Can someone help me please - Velocity and GPE and Forces

In summary: So can someone tell me if what I'm thinking makes any sense and if not can someone explain it to me in simpler terms?In summary, the velocity at the time of impact was much higher than the average because the rocket was freefalling and the velocity increased with time. The actual gravitational potential energy will in reality be much higher than this as the rocket had more momentum due to the high initial velocity. Increasing the time of impact would mean the rocket was less likely to get damaged upon impact. The force exerted by the ground on the rocket would be the same exerted by the ground but in the opposite direction.
  • #1
AdeptDz
8
0

Homework Statement



Explain why the velocity at time of impact was much higher than the average velocity

Explain why the actual gravitational potential energy will in reality be much higher than this

Explain why increasing the time of impact would mean the rocket was less likely to get damaged upon impact

Force = Change in momentum/time taken for change

Homework Equations

The Attempt at a Solution



[/B]
Explain why the velocity at time of impact was much higher than the average velocity
The velocity at the time of impact was higher than the average because

Explain why the actual gravitational potential energy will in reality be much higher than this
The GPE will be much higher in reality as there is

Explain why increasing the time of impact would mean the rocket was less likely to get damaged upon impact
Force = Change in momentum/time taken for change
Increasing the time of impact (time taken for change) increases the time taken to change the momentum which reduces the forces. It will take a lot of force to damage the rocket, so reducing the force will damage the rocket less.
 
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  • #2
Hello Dz, :welcome:

Here at PF we will try to help you as best we can, but you have to make it possible for us to do so. There are some guidelines that can help you post successful threads. (here: re-read your writing before you post and try to imagine what someone who tries to help you needs).

In both your threads the problem statement is incomplete and rewriting in bold face doesn't make it clearer to us helpers.

For this thread:
What on Earth (or in space :smile:) is this about? The full story (literally if you can't come up with a concise and complete abstract)
 
  • #3
AdeptDz said:
Explain why the velocity at time of impact was much higher than the average velocity

How does the velocity of a free falling object vary with respect to time? It may help to sketch a rough velocity versus time graph.
 
  • #4
Sorry about the first place

You are a Physicist who works for NASA and your job is to evaluate the crash landing of a test rocket. Complete a summary assessment of the rocket’s descent and landing using the data below. You will then need to write a formal report. - Task (I've done most of it)

Mass of the Rocket = 1500kg
Duration of landing impact = 0.04 s
Velocity at time of impact = 175 m/s
Distance traveled = 30km
Time taken to complete journey = 6 minutesAcceleration due to gravity = 9.8 m/s2


Calculations

Average velocity of the rocket’s journey (m/s)
Average velocity = Initial velocity+Final velocity/2
Final velocity = 175 m/s
Initial velocity = u=v-at → u=175-(9.8*360) = -3353
Average velocity = -3353+175/2 = -1589m/s
Momentum of the rocket just before landing (kg m/s)
Momentum = mass * velocity
Distance = 30km (30000m) Time = 6 minutes (360 seconds)
Velocity = Distance/Time Velocity = 30000/360 = 83.333..m/s
1500kg * 83.333..m/s = 125,000kg m/s
Kinetic energy of the rocket just before landing (J)
0.5 x mv2
0.5*(1500*83.333..)2 = 7812500000 J
Gravitational potential energy at start of descent (J)
GPE = mgh
1500 * 9.8 * 30000 = 441000000 J
Force exerted by the ground on the rocket (N)
F = mass*acceleration
Acceleration = change in velocity/time = Final Vel.-Initial Vel./0.04 = 175-(-3353)/0.04=84000m/s2
F = 1500*84000 = 126000000 N
Use the idea of interaction pairs to state what the force exerted by the rocket on the ground would be.
Equal and opposite ground reaction force is exerted by the ground on the rocket.
The force exerted by the rocket on the ground would be the same exerted by the ground but in the opposite direction. It would be 126000000 N.

Analysis and Evaluation
Explain why the velocity at time of impact was much higher than the average velocity
The velocity at the time of impact was higher than the average because May you help me here pleaseExplain why the actual gravitational potential energy will in reality be much higher than this
The GPE will be much higher in reality as there is May you also help me here please


Explain why increasing the time of impact would mean the rocket was less likely to get damaged upon impact

Force = Change in momentum/time taken for change
Increasing the time of impact (time taken for change) increases the time taken to change the momentum which reduces the forces. It will take a lot of force to damage the rocket, so reducing the force will damage the rocket less.
 
