A revolutionary war cannon, with a mass of 2040 kg, fires a 16.2 kg ball horizontally. The cannonball has a speed of 112m/s after it has left the barrel.. The cannon carriage is on a flat platform and is free to roll horizoni tally. What is the speed of the cannon immedi- ately after it was fired? Answer in units of m/s Answer in units of m/s

We can solve this problem using the principle of conservation of momentum. Since the system (cannon + cannonball) is initially at rest, the total momentum before firing is zero. After firing, the total momentum must still be zero. The momentum of the cannonball is in one direction, and the momentum of the cannon is in the opposite direction.<br /><br />* **Momentum of Cannonball:** The momentum of the cannonball is its mass multiplied by its velocity:<br /><br /> Momentum_cannonball = (mass_cannonball) * (velocity_cannonball)<br /> Momentum_cannonball = (16.2 kg) * (112 m/s)<br /> Momentum_cannonball = 1814.4 kg*m/s<br /><br />* **Momentum of Cannon:** The momentum of the cannon must be equal in magnitude and opposite in direction to the momentum of the cannonball:<br /><br /> Momentum_cannon = - Momentum_cannonball<br /> Momentum_cannon = -1814.4 kg*m/s<br /><br />* **Velocity of Cannon:** We can find the velocity of the cannon by dividing its momentum by its mass:<br /><br /> Velocity_cannon = Momentum_cannon / mass_cannon<br /> Velocity_cannon = (-1814.4 kg*m/s) / (2040 kg)<br /> Velocity_cannon ≈ -0.889 m/s<br /><br />The negative sign indicates that the cannon recoils in the opposite direction of the cannonball.<br /><br />Therefore, the speed of the cannon immediately after it was fired is approximately $\boxed{0.889}$ m/s.<br />