Question

a battleship that is ( 4.00\times10^{7}) kg and is originally at rest fires a 1100 - kg artillery shell horizontally with a velocity of 675 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.

m/s
(b) calculate the increase in internal kinetic energy in joules (that is, for the ship and the shell). this energy is less than the energy released by the gun powder - significant heat transfer occurs.

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Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the system (battleship + shell) is \(P_{i}=0\) (since both are at rest). The final momentum \(P_{f}=m_{s}v_{s}+M_{b}v_{b}\), where \(m_{s} = 1100\space kg\), \(v_{s}=675\space m/s\), \(M_{b}=4.00\times 10^{7}\space kg\) and \(v_{b}\) is the recoil velocity of the battleship.
By conservation of momentum \(P_{i} = P_{f}\), so \(0=m_{s}v_{s}+M_{b}v_{b}\).
Solving for \(v_{b}\): \(v_{b}=-\frac{m_{s}v_{s}}{M_{b}}\)
Substitute the values: \(v_{b}=-\frac{1100\times675}{4.00\times 10^{7}}\)

$$v_{b}=-\frac{742500}{4.00\times 10^{7}}=- 0.0185625\space m/s$$

The negative sign indicates the direction (opposite to the shell's velocity). The magnitude of the recoil velocity is \(v_{b}=0.0186\space m/s\) (rounded to three significant figures)

Step2: Calculate the initial and final kinetic energies

The initial kinetic energy \(K_{i} = 0\) (since both are at rest).
The final kinetic energy \(K_{f}=\frac{1}{2}m_{s}v_{s}^{2}+\frac{1}{2}M_{b}v_{b}^{2}\)
Substitute \(m_{s} = 1100\space kg\), \(v_{s}=675\space m/s\), \(M_{b}=4.00\times 10^{7}\space kg\) and \(v_{b}=0.0185625\space m/s\)
\(\frac{1}{2}m_{s}v_{s}^{2}=\frac{1}{2}\times1100\times(675)^{2}=\frac{1100\times455625}{2}=249593750\space J\)
\(\frac{1}{2}M_{b}v_{b}^{2}=\frac{1}{2}\times4.00\times 10^{7}\times(0.0185625)^{2}\)

$$=\frac{4.00\times 10^{7}\times0.00034453125}{2}=6890.625\space J$$

\(K_{f}=249593750 + 6890.625=249600640.625\space J\)
The increase in internal kinetic energy \(\Delta K=K_{f}-K_{i}=249600640.625\space J\approx2.50\times 10^{8}\space J\)

Answer:

(a) \(0.0186\space m/s\)
(b) \(2.50\times 10^{8}\space J\)