(a) A rocket is travelling out of the earth’s atmosphere. During flight it burns fuel at a rate of 80kg/s producing a constant thrust of 500kN which drives the rocket in a vertical trajectory. It has a mass of 3500Kg when it reaches a speed of 1220m/s. At this point in the rocket’s trajectory the gravitational pull of the earth is negligible and air resistance can be also be neglected.
(i) Determine the rocket’s velocity 10s after it has reached a speed of 1220m/s
(ii) What would the error in the velocity calculation have been if the mass of the rocket had been assumed constant over the 10s?
Let time t=0 when mass is 3500 and speed is 1220 m/s.
then
F(t)=500 kN
m(t)=3500-80t
a(t)=(500*1000/m(t))=acceleration
=>
(i)
v(10)=v(0)+∫(500000/(3500-80t))dt
for t=0 to 10.
=1220-500000[ln(3500-80t)]/80
=1220+1621.945
=2841.945 m/s
(ii) using initial mass
v(10)=v(0)+(500000/3500*10)
=1220+1428.57
=2648.57 m/s
So error of simplification = (2648.57-2841.945)/2941.945= -6.8%
To determine the rocket's velocity 10 seconds after it has reached a speed of 1220 m/s, we need to consider the rocket's acceleration and how it changes over time.
1) Determine the acceleration of the rocket:
Acceleration is given by Newton's second law of motion: F = ma, where F is the force acting on the rocket and m is its mass. The force on the rocket is the thrust minus the weight: F = T - mg, where T is the thrust and g is the acceleration due to gravity. Since the gravitational pull of the Earth is negligible, the weight can be neglected.
T = 500 kN = 500,000 N
m = 3500 kg
Using F = ma, we can calculate the acceleration:
a = F / m
= 500,000 N / 3500 kg
= 142.86 m/s^2
2) Calculate the change in velocity over 10 seconds:
The change in velocity (Δv) is given by the equation:
Δv = a * t
where t is the time interval. In this case, t = 10 s.
Δv = 142.86 m/s^2 * 10 s
= 1428.6 m/s
3) Determine the rocket's velocity 10 seconds after reaching a speed of 1220 m/s:
The rocket's velocity at this point is 1220 m/s. To find the final velocity (v_f), we add the change in velocity (Δv) to the initial velocity (v_i):
v_f = v_i + Δv
= 1220 m/s + 1428.6 m/s
= 2648.6 m/s
Therefore, the rocket's velocity 10 seconds after reaching a speed of 1220 m/s is 2648.6 m/s.
Now, let's move on to part (ii) of the question:
If the mass of the rocket had been assumed constant over the 10-second interval, there would be an error in the velocity calculation.
When the rocket burns fuel, its mass decreases over time. Consequently, the force produced by the engine (thrust) also changes. Accounting for this changing mass and thrust is crucial in accurately calculating the rocket's velocity.
Assuming the rocket's mass remains constant would lead to an incorrect value for the force acting on the rocket (F). This, in turn, would result in an inaccurate acceleration calculation and hence, an error in the velocity calculation.
To avoid such errors, it is essential to consider the changing mass and thrust of the rocket when calculating its velocity over a given time interval.