What is the final velocity and direction of motion?

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Homework Help Overview

The problem involves a spaceship with a mass of 2,000,000 kg moving at a constant velocity of 3.0 x 10^4 m/s, which is subjected to a force of 8.0 x 10^10 N acting in the opposite direction for 10 seconds. Participants are discussing the final velocity and direction of motion after this force is applied.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between force, momentum, and velocity, with some attempting to calculate the change in momentum and final velocity using the impulse-momentum theorem. Questions arise regarding the sign of the force and its implications for the direction of motion.

Discussion Status

There is an ongoing examination of the calculations and assumptions regarding the direction of the force and its effect on the spaceship's velocity. Some participants suggest re-evaluating the initial conditions and the signs used in the calculations, while others provide feedback on the methods employed.

Contextual Notes

Participants are navigating the implications of choosing positive or negative values for velocity and force, which affects the interpretation of the final motion direction. There is a recognition of the need to clarify these assumptions in the context of the problem.

looi76
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Homework Statement


A spaceship of mass 2000000kg is moving with a constant velocity of 3.0 \times 10^4 ms^{-1}. It is suddenly hit by a mysterious force of 8.0 \times 10^{10}N for 10s in the opposite direction to its motion. What is the final velocity and direction of motion?

Homework Equations


Force = \frac{Change \ in \ momentum}{Time}

The Attempt at a Solution



m = 2000000kg \ , \ u = 3.0 \times 10^4 ms^{-1} \ , \ F = 8.0 \times 10^8 \ , \ t = 10s

Change \ in \ momentum = Force \times Time

mv - mu = F \times t

2000000 \times v - 2000000 \times 3.0 \times 10^4 = 8.0 \times 10^8 \times 10

2000000v = 8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4

v = \frac{8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{<br /> 2000000}

v = 3.4 \times 10^4 ms^{-1}

and the direction of motion is in the opposite direction because momentum after collision is greater then momentum before collision.

Are my answers correct?
 
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Notice that the force acts in the opposite direction to the motion. Therefore, if you chose your initial velocity to be positive, then the force must be negative.
 
\mbox{Change in momentum = Final momentum - Initial momentum}

m = 2000000kg \ , \ v = 3.4 \times 10^4 ms^{-1} \ , \ u = 3.0 \times 10^4 ms^{-1}

= mv - mu

= 2000000 \times 3.4 \times 10^4 - 2000000 \times (-3.0 \times 10^4)
= 1.28 \times 10^{11} N

Is this right?

I'm getting a positive momentum, so it is in the same direction?
 
looi76 said:
\mbox{Change in momentum = Final momentum - Initial momentum}

m = 2000000kg \ , \ v = 3.4 \times 10^4 ms^{-1} \ , \ u = 3.0 \times 10^4 ms^{-1}

= mv - mu

= 2000000 \times 3.4 \times 10^4 - 2000000 \times (-3.0 \times 10^4)
= 1.28 \times 10^{11} N

Is this right?

I'm getting a positive momentum, so it is in the same direction?

I don't think you understand what I'm getting at. In reference to my previous post,
Hootenanny said:
Notice that the force acts in the opposite direction to the motion. Therefore, if you chose your initial velocity to be positive, then the force must be negative.
This means that,
looi76 said:
F = 8.0 \times 10^8
Should actually be F = -8.0x108 N.

Do you follow?
 
Understood, velocity should be negative because the force is negative because it is in the opposite direction. So, the spaceship will move in the same direction?
 
looi76 said:
Understood, velocity should be negative because the force is negative because it is in the opposite direction. So, the spaceship will move in the same direction?
looi76, you need to re-work your solution with the negative value for the force. You're answer is wrong as it stands.
 
Thnx Hootenanny,

m = 2000000kg \ , \ u = 3.0 \times 10^4 ms^{-1} \ , \ F = -8.0 \times 10^8 \ , \ t = 10s

Change \ in \ momentum = Force \times Time

mv - mu = F \times t

2000000 \times v - 2000000 \times 3.0 \times 10^4 = -8.0 \times 10^8 \times 10

2000000v = -8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4

v = \frac{-8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{<br /> 2000000}

v = 2.6 \times 10^4 ms^{-1}

Change \ in \ momentum = mv-mu
= 2000000 \times 2.6 \times 10^4 - 2000000 \times -(3.0 \times 10^4)
= 1.1 \times 10^{11} N

So, the spaceship will move in the same direction?
 
looi76 said:
m = 2000000kg \ , \ u = 3.0 \times 10^4 ms^{-1} \ , \ F = -8.0 \times 10^8 \ , \ t = 10s

Change \ in \ momentum = Force \times Time

mv - mu = F \times t

2000000 \times v - 2000000 \times 3.0 \times 10^4 = -8.0 \times 10^8 \times 10

2000000v = -8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4

v = \frac{-8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{<br /> 2000000}

v = 2.6 \times 10^4 ms^{-1}
I've not checked your arithmetic, but your method looks better :approve:
looi76 said:
Change \ in \ momentum = mv-mu
= 2000000 \times 2.6 \times 10^4 - 2000000 \times {\color{red}-}(3.0 \times 10^4)
= 1.1 \times 10^{11} N
I don't know why your calculating this, the final velocity gives you all the information you require. Eitherway, the intial velocity should be positive (you need to remove the highlighted negative sign). Note that your change in momentum should be identical to the impulse.
 

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