Question

1. a vehicle with a rocket engine is being tested on a smooth track. starting from rest the engine is fired for a short period of time, releasing 4.5 x 10² kg of gases. it is estimated that the average velocity of the gases is 1.4 x 10³ m/s to the right, and that the maximum velocity of the vehicle is 45 m/s left. what is the mass of the vehicle? (1.4x10⁴ kg)

Explanation:

Step1: Apply the principle of conservation of momentum

The initial momentum of the system (vehicle + gas) is \(0\) (since it starts from rest). Let \(m_{gas}=4.5\times 10^{2}\space kg\), \(v_{gas}=1.4\times 10^{3}\space m/s\) and \(m_{vehicle}\) be the mass of the vehicle, \(v_{vehicle} = 45\space m/s\). According to the law of conservation of momentum \(m_{gas}v_{gas}-m_{vehicle}v_{vehicle}=0\) (the negative sign for the vehicle's velocity is because it moves in the opposite direction of the gas). So \(m_{vehicle}=\frac{m_{gas}v_{gas}}{v_{vehicle}}\).

Step2: Substitute the values

Substitute \(m_{gas}=4.5\times 10^{2}\), \(v_{gas}=1.4\times 10^{3}\) and \(v_{vehicle} = 45\) into the formula \(m_{vehicle}=\frac{4.5\times 10^{2}\times1.4\times 10^{3}}{45}\).
First, calculate \(4.5\times 10^{2}\times1.4\times 10^{3}=(4.5\times1.4)\times10^{2 + 3}=6.3\times 10^{5}\).
Then, \(\frac{6.3\times 10^{5}}{45}=\frac{6.3}{45}\times10^{5}\). Since \(\frac{6.3}{45}=0.14\), so \(m_{vehicle}=0.14\times 10^{5}=1.4\times 10^{4}\space kg\).

Answer:

\(1.4\times 10^{4}\space kg\)