Answer: We can use conservation of momentum to find the initial speed of the bullet. Before the collision, the bullet has momentum p1 and after the collision, the bullet and block together have momentum p2. Assuming there are no external forces acting on the system, we can say that p1 = p2.
Let v be the initial speed of the bullet, m1 be the mass of the bullet, m2 be the mass of the block, and f be the coefficient of kinetic friction between the block and the surface.
The momentum of the bullet before the collision is given by:
p1 = m1 * v
The momentum of the block and bullet after the collision is given by:
p2 = (m1 + m2) * vf
where vf is the final velocity of the block and bullet together.
Since momentum is conserved, we can equate p1 and p2:
m1 * v = (m1 + m2) * vf
We also know that the block and bullet slide a distance of 0.230 m before stopping. Using this, we can find the final velocity of the block and bullet together using the equation for kinetic friction:
f * (m1 + m2) * g * d = (m1 + m2) * vf^2 / 2
where g is the acceleration due to gravity and d is the distance the block and bullet slide.
Simplifying this equation, we get:
vf^2 = 2 * f * g * d
Substituting this expression for vf into the equation for conservation of momentum, we get:
m1 * v = (m1 + m2) * sqrt(2 * f * g * d)
Solving for v, we get:
v = (m1 + m2) / m1 * sqrt(2 * f * g * d)
Substituting the given values, we get:
v = (0.00700 + 1.17) / 0.00700 * sqrt(2 * 0.240 * 9.81 * 0.230)
which gives v = 348 m/s (to three significant figures).
Therefore, the initial speed of the bullet was approximately 348 m/s.