A small 120 kg rocket is fired vertically, proppelled by an engine with a thrust of 6500N. THe engine ejects gas at the rate of 4.5kg/s. Calculate the change in velocity of gases relative to the engine. You must use momentum as a last step to answer this question.

I am confused on how to answer this question because I am not sure if I should subtract 120-4.5 for mass. In addition, I am not sure if there is any Force of friction. THis is what I have done so far.

F delta T= m delta V
6500 (1)= 115.5 (Vf - 0)
Vf= 56 m/s

I would start with the concept that momentum is conserved, so


m_r \Delta v_r = m_g \Delta v_g


In other words the momentumn of the gas (g) goint out the back equals the momentumn of the rocket (r) moving forward. You also know


m_r \Delta v_r = F \Delta t \\


Put these together to get:


m_g \Delta v_g = F \Delta t


Now you should have it.