To solve this problem, we need to use the principle of conservation of momentum. Given the parameters, we need to determine the velocity of the cannonball when the cannon fires it.
Here are the given values:
- Mass of the cannon ([tex]\( m_{\text{cannon}} \)[/tex]) = 2,000 kg
- Mass of the shell ([tex]\( m_{\text{shell}} \)[/tex]) = 10 kg
- Recoil velocity of the cannon ([tex]\( v_{\text{cannon}} \)[/tex]) = 0.5 m/s
The law of conservation of momentum states that the total momentum before and after the event must be equal. Initially, both the cannon and the shell are stationary, so the initial momentum is zero.
When the cannon fires the shell, the system's final momentum must also be zero because there are no external horizontal forces. So we can set up the equation:
[tex]\[
m_{\text{cannon}} \cdot (-v_{\text{cannon}}) + m_{\text{shell}} \cdot v_{\text{shell}} = 0
\][/tex]
Here, [tex]\( -v_{\text{cannon}} \)[/tex] represents the recoil velocity of the cannon in the opposite direction to the shell's velocity [tex]\( v_{\text{shell}} \)[/tex].
Solving for [tex]\( v_{\text{shell}} \)[/tex]:
[tex]\[
m_{\text{cannon}} \cdot (-v_{\text{cannon}}) + m_{\text{shell}} \cdot v_{\text{shell}} = 0
\][/tex]
[tex]\[
2000 \cdot (-0.5) + 10 \cdot v_{\text{shell}} = 0
\][/tex]
[tex]\[
-1000 + 10 \cdot v_{\text{shell}} = 0
\][/tex]
[tex]\[
10 \cdot v_{\text{shell}} = 1000
\][/tex]
[tex]\[
v_{\text{shell}} = \frac{1000}{10}
\][/tex]
[tex]\[
v_{\text{shell}} = 100 \text{ m/s}
\][/tex]
Thus, the velocity of the cannonball is:
[tex]\[
\boxed{100 \text{ m/s}}
\][/tex]
Given the available choices, the correct answer is:
c. 100 m/s
