Final answer:
To find the initial speed of the bullet, we can use the principles of conservation of momentum and work-energy. By applying the conservation of momentum, we can determine the final velocity of the block-bullet system. Then, using the work-energy principle and the compression distance of the spring, we can find the initial speed of the bullet.
Step-by-step explanation:
To determine the initial speed of the bullet, we can use the principle of conservation of momentum. Initially, the bullet is traveling horizontally with a mass of 0.050 kg and the wooden block is at rest with a mass of 0.600 kg. After the collision, the bullet embeds itself into the block and both objects move together. Let's denote the initial speed of the bullet as vb and the final velocity of the block-bullet system as vf.
By applying the conservation of momentum, we have:
mbvb + 0 = (mb + mwb)vf
Substituting the given values, we have:
0.050 kg \cdot vb = (0.050 kg + 0.600 kg) \cdot vf
Simplifying the equation, we get:
vf = \frac{0.050 kg \cdot vb}{0.650 kg}
Next, we can use the work-energy principle to find the initial speed of the bullet. When the block hits the wall and compresses the spring, the work done by the block is equal to the change in potential energy of the spring. The potential energy stored in the spring is given by:
\frac{1}{2} kx^2
where k is the spring constant and x is the compression distance. Setting the work done by the block equal to the change in potential energy, we have:
F \cdot x = \frac{1}{2} kx^2
mgx = \frac{1}{2} kx^2
Simplifying the equation, we get:
vf2 - v02 = 2a d
0 - vb2 = 2 \left(-\frac{k}{mf + mwb}\right) \cdot \frac{1}{2} kx^2
Simplifying the equation, we get:
vb2 = \left(-\frac{kx^2}{mf + mwb}\right)
Solving for vb,
vb = \sqrt{\left(-\frac{kx^2}{mf + mwb}\right)}
Substituting the given values, we have:
vb = \sqrt{\left(-\frac{(2500 N/m)(0.15 m)^2}{0.600 kg + 0.050 kg}\right)}