91Ó°ÊÓ

Kinetic energy is the energy an object has due to its motion. It's an important concept when dealing with moving objects. In the problem, we examine two scenarios where a cannon fires a shell. The main point is that, whether the cannon is bolted or free, the kinetic energy of the gunpowder remains the same. This is because the energy from the gunpowder converts into the kinetic energy of the moving cannon and shell, whether the cannon itself is allowed to move or not.

The formula for kinetic energy is:

• \( K = \frac{1}{2} m v^2 \)

where \( m \) is the mass of the object and \( v \) is its velocity.

For the bolted cannon scenario, the shell moves with a velocity of 551 m/s. Using the kinetic energy formula, we can calculate the energy provided to the shell by the gunpowder, which was approximately \( 1.29 \times 10^7 \) Joules. This calculation sets the foundation for understanding how energy transfers in physical systems.

Conservation of linear momentum is a principle stating that the total momentum of a system remains constant if no external forces act upon it. This is especially useful in understanding what happens when our cannon is unbolted. Previously, the whole momentum of the system was carried by the shell. With the cannon free, both the shell and cannon move in opposite directions to preserve momentum. This means when the cannon recoils, it absorbs some of the momentum transferred to the shell.

The momentum, \( p \), of an object is calculated by:

• \( p = mv \)

For the system of the cannon and shell:

• \( m_s v_s = m_c v_c \)

where \( m_s \) and \( v_s \) are the mass and velocity of the shell, while \( m_c \) and \( v_c \) are for the cannon. This equation shows the dependency of both the shell's and cannon's velocities on the relative masses. Their relationship through momentum conservation helps us find the shell's new velocity when the cannon is free to move.

Calculating velocities in the context of energy and momentum conservation requires balancing both concepts. In our exercise, we start with knowing the initial kinetic energy when the cannon was bolted. Now, when the cannon is unbolted, the kinetic energy is shared between the moving shell and the recoiling cannon.

To find the velocity of the shell fired from the unbolted cannon:

• We know the total kinetic energy is divided between the two objects.
• Using the expression of kinetic energy for both objects: \[ K = \frac{1}{2} m_s v_s^2 + \frac{1}{2} m_c v_c^2 \]
• And expressing \( v_c \) in terms of \( v_s \) from momentums \( v_c = \frac{m_s}{m_c} v_s \).
• We simplify and solve for \( v_s \) to find the new shell velocity.

After correctly substituting the known quantities and solving the equation, we find that the velocity of the shell is approximately 546.6 m/s in the unbolted scenario. This process not only involves plugging into equations but understanding how energy and momentum balance each other in physical systems.