Consider a wheeled, 537 kg cannon firing a 2 kg cannonball
The correct answer and explanation is:
To analyze this scenario, we can apply conservation of momentum.
Correct Answer:
When the 2 kg cannonball is fired from a 537 kg wheeled cannon (initially at rest), the cannon will recoil in the opposite direction due to the law of conservation of momentum.
Explanation (300 words):
The law of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. Initially, both the cannon and the cannonball are at rest, so the total momentum of the system is zero.
Let’s define:
- Mass of cannon: M=537 kgM = 537 \, \text{kg}
- Mass of cannonball: m=2 kgm = 2 \, \text{kg}
- Velocity of cannonball after firing: vv
- Recoil velocity of cannon: VV
After the cannon fires the cannonball, the system’s total momentum must still be zero: Total initial momentum=0\text{Total initial momentum} = 0 Total final momentum=mv+MV\text{Total final momentum} = mv + MV
Since momentum is conserved: mv+MV=0⇒MV=−mv⇒V=−mvMmv + MV = 0 \Rightarrow MV = -mv \Rightarrow V = -\frac{mv}{M}
This equation shows that the cannon moves in the opposite direction of the cannonball’s motion, and with a much smaller velocity, due to its much greater mass.
Key Concepts:
- Recoil: When the cannon fires, the explosion pushes the cannonball forward and pushes the cannon backward (Newton’s Third Law).
- Velocity relationship: Because m≪Mm \ll M, the cannon’s recoil speed is much less than the cannonball’s speed.
- Practical implication: Cannons are mounted on wheels or platforms to absorb or allow controlled recoil motion.
Example Calculation:
If the cannonball leaves the cannon at 300 m/s: V=−2×300537≈−1.12 m/sV = -\frac{2 \times 300}{537} \approx -1.12 \, \text{m/s}
So the cannon recoils at approximately 1.12 m/s in the opposite direction.
This situation illustrates Newton’s third law and conservation of momentum in real-world mechanics.