Question:
A model rocket of mass 0.350 kg is launched vertically. The rocket has an engine that is ignited at time t
= 0, as shown below, and the engine fires for 2.50 s, providing a thrust of 14.6 N. When the rocket
reaches its maximum height, a parachute is deployed, and the rocket then descends vertically to the
ground.
a.
On the figures below, draw and label a free-body diagram for the rocket during each of the
following intervals:
(i) while the engine
is firing
(ii)
after engine shuts down, but
before the parachute is
deployed
(iii)
the moment the
parachute is deployed
b.
Determine the magnitude of the average acceleration of the rocket while the engine is firing.
c.
Determine the velocity of the rocket at the end of the period when the engine fires.
d.
A person stands on a bathroom scale in a full-sized rocket that is being launched vertically.
Consider the time interval when the rocket engine is firing. Is the scale reading during this time interval
less than, equal to, or greater than the scale reading when the rocket is at rest? Justify your answer.
Answer:
a.
(i) While the engine is firing, there are two forces acting on the rocket: the upward thrust force of
the engine and the downward weight of the rocket.
(ii) After the engine shuts down, but before the parachute is deployed, there is only one force acting on
the rocket: the downward weight of the rocket.
(iii) The moment the parachute is deployed, there are two forces acting on the rocket: the upward
parachute force and the downward weight of the rocket. At first, the parachute slows the descent of the
rocket, so Fparachute is initially greater than Fg.
The free-body diagrams for the rocket during the three intervals are as follows.
(The vector for the weight, Fg, also could be labeled mg.)
b.
Using Newton’s second law:
First, find the net force:
Fnet = Fengine – Fg
Given
Fengine = 14.6 N
m = 0.350 kg
Fnet = Fengine – mg
c.
Use an equation of motion:
vf – vi = at, where a is the average acceleration determined in part b.
Using substitution
vf = vi + at
vf = vi + a(tf – ti)
Given
a = 31.9 m/s2
vi = 0 m/s (The rocket is launched from rest.)
tf = 2.50 s
ti = 0 s
Then
vf = 0 m/s + (31.9 m/s2)(2.50 s – 0 s)
vf = 79.8 m/s
d. While the rocket engine is firing, the rocket is accelerating upward. Consider the forces on the person.
The scale reading is equal to the upward force exerted by the scale, Fscale.
Fnet = ma
Fnet = Fscale – Fg
Fscale = Fnet + Fg
When the rocket is at rest, a = 0, so Fnet = 0.
In this case, the scale reading shows the true weight of the person:
(Fscale)at rest = Fg = mg
When the rocket is accelerating upward, a  0.
In this case, the scale reading shows the apparent weight of the person:
Fscale = Fnet + Fg
(Fscale)accelerating = ma + mg
Hence
(Fscale)accelerating  (Fscale)at rest
The scale reading while the engine is firing is greater than the scale reading when the rocket is at rest.
Question:
An astronaut is preparing calculations for a flight to the Moon. The combined mass of the crew, all
equipment, fuel, and the rocket is 2.8106 kg on the launch pad.
a.
The rocket’s engines produce a combined 35106 N of thrust. Is this enough to lift the rocket and
its payload? What if the engines produced 30106 N of thrust? Would 25106 N be enough?
b.
What is the minimum force the rocket needs to exert in order to lift off the ground and accelerate
upward?
c.
To return to Earth, the crew must take off from the Moon, where gravity is one-sixth that of
Earth. Since most of the mass of the rocket is fuel, assume that three-fourths of the mass was lost in the
journey to the Moon. How much force does the rocket need to exert in order to lift off the surface of the
Moon?
Answer:
a.
To get off the launch pad, the acceleration upward must be greater than the acceleration
downward due to the pull of gravity, namely
athrust = agravity = 9.80 m/s2.
Since the acceleration due to the forces of 35106 N and 30106 N is greater than g, either of these
forces are sufficient to lift the rocket off the ground. However, the acceleration due to a force of 25106
N results in a net acceleration downward and does not move the rocket and its payload in a positive
(upward) direction.
b.
To balance the acceleration of gravity, the acceleration of the rocket should be equal and
opposite, a = 19.80 m/s2.
F = ma
= (2.8106 kg)(9.80 m/s2)
= 27106 N
c.
First, find the acceleration due to gravity on the Moon:
Next, calculate the mass of the rocket on the Moon after it has lost three-fourths of its mass due to fuel
consumption:
To balance the acceleration of gravity, the acceleration of the rocket should be equal and opposite,
a = +1.63 m/s2.
F = ma
= (7.0105 kg)(1.63 m/s2)
= 1.1106 N