In a rocket, fuel burns at the rate of \(1 \mathrm{~kg} / \mathrm{s}\). This fuel is ejected from the rocket…

In a rocket, fuel burns at the rate of \(1 \mathrm{~kg} / \mathrm{s}\). This fuel is ejected from the rocket with a velocity of \(60 \mathrm{~km} / \mathrm{s}\). The force exerted on the rocket by this is
  1. \(60 \mathrm{~N}\)
  2. \(600 \mathrm{~N}\)
  3. \(6000 \mathrm{~N}\)
  4. \(60000 \mathrm{~N}\)

Solution

For a rocket, \(\frac{d m}{d t}=1 \mathrm{~kg} / \mathrm{s}\) Velocity of fuel ejected by rocket, \(v=60 \mathrm{~km} / \mathrm{s}=60000 \mathrm{~m} / \mathrm{s}\) \(\therefore\) Force exerted on the rocket is given as \(F=v \frac{d m}{d t}=60000 \times 1=60000 \mathrm{~N}\)

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

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