(Solved): The \( 0.04 \mathrm{~kg} \) bullet is traveling at \( 450 \mathrm{~m} / \mathrm{s} \) when it becom ...

The \( 0.04 \mathrm{~kg} \) bullet is traveling at \( 450 \mathrm{~m} / \mathrm{s} \) when it becomes embedded in the \( 3 \mathrm{~kg} \) stationary block. The coefficient of kinetic friction between the block and plane is \( 0.2 \). A. Determine the velocity of the bullet and block just after the collision. B. Find an expression for the velocity of the bullet/ block system as it slides along the plane. C. Use the expression from Part B to determine the distance that the bullet/ block system will slide before it stops.