A rocket at rest in space, where there is virtually no gravity, has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\), of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The engine consumes fuel at the rate of \(480 \mathrm{~kg} / \mathrm{s}\), and the exhaust speed is \(3.27 \mathrm{~km} / \mathrm{s}\). The engine is fired for \(250 \mathrm{~s}\). (a) Find the thrust of the rocket engine. \((b)\) What is the mass of the rocket after the engine burn? \((c)\) What is the final speed attained?

We can calculate the thrust of the rocket by considering the momentum of outflowing exhaust gases. The speed of the exhaust gases is given as 3.27 km/s (or 3270 m/s to convert to SI units). The thrust \(T\) (which is a force) is the rate of change of momentum and can be found using the formula \(T = dm/dt \times v\), where \(dm/dt\) is the rate of fuel consumption (480 kg/s), and \(v\) is the speed of exhaust gases. Plugging in the given values, we get, \(T = 480 \times 3270 = 1.57 \times 10^{6} N\).

The mass of the rocket after the engine burn can be found by subtracting the mass of the fuel consumed by the rocket from its initial mass. The rate of fuel consumption is 480 kg/s and the engine is fired for 250 seconds. So the total fuel consumed is \(480 \times 250 = 12 \times 10^{4} kg\). The initial mass of the rocket is \(2.55 \times 10^{5} kg\). So the final mass of the rocket after the engine burn is \((2.55 - 1.2) \times 10^{5} kg = 1.35 \times 10^{5} kg\).

Using the law of conservation of momentum, the final speed of the rocket can be calculated. We know that the initial momentum (when the rocket was at rest) is 0. If we consider the momentum of the rocket after consuming some amount of fuel, the total momentum would be the momentum of the rocket (mass times velocity) and the momentum of the outflowing exhaust gases (fuel consumed times exhaust speed). By the conservation of momentum, these should sum up to 0. This gives us the equation: \(v \times m_{final} - v_{exhaust} \times m_{fuel} = 0\). From this, we can express the speed of the rocket as \(v = v_{exhaust} \times (m_{fuel}/ m_{final})\). Substituting the known values, we get \(v = 3270 \times (1.2 \times 10^{5} / 1.35 \times 10^{5}) = 2.9 \times 10^{3} m/s\) or 2.9 km/s.