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.

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the system (battleship + shell) is \(P_{i}=0\) (since the system is initially at rest, \(m_{1}v_{1i}+m_{2}v_{2i} = 0\), where \(m_{1}\) is the mass of the battleship, \(v_{1i}\) is the initial velocity of the battleship, \(m_{2}\) is the mass of the shell, and \(v_{2i}\) is the initial velocity of the shell).

The final momentum of the system is \(P_{f}=m_{1}v_{1f}+m_{2}v_{2f}\).

According to the law of conservation of momentum \(P_{i} = P_{f}\), so \(0=m_{1}v_{1f}+m_{2}v_{2f}\).

We can solve for \(v_{1f}\): \(v_{1f}=-\frac{m_{2}v_{2f}}{m_{1}}\)

Step2: Substitute the given values

We are given that \(m_{1}=6.00\times 10^{7}\space kg\), \(m_{2} = 1100\space kg\), and \(v_{2f}=- 450\space m/s\) (negative because the shell is fired aft).

Substitute these values into the formula for \(v_{1f}\):

\(v_{1f}=-\frac{1100\times(- 450)}{6.00\times 10^{7}}\)

First, calculate the numerator: \(1100\times450 = 495000\)

Then, \(v_{1f}=\frac{495000}{6.00\times 10^{7}}\)

\(v_{1f}=0.00825\space m/s\)

Answer:

\(0.00825\space m/s\)