Question

a battleship that is ( 4.90 \times 10^{7} mathrm{~kg} ) and is originally at rest fires a ( 1150-mathrm{kg} ) artillery shell horizontally with a velocity of ( 450 mathrm{~m} / mathrm{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. ( mathrm{m} / mathrm{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. j

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the system (ship + shell) is \(P_{i}=0\) (since both are at rest initially). The final momentum \(P_{f}=m_{shell}v_{shell}+m_{ship}v_{ship}\). According to the law of conservation of momentum \(P_{i} = P_{f}\), so \(0=m_{shell}v_{shell}+m_{ship}v_{ship}\). We can solve for \(v_{ship}\): \(v_{ship}=-\frac{m_{shell}v_{shell}}{m_{ship}}\). Given \(m_{shell} = 1100\space kg\), \(v_{shell}=450\space m/s\), \(m_{ship}=4.90\times 10^{7}\space kg\).

$$v_{ship}=-\frac{1100\times450}{4.90\times 10^{7}}$$

$$v_{ship}=-\frac{495000}{4.90\times 10^{7}}\approx - 0.0101\space m/s$$

The negative sign indicates the direction of the ship's recoil (opposite to the shell's motion). The magnitude of the recoil velocity is \(|v_{ship}| = 0.0101\space m/s\)

Step2: Calculate the increase in internal kinetic energy

The internal kinetic energy \(K=\frac{1}{2}m_{shell}v_{shell}^{2}+\frac{1}{2}m_{ship}v_{ship}^{2}\)
Substitute \(m_{shell} = 1100\space kg\), \(v_{shell}=450\space m/s\), \(m_{ship}=4.90\times 10^{7}\space kg\), \(v_{ship}=0.0101\space m/s\)

$$K=\frac{1}{2}(1100)(450)^{2}+\frac{1}{2}(4.90\times 10^{7})(0.0101)^{2}$$

First term: \(\frac{1}{2}(1100)(450)^{2}=\frac{1100\times202500}{2}= 1.11375\times 10^{8}\space J\)
Second term: \(\frac{1}{2}(4.90\times 10^{7})(0.0101)^{2}=\frac{4.90\times 10^{7}\times0.00010201}{2}\approx2500\space J\)

$$K\approx1.11375\times 10^{8}+ 2500\approx1.11\times 10^{8}\space J$$

Answer:

(a) The recoil velocity of the ship is \(0.0101\space m/s\)
(b) The increase in internal kinetic energy is \(1.11\times 10^{8}\space J\)