Question

a battleship that is ( 6.00\times10^{7}) kg and is originally at rest fires a 1100 - kg artillery shell horizontally with a velocity of 450 m/s
(a) if the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ships recoil. calculate its recoil velocity in meters per second
0.0101 × m/s

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum \(p_i = 0\) (since the system is at rest initially). According to the law of conservation of momentum \(p_i=p_f\), where \(p_f = m_{shell}v_{shell}+m_{ship}v_{ship}\)

So, \(0 = m_{shell}v_{shell}+m_{ship}v_{ship}\)

Step2: Solve for the recoil velocity of the ship

We can re - arrange the equation \(v_{ship}=-\frac{m_{shell}v_{shell}}{m_{ship}}\)

Given \(m_{shell} = 1100\space kg\), \(v_{shell}=450\space m/s\), \(m_{ship}=6.00\times 10^{7}\space kg\)

Substitute the values: \(v_{ship}=-\frac{1100\times450}{6.00\times 10^{7}}\)

\(v_{ship}=-\frac{495000}{6.00\times 10^{7}}=- 0.00825\space m/s\)

The negative sign indicates the direction of the ship's recoil (opposite to the direction of the shell's motion)

Answer:

\(-0.00825\space m/s\)