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: Determine the Common Velocity After the 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): p1 = m1 * v1
- Where:
- m1 = mass of the bullet = 0.25 kg
- v1 = velocity of the bullet = 302 m/s
- Momentum of the block (m2): p2 = m2 * v2
- Where:
- m2 = mass of the block = 37.5 kg
- v2 = initial velocity of the block = 0 m/s (since it is at rest)
Now, calculating the initial momentum:
p1 = 0.25 kg * 302 m/s = 75.5 kg·m/s
p2 = 37.5 kg * 0 m/s = 0 kg·m/s
The total initial momentum (P_initial) is:
P_initial = p1 + p2 = 75.5 kg·m/s + 0 kg·m/s = 75.5 kg·m/s
After the collision, the bullet and the block move together with a common velocity (V). The total mass after the collision is:
Total mass = m1 + m2 = 0.25 kg + 37.5 kg = 37.75 kg
Using the conservation of momentum:
P_initial = P_final
75.5 kg·m/s = (37.75 kg) * V
Now, solving for V:
V = 75.5 kg·m/s / 37.75 kg ≈ 2.00 m/s
Step 2: Analyzing the Motion of m1 and m2
Now that we have the common velocity after the collision, we need to determine how far the blocks will move together before coming to a stop due to friction.
The force of friction (F_friction) acting on the combined mass (m1 + m2) can be calculated using the formula:
F_friction = μ * N
Where:
- μ = coefficient of friction = 0.5
- N = normal force = (m1 + m2) * g
- g = acceleration due to gravity ≈ 9.81 m/s²
Calculating the normal force:
N = (37.75 kg) * (9.81 m/s²) ≈ 370.5 N
Now, calculating the frictional force:
F_friction = 0.5 * 370.5 N ≈ 185.25 N
Step 3: Finding the Deceleration Due to Friction
The deceleration (a) caused by the frictional force can be determined using Newton's second law:
F = m * a
Rearranging gives:
a = F_friction / (m1 + m2)
Substituting the values:
a = 185.25 N / 37.75 kg ≈ 4.91 m/s²
Step 4: Calculating the Displacement
To find the displacement (d) while the blocks come to a stop, we can use the kinematic equation:
V² = U² + 2a * d
Where:
- V = final velocity = 0 m/s (when they stop)
- U = initial velocity = 2.00 m/s
- a = -4.91 m/s² (deceleration is negative)
Rearranging the equation to solve for d:
0 = (2.00 m/s)² + 2 * (-4.91 m/s²) * d
0 = 4 + (-9.82) * d
Solving for d gives:
9.82 * d = 4
d = 4 / 9.82 ≈ 0.41 m
Summary of Results
In conclusion, the common velocity of the blocks after the bullet embeds itself is approximately 2.00 m/s, and the displacement of both blocks before coming to a stop is approximately 0.41 m.
