Question

1. a group of pirates are firing a cannon during a naval battle. the cannon has a mass of 360 kg and the cannon balls each have a mass of 12 kg. when the cannon is fired, it recoils with a velocity of 5.4 m/s.

a) draw a picture showing the situation described before and after the explosion. draw arrows showing the velocities of each object.
b) calculate the velocity of a cannon ball fired from the cannon.

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(p_{initial}=p_{final}\). Initially, the cannon and cannon - ball are at rest, so \(p_{initial} = 0\). Let \(m_1 = 360\space kg\) (mass of the cannon), \(v_1=- 5.4\space m/s\) (recoil velocity of the cannon, negative because it moves in the opposite direction of the cannon - ball), and \(m_2 = 12\space kg\) (mass of the cannon - ball) and \(v_2\) be the velocity of the cannon - ball.
According to \(p_{initial}=p_{final}\), we have \(0=m_1v_1 + m_2v_2\).

Step2: Solve for \(v_2\)

Rearrange the equation \(0=m_1v_1 + m_2v_2\) to get \(v_2=-\frac{m_1v_1}{m_2}\).
Substitute \(m_1 = 360\space kg\), \(v_1=-5.4\space m/s\) and \(m_2 = 12\space kg\) into the formula:
\(v_2=-\frac{360\times(- 5.4)}{12}\)
First, calculate \(360\times(-5.4)=-1944\). Then \(\frac{-1944}{12}=- 162\). But since we have two negative signs (\(-\frac{-1944}{12}\)), \(v_2 = 162\space m/s\).

Answer:

The velocity of the cannon - ball is \(162\space m/s\).