Question

1. a rail 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 10³m/s to the right, and that the maximum velocity of the vehicle is 45 m/s left. what is the mass of the vehicle?

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum of the system (vehicle + gas) is \(0\) (since it starts from rest). According to the law of conservation of momentum \(m_1v_1 + m_2v_2=0\), where \(m_1\) is the mass of the gas, \(v_1\) is the velocity of the gas, \(m_2\) is the mass of the vehicle, and \(v_2\) is the velocity of the vehicle.
We can rewrite the equation as \(m_2=\frac{-m_1v_1}{v_2}\).

Step2: Substitute the given values

Given \(m_1 = 4.5\times10^{2}\space kg\), \(v_1=1.4\times 10^{3}\space m/s\) (take right - direction as positive), and \(v_2=- 45\space m/s\) (left - direction is negative).
Substitute into the formula: \(m_2=\frac{-4.5\times 10^{2}\times1.4\times 10^{3}}{-45}\)
First, calculate the numerator: \(-4.5\times 10^{2}\times1.4\times 10^{3}=-6.3\times10^{5}\)
Then, \(m_2=\frac{-6.3\times 10^{5}}{-45}=1.4\times10^{4}\space kg\)

Answer:

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