  • #5
David Lewis said:
How does the velocity of a free falling object vary with respect to time? It may help to sketch a rough velocity versus time graph.

Velocity increases as time increases right?

Sorry I'm kind of new to all this stuff, just starting physics a-levels and tbh I've forgotten most of GCSE physics and threw away all my books so all i have is the internet and i can't seem to find the answer.
 
  • #6
I'm going to assume motors were turned off and air resistance negligible.
We know the initial velocity at time = 0 s
We know final velocity at time = 360 s - 0.04 s
We know velocity at time = 360 s
Plot those 3 points and connect the dots.
 
  • #7
AdeptDz said:
0.5*(1500*83.333..)2 = 7812500000 J
0.5 * 1500 kg * (83.333 m/s)2 = 5.21 MJ
Note that all units of measure are stated, mass is not squared, the answer is rounded off to 3 significant figures, and a unit prefix (mega-) was chosen to make the number manageable.
 
  • #8
Was there a diagram provided with this problem?

How was the 30km distance measured?
 
  • #9
While calculating momentum you have calculated Velocity is distance / time. This formula can be applied when the velocity is uniform and there is no acceleration. But the falling rocket has an acceleration = 9.8 m/s^2 which you have used in the first part of the calculation.You have to make changes there.
 
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Likes AdeptDz
  • #10
There is still a lot of information missing. Usually a test flight with a rocket has at least two phases: a launch phase where the thing is propelled up until the fuel is exhausted, and a free fall phase that can end with a crash or go over into a slower descent with a parachute. The launch can be vertical (did the thing reach a height of 30 km ? 15 km ?) or at an angle (did it travel 30 km horizontally ?). Your GPE calculation suggests it went 30 km high, but: did it ?

You are clearly informed that the velocity at time of impact is 175 m/s. Why calculate momentum and kinetic energy "just before landing" on the basis of an average velocity of 83 m/s ? If the thing is really shot straight up the average velocity is zero: after these 6 minutes it's back where it started !

David Lewis said:
Plot those 3 points and connect the dots
The way I described the usual test rocket flight needs one more point (when the fuel is spent). Sure makes a difference :smile:.
 
  • #11
Thanks. Point well taken. For simplicity I assumed motors were turned off during the portion of the flight being analyzed. Whether my assumption is valid perhaps OP can confirm or deny for us.
 
  • #12
Sorry, ill get back to this thread next week and reply, but thanks for all the help, just getting a lot of physics homework recently so this task is not due until next wekk
I appreciate all the help.

And there is no diagram + ill fix the mistakes
 
  • #13
Without more info I think it's very difficult to answer the questions. For example you state that the time for the journey is 6 mins. So is that the time for the whole flight (launch to landing) or just the descent phase? You appear to have assumed the latter.
 
  • #14
All the information I've given is for the descent of the rocket, the descent took 6minutes, sorry if it wasn't clear
 
  • #15
And the "distance travelled" is that horizontal, vertical, or ?
 
  • #16
vertical
 

Related to Can someone help me please - Velocity and GPE and Forces

1. What is velocity and how is it related to forces and GPE?

Velocity is a measure of an object's speed and direction. It is related to forces through Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. GPE, or gravitational potential energy, is related to velocity through the equation GPE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

2. How does an increase in velocity affect an object's GPE?

An increase in velocity will not directly affect an object's GPE. However, if the object is moving horizontally, an increase in velocity will result in an increase in kinetic energy, which can be converted into GPE if the object is lifted to a higher height.

3. Can forces affect an object's velocity?

Yes, forces can affect an object's velocity. If a force is applied to an object, it will cause the object to accelerate and therefore change its velocity. This can be seen in everyday examples, such as pushing a car to make it move faster.

4. How is GPE related to potential energy?

GPE is a type of potential energy. Potential energy is the energy that an object has due to its position or state. GPE specifically refers to the potential energy an object has due to its position in a gravitational field.

5. What is the relationship between forces and GPE?

Forces and GPE are related through the concept of conservation of energy. When a force is applied to an object, it can do work on the object and change its GPE. However, the total energy of the system (including both the object and the force acting on it) will remain constant. This means that the increase in GPE will be equal to the work done by the force.

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