A 1200-kg cannon suddenly fires a $\text { 100-kg }$ cannonball at $35 \mathrm {~m} / \mathrm { s }$ What is the recoil speed of the cannon? Assume that frictional forces are negligible and the cannon is fired horizontally.
A)2.9 m/s
B)35 m/s
C)3.5 m/s
D)3.2 m/s

Question

Quizplus · Accepted Answer

The recoil speed of the cannon can be calculated using the conservation of momentum principle. Before the cannon fires, the total momentum of the system (cannon + cannonball) is 0 because neither is moving. After firing, the momentum of the cannonball is $100 \, \text{kg} \times 35 \, \text{m/s} = 3500 \, \text{kg} \cdot \text{m/s}$. To conserve momentum, the cannon must have an equal and opposite momentum. Thus, the recoil speed $v$ of the cannon is given by $1200 \, \text{kg} \times v = 3500 \, \text{kg} \cdot \text{m/s}$, solving for $v$ gives $v = 2.9 \, \text{m/s}$.

A 1200-Kg Cannon Suddenly Fires A 100-kg \text { 100-kg } 100-kg Cannonball At

A 1200-Kg Cannon Suddenly Fires A $\text { 100-kg }$ Cannonball At