A rocket of mass 10,000 kg uses 5.0 kg of fuel and oxygen to produce exhaust gases ejected at 5000 m/s. Calculate the increase in its velocity.
Answers
"A rocket of mass 10000 kg uses 5.0 kg of fuel and oxygen to produce exhaust gas ejected at 5000 m/s. Calculate the increase in its velocity."
ΔV = Vex * ln(initial mass / final mass)
ΔV = 5000m/s * ln(10000kg / 9995kg) = 2.5 m/s
Answer: 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 veloci (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 factorize)
dP = m (v - u)
dP = m * dv (change in velocity)
dP = 5 * 5000 (5000 - 0)
dP = 25000 kg m/s
We know that
impusle = 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
Bacward 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’ (factorize)
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
That’s it. Hope you founded it helpful and easier to understand than other complex and unexplained answers.