Question

A rocket of mass 10000kg uses 5.0kg of fuel and oxygen
to produce exhaust gases ejected at 5000 m/s. Calculate the
increase in its velocity.
please give brief explanation.​

Answer 1

Answer: dv’ = 2.5 m/s

Explanation:

We know that the forward force of the engine is equal to the backward force pushing back out of the exhaust gas (Newton’s third law). So we can calculate the force of the rocket/engine by calculating the backward force of the exhaust gas.

In 1 second, 5.0 Kg of fuel increases it velocity (gas) from 0 m/s to 5000 m/s

we assume that rocket is using 5.0 Kg of fuel and 5000 m/s remains constant.

d = delta/change in .....

P = momentum

m = mass of the gas

v = final velocity

u = initial velocity

(* = multiply)

change in momentum = mass * change in velocity

dP = mv - mu (we can factorise)

dP = m (v - u)

dP = m * dv (change in velocity)

dP = 5 * 5000 (5000 - 0)

dP = 25000 kg m/s

We know that

impulse = change in momentum

J            = dP

Ft          = dP

rearrange to get F

F  = dP/t

F = ?

dP = 25000 kg m/s

t = 1 s (every second 5 kg of fuel is ejected at 5000 m/s - refer to previous question)

F = 25000 / 1

F = 25000 N

Backward force (exhaust gas) = Forward force (engine/rocket)                  (Newton’s third law)

25000 N  = 25000 N

m’ = mass rocket

v’ = final velocity of rocket

u’  = initial velocity of rocket

The rocket basic mass does not change - dm’ = 0 kg so

using the formula

impulse = dP

Ft = dP

Ft = m’v’ - m’u’ (factorise)

Ft = m’ (v’ - u’)

Ft = m’ * dv’ (change in rocket‘s velocity)

F = 25000 N

t = 1 s            

m’ = 10,000  (basic mass remains constant)

dv’ = ?

(25000)(1) = (10000)(dv’)

25000/10000 = dv’

2.5 = dv’

so increase in velocity is equal to 2.5 m/s

dv’ = 2.5 m/s

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