# What is the final velocity and direction of motion?

1. May 14, 2008

### looi76

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$$Force = \frac{Change \ in \ momentum}{Time}$$

3. The attempt at a solution

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

$$Change \ in \ momentum = Force \times Time$$

$$mv - mu = F \times t$$

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

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

$$v = \frac{8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{ 2000000}$$

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

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

2. May 14, 2008

### Hootenanny

Staff Emeritus
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. May 14, 2008

### looi76

$$\mbox{Change in momentum = Final momentum - Initial momentum}$$

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

$$= mv - mu$$

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

Is this right?

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

4. May 14, 2008

### Hootenanny

Staff Emeritus
I don't think you understand what I'm getting at. In reference to my previous post,
This means that,
Should actually be F = -8.0x108 N.

Do you follow?

5. May 14, 2008

### looi76

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. May 14, 2008

### Hootenanny

Staff Emeritus
looi76, you need to re-work your solution with the negative value for the force. You're answer is wrong as it stands.

7. May 14, 2008

### looi76

Thnx Hootenanny,

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

$$Change \ in \ momentum = Force \times Time$$

$$mv - mu = F \times t$$

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

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

$$v = \frac{-8.0 \times 10^8 \times 10 + 2000000 \times 3.0 \times 10^4}{ 2000000}$$

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

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

So, the spaceship will move in the same direction?

8. May 14, 2008

### Hootenanny

Staff Emeritus