = 50 \, \text{kg}
Now, using the friction force to find acceleration:
183.9375 \, \text{N} = 50 \, \text{kg} × a
Solving for a gives:
a = 183.9375 \, \text{N} / 50 \, \text{kg} ≈ 3.67875 \, \text{m/s²}
Step 4: Determine the Displacement of the Blocks
To find the displacement (d) of the blocks while they are moving together, we can use the kinematic equation:
d = v² / (2a)
Substituting the values we have:
d = (2 \, \text{m/s})² / (2 × 3.67875 \, \text{m/s²})
Calculating this gives:
d = 4 \, \text{m²/s²} / 7.3575 \, \text{m/s²} ≈ 0.544 \, \text{m}
Summary of Results
In summary, after the bullet embeds into the block:
- The common velocity of m1 and m2 is approximately 2 m/s.
- The displacement of both blocks while moving together is approximately 0.544 m.
This analysis illustrates how momentum conservation and frictional forces interact in a system involving a bullet and blocks. If you have any further questions or need clarification on any part, feel free to ask!
