The speed of the bullet immediately before impact is approximately \(1298.78 \, \text{m/s}\).
Let's denote the speed of the bullet before impact as \(v\) and the speed of the block-bullet combination after the impact as \(V\).
The conservation of linear momentum can be applied in the horizontal direction before and after the collision:
\[m_{\text{bullet}} \cdot v_{\text{bullet, initial}} = (m_{\text{bullet}} + m_{\text{block}}) \cdot V_{\text{final}}\]
Given that the bullet becomes embedded in the block and both move together after the collision, the final velocity of the block-bullet combination (\(V_{\text{final}}\)) is the same for both.
Now, let's find \(V_{\text{final}}\):
\[m_{\text{bullet}} \cdot v = (m_{\text{bullet}} + m_{\text{block}}) \cdot V_{\text{final}}\]
\[11.6 \, \text{g} \cdot v = (11.6 \, \text{g} + 105 \, \text{g}) \cdot V_{\text{final}}\]
Convert the masses to kilograms:
\[0.0116 \, \text{kg} \cdot v = (0.0116 \, \text{kg} + 0.105 \, \text{kg}) \cdot V_{\text{final}}\]
Now, solve for \(V_{\text{final}}\).
\[V_{\text{final}} = \frac{0.0116 \, \text{kg} \cdot v}{0.1166 \, \text{kg}}\]
\[V_{\text{final}} = 0.001 \, v\]
The work done by friction during the sliding motion can be expressed as:
\[W_{\text{friction}} = \mu_k \cdot m_{\text{eff}} \cdot g \cdot d\]
where
\(\mu_k\) is the coefficient of kinetic friction,
\(m_{\text{eff}}\) is the effective mass of the block-bullet system, and
\(d\) is the distance traveled.
The effective mass can be calculated as the sum of the masses:
\[m_{\text{eff}} = m_{\text{bullet}} + m_{\text{block}}\]
Now, the work done by friction is equal to the initial kinetic energy of the block-bullet system:
\[W_{\text{friction}} = \frac{1}{2} \cdot m_{\text{eff}} \cdot V_{\text{final}}^2\]
Substitute the expression for \(V_{\text{final}}\) and solve for \(v\):
\[\frac{1}{2} \cdot (0.0116 \, \text{kg} + 0.105 \, \text{kg}) \cdot (0.001 \, v)^2 = \mu_k \cdot (0.0116 \, \text{kg} + 0.105 \, \text{kg}) \cdot 9.8 \, \text{m/s}^2 \cdot 6.5 \, \text{m}\]
Solve for \(v\).
\[0.5 \cdot 0.1166 \cdot (0.001 \, v)^2 = 0.75 \cdot 0.1166 \cdot 9.8 \cdot 6.5\]
\[0.0583 \cdot (0.000001 \, v^2) = 0.75 \cdot 0.1166 \cdot 9.8 \cdot 6.5\]
\[0.0000000583 \, v^2 = 0.75 \cdot 0.1166 \cdot 9.8 \cdot 6.5\]
Now, solve for \(v\):
\[v^2 = \frac{0.75 \cdot 0.1166 \cdot 9.8 \cdot 6.5}{0.0000000583}\]
\[v^2 \approx 1687685.06\]
\[v \approx 1298.78 \, \text{m/s}\]
Therefore, the speed of the bullet immediately before impact is approximately \(1298.78 \, \text{m/s}\).
The probable question may be:
A 105 g wooden block is initially at rest on a rough horizontal surface when a 11.6 g bullet is fired horizontally into (but does not go through) it. After the impact, the block-bullet combination slides 6.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet (in m/s) immediately before impact.