Question

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600 kg of fuel, which it burns in 30 s. What is the rocket’s speed 10 s, 20 s, and 30 s after launch?

Short answer

At t = 10s, speed is equal to 775.5 m/s

At t = 20s, speed is equal to 1905.3 m/s

At t = 30s, speed = v_max = 4023.5 m/s

Step-by-step solution

Step 1. Given information is Empty mass of Rocket mR = 150 kgMass of fuel mFO = 600 kgTime taken to burn the fuel∆t=30 sExhaust speed vex = 2500 m/s

We need to calculate the rocket’s speed 10 s, 20 s, and 30 s after launch.

Step 2. Using the rate of burning fuel to find out the rocket's speed after launch.

Rate of burning fuel :

$$\begin{gathered} R=\frac{dm}{dt}=\frac{600}{30}=20kg/s \\ Massoffuelburnedingiventimeinterval: \\ m\text{'}={\int }_0^{t}R.dt=Rt \\ Massoffuelremainingafterthattime: \\ m\text{'}\text{'}=m_{FO}-RT \\ Hence,thetotalweightoftherocketis, \\ m=m\text{'}\text{'}+m_R \\ {m}_0=m_R+m_{FO} \\ =(150+600)kg=750kg \\ Weknowthat, \\ v=v_{ex}\ln \frac{m_0}{m} \\ Fort=10s \\ m\text{'}\text{'}=m_{FO}-Rt=600-(20\times 10)=400kg \\ m=m\text{'}\text{'}+m_R=(400+150)kg=550kg \\ v=2500\ln \frac{750}{550} \\ v=775.5m/s \\ Fort=20s \\ m\text{'}\text{'}=m_{FO}-Rt=600-20\times 20=200kg \\ m=m\text{'}\text{'}+m_R=(200+150)kg=350kg \\ v=2500\ln \frac{750}{350} \\ v=1905.3m/s \\ Fort=30s \\ m\text{'}\text{'}=m_{FO}-Rt=600-20\times 30=0 \\ Itmeanstherocketwillachievemaximumvelocity \\ U\sin gformula, \\ {v}_{max}=v_{ex}\ln \frac{m_{FO}+m_R}{m_R} \\ {v}_{max}=2500\ln \frac{750}{150}=4023.5m/s \end{gathered}$$