Spaceship hit the ground with a velocity of 100 m/s

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Discussion Overview

The discussion centers around the physics of a spaceship impacting the ground at a velocity of 100 m/s, along with related concepts of force, mass, and acceleration. Participants explore the implications of these concepts in the context of both the spaceship and a ball thrown upwards, examining how forces and accelerations interact during motion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions how to apply F=ma to the impact force of the spaceship, suggesting that the velocity implies a greater force than calculated based on gravity alone.
  • Another participant clarifies that the negative sign in acceleration indicates direction, with upward being positive and downward negative, which applies to forces and velocities as well.
  • A different participant explains that when a ball is thrown upwards, it experiences a downward acceleration of -9.8 m/s², indicating it is slowing down until it reaches its highest point.
  • Some participants note that to determine the force of impact, one must know the change in velocity during the impact and the duration of that impact.
  • There is a suggestion that momentum (p=mv) may also be relevant to understanding the forces involved in these scenarios.
  • One participant emphasizes that if the mass of the spaceship were known, the force of impact could be calculated using F=ma, with the acceleration being the gravitational force.
  • Another participant elaborates on the relationship between velocity and acceleration, explaining that the upward velocity decreases due to the negative acceleration until it becomes zero at the peak height, after which it begins to increase downward.

Areas of Agreement / Disagreement

Participants express various viewpoints on the application of F=ma and the interpretation of acceleration and velocity. There is no consensus on the implications of these concepts for the spaceship's impact force, and multiple competing views remain regarding the relationship between velocity, acceleration, and force.

Contextual Notes

Participants acknowledge that additional information, such as the mass of the spaceship, is necessary to fully resolve the questions posed. The discussion also highlights the dependence on conventions for defining positive and negative directions in physics.

szekely
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Ok so I'm in 7th grade and I think I'm pretty good in science class. The other day we had some math questions involving some equations. One question was something like a spaceship hit the ground with a velocity of 100 m/s and then it asked like what was its mass. This question also included the gravity of the planet (like 7.8 m/s squared). The next question was what was the force of the impact. The answer was something low because the force of gravity was 7.8. I was wondering how F=ma would work here because the velocity makes it seem like the impact would have much more force than what it was.

Then i started thinking about if you threw a ball straight up. As soon as it left your hand the acceleration would be downward at 9.8 because of gravity. So the acceleration upward is -9.8 even though its still going up. I know that if it hits something while going up its going to have a force so i don't see how F=ma works if acceleration is negative.

Just today i started wondering if this was because of momentum and if you would somehow get p=mv into there.

Thanks for all of the responses. I know this should be extremely easy for you guys.
 
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in these kinds of problems the negative sign is only there to denote direction, with up being the positive direction, and down being the negative(which I am assuming is the convention being used for your class). This applies for forces, momentum, accelerations, velocities, etc.
 


When the ball leaves your hand, a=-9.8m/ss. So it is slowing down. If it hits something on the way up, it will slow down even more quickly, so its deceleration increases to a>-9.8m/ss. This is due to whatever it hit exerting a force on the ball. By Newton's third law, the ball will exert the a force of the same magnitude on whatever it hit. To find out the force, one has to know the change in velocity during the impact, and the duration of the impact.
 


szekely said:
Ok so I'm in 7th grade and I think I'm pretty good in science class. The other day we had some math questions involving some equations. One question was something like a spaceship hit the ground with a velocity of 100 m/s and then it asked like what was its mass.
There must have been more to this question, you do not have enough information to find the mass.
This question also included the gravity of the planet (like 7.8 m/s squared). The next question was what was the force of the impact. The answer was something low because the force of gravity was 7.8. I was wondering how F=ma would work here because the velocity makes it seem like the impact would have much more force than what it was.
If you knew the mass then the force of the impact would indeed be just [tex]F = ma[/tex] with the acceleration being that given.
Then i started thinking about if you threw a ball straight up. As soon as it left your hand the acceleration would be downward at 9.8 because of gravity. So the acceleration upward is -9.8 even though its still going up. I know that if it hits something while going up its going to have a force so i don't see how F=ma works if acceleration is negative.

Just today i started wondering if this was because of momentum and if you would somehow get p=mv into there.

Thanks for all of the responses. I know this should be extremely easy for you guys.

When you throw the ball up it has velocity of +V and acceleration of -g that means (on earth) 9.81 ms-2 down. Since it is accelerating downward its upward velocity is always getting smaller, it is slowing down. When the velocity reaches zero the ball has reached its highest point. It still has a acceleration of -g, so now its velocity begins to increase downward. As it falls in the negative direction (down) its velocity increases due to the acceleration in the negitive direction. When it reaches the point of release its downward speed will equal the speed at which it was released initially. I carefully used the word speed here, since the velocities are not equal, Vup = - Vdown

Does this help?
 

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