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question 14 of 22
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question 14
a 100 - kg cannon at rest contains a 10 - kg cannon ball. when fired, the cannon ball leaves the cannon with a speed of 90 m/s. what is the recoil speed of the cannon?
4.5 m/s
45 m/s
9 m/s
zero m/s
90 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the system (cannon + cannon - ball) is \(P_i = 0\) (since both are at rest, \(v_{i,cannon}=v_{i,ball} = 0\)). According to the law of conservation of momentum \(P_i=P_f\), where \(P_f=m_{cannon}v_{cannon}+m_{ball}v_{ball}\). So, \(0 = m_{cannon}v_{cannon}+m_{ball}v_{ball}\).

Step2: Solve for the recoil speed of the cannon

We can rewrite the equation \(0 = m_{cannon}v_{cannon}+m_{ball}v_{ball}\) as \(v_{cannon}=-\frac{m_{ball}v_{ball}}{m_{cannon}}\). Given \(m_{cannon} = 100\space kg\), \(m_{ball}=10\space kg\), and \(v_{ball}=90\space m/s\). Substitute the values: \(v_{cannon}=-\frac{10\times90}{100}\space m/s=- 9\space m/s\). The negative sign indicates the direction opposite to the cannon - ball's motion, and the magnitude of the recoil speed is \(9\space m/s\).

Answer:

9 m/s