Question

A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its velocity?​

Answer 1

Approximately
`\rm 2.5\; m \cdot s^(-1)`.

Step-by-step explanation:

Let the increase in the rocket's velocity be
`\Delta v`. Let
`v_0` represent the initial velocity of the rocket. Note that for this question, the exact value of
`v_0` doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

• Mass of the rocket with the 5 kg of fuel:
  `1000`.
• Initial velocity of the rocket and the fuel:
  `v_0`.
• Hence the initial momentum of the rocket:
  `1000\,v_0`.

• Mass of the rocket without that 5 kg of fuel:
  `1000 - 5 = 995`.
• Final velocity of the rocket:
  `v_0 + \Delta v`.
• Hence the final momentum of the rocket:
  `995\,(v_0 + \Delta v)`.

• Mass of the 5 kg of fuel:
  `5`.
• Final velocity of the fuel:
  `v_0 - 500` (assuming that the the 500 m/s in the question takes the rocket as its reference.)
• Hence the final momentum of the fuel:
  `5\,(v_0 - 500)`.

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

`1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500)`.

Note that
`1000\, v_0` appears on both sides of the equation. These two terms could hence be eliminated.

`0 = 995\, \Delta v - 5* 500`.

`\displaystyle \Delta v = (5)/(995)* 500 \approx \rm 2.5\; m \cdot s^(-1)`.

Hence, the velocity of the rocket increased by around 2.5 m/s.