Cop

Sure, thank you for your question.

The key concept to use in this question is conservation of energy. As the same explosive charge is used, we can assume that the total energy added to the system (through chemical potential energy of the explosives) is the same.

So, for the first scenario (when the cannon was not mounted rigidly), the chemical potential energy has been converted to kinetic energy of the cannon and the kinetic energy of the cannonball. On the other hand, for the second scenario (when the cannon was mounted rigidly), the chemical potential energy has been converted to only the kinetic energy of the cannonball. Here is how I would set the equation.

∑Ei = ∑Ef

KEc + KEb1 = KEb2

0.5Mc(Vc)^2 + 0.5Mb(Vb1)^2 = 0.5Mb(Vb2)^2

Mc(Vc)^2 + Mb(Vb1)^2 = Mb(Vb2)^2

2080(0.885528846)^2 + 16.3(113)^2 = 16.3(Vb2)^2

Vb2 = 113.4419003701 m/s

As the question is asking how much faster would the ball travel, the answer would be 113.4419003701 m/s - 113 m/s = 0.4419003701 m/s.

Does this answer your question?