Question

1. find the final velocity of the 500 kg cannon after a 100 kg person exits at a speed of 15m/s.

Explanation:

Step1: Apply conservation of momentum

Initially, total momentum is 0, so $m_1v_1 + m_2v_2 = 0$

Step2: Define variables and rearrange

Let $m_1=500\ \text{kg}$ (cannon), $v_1=v$, $m_2=100\ \text{kg}$, $v_2=15\ \text{m/s}$. Rearrange to solve for $v$:
$v = -\frac{m_2v_2}{m_1}$

Step3: Substitute values

$v = -\frac{100\ \text{kg} \times 15\ \text{m/s}}{500\ \text{kg}}$

Step4: Calculate final value

$v = -3\ \text{m/s}$
The negative sign indicates direction opposite to the person.

Answer:

The final velocity of the cannon is 3 m/s in the direction opposite to the exiting person (or -3 m/s if using the coordinate system in the diagram).