Question

(II) Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of 1300 kg/s with a speed of \(4.5 \times {10^4}\;{\rm{m/s}}\).

Short answer

The exerted force on the rocket is \(5.9 \times {10^7}\;{\rm{N}}\).

Step-by-step solution

Given data

The rate of expelled gas is \(\frac{{\Delta m}}{{\Delta t}} = 1300\;{\rm{kg/s}}\).

The speed of the expelled gas is \(v = 4.5 \times {10^2}\;{\rm{m/s}}\).

Let F be the force exerted on the rocket.

Calculation of the force on the rocket

The magnitude of the force is equal to the rate of change of momentum.

From the impulse-momentum theorem, the force on the rocket is equal to the rate of change of momentum of the expelled gas.

Therefore,

Now substituting the values in the above equation of the force,

\(\begin{array}{c}F = \left( {1300\;{\rm{kg/s}}} \right) \times \left( {4.5 \times {{10}^2}\;{\rm{m/s}}} \right)\\ = 5.9 \times {10^7}\;{\rm{N}}\end{array}\)

Hence, exerted force on the rocket is \(5.9 \times {10^7}\;{\rm{N}}\).