What is the final velocity and direction of motion?

In summary, 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. The final velocity and direction of motion is 3.4 \times 10^4 ms^{-1} in the opposite direction.f
  • #1
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Homework Statement


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

Homework Equations


[tex]Force = \frac{Change \ in \ momentum}{Time}[/tex]

The Attempt at a Solution



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

[tex]Change \ in \ momentum = Force \times Time[/tex]

[tex]mv - mu = F \times t[/tex]

[tex]2000000 \times v - 2000000 \times 3.0 \times 10^4 = 8.0 \times 10^8 \times 10[/tex]

[tex]2000000v = 8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4[/tex]

[tex]v = \frac{8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{
2000000}[/tex]

[tex]v = 3.4 \times 10^4 ms^{-1}[/tex]

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

Are my answers correct?
 
  • #2
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.
 
  • #3
[tex]\mbox{Change in momentum = Final momentum - Initial momentum}[/tex]

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

[tex]= mv - mu[/tex]

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

Is this right?

I'm getting a positive momentum, so it is in the same direction?
 
  • #4
[tex]\mbox{Change in momentum = Final momentum - Initial momentum}[/tex]

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

[tex]= mv - mu[/tex]

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

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,
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,
[tex]F = 8.0 \times 10^8[/tex]
Should actually be F = -8.0x108 N.

Do you follow?
 
  • #5
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?
 
  • #6
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.
 
  • #7
Thnx Hootenanny,

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

[tex]Change \ in \ momentum = Force \times Time[/tex]

[tex]mv - mu = F \times t[/tex]

[tex]2000000 \times v - 2000000 \times 3.0 \times 10^4 = -8.0 \times 10^8 \times 10[/tex]

[tex]2000000v = -8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4[/tex]

[tex]v = \frac{-8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{
2000000}[/tex]

[tex]v = 2.6 \times 10^4 ms^{-1}[/tex]

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

So, the spaceship will move in the same direction?
 
  • #8
[tex]m = 2000000kg \ , \ u = 3.0 \times 10^4 ms^{-1} \ , \ F = -8.0 \times 10^8 \ , \ t = 10s[/tex]

[tex]Change \ in \ momentum = Force \times Time[/tex]

[tex]mv - mu = F \times t[/tex]

[tex]2000000 \times v - 2000000 \times 3.0 \times 10^4 = -8.0 \times 10^8 \times 10[/tex]

[tex]2000000v = -8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4[/tex]

[tex]v = \frac{-8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{
2000000}[/tex]

[tex]v = 2.6 \times 10^4 ms^{-1}[/tex]
I've not checked your arithmetic, but your method looks better :approve:
[tex]Change \ in \ momentum = mv-mu[/tex]
[tex]= 2000000 \times 2.6 \times 10^4 - 2000000 \times {\color{red}-}(3.0 \times 10^4)[/tex]
[tex]= 1.1 \times 10^{11} N[/tex]
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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