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

Respuesta :

Answer:

Approximately [tex]\rm 2.5\; m \cdot s^{-1}[/tex].

Explanation:

Let the increase in the rocket's velocity be [tex]\Delta v[/tex]. Let [tex]v_0[/tex] represent the initial velocity of the rocket. Note that for this question, the exact value of  [tex]v_0[/tex] 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: [tex]1000[/tex].
  • Initial velocity of the rocket and the fuel: [tex]v_0[/tex].
  • Hence the initial momentum of the rocket: [tex]1000\,v_0[/tex].
  • Mass of the rocket without that 5 kg of fuel: [tex]1000 - 5 = 995[/tex].
  • Final velocity of the rocket: [tex]v_0 + \Delta v[/tex].
  • Hence the final momentum of the rocket: [tex]995\,(v_0 + \Delta v)[/tex].
  • Mass of the 5 kg of fuel: [tex]5[/tex].
  • Final velocity of the fuel: [tex]v_0 - 500[/tex] (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: [tex]5\,(v_0 - 500)[/tex].

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

Sum of initial momentum = Sum of final momentum.

[tex]1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500)[/tex].

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

[tex]0 = 995\, \Delta v - 5\times 500[/tex].

[tex]\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}[/tex].

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

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