# Conservation of momentum question. (which is the correct method)

1. Apr 9, 2013

### chestycougth

I am trying to revise this subject but I am unsure which of the two method's I've used is the correct one. Can someone help me?

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

two trucks travel towards each other. truck A has a velocity of 10 m/s and truck B has a velocity of 3 m/s. Truck A has a mass of 1000 kg and truck B has a mss of 1500 kg. Find the velocity of B after the collision if A has a velocity of 1 m/s afterwards in the same direction has before the collision.

2. relevant equations

total momentum before collision = total momentum after collision
p = mv

3. The attempt at a solution

method 1

total momentum = (1000 * 10) + (1500 * 3) = 14500 kg m/s

momentum of A after collision = 1000 * 1 = 1000 kg m/s

momentum of B after collision = 14500 - 1000 = 13500 kg m/s

velocity of B:
v = p/m
v = 13500 / 1500 = 9 m/s

method 2

1 = (1000 * 10) - 1500v
1500v = 1000 * 10
v = 1000 *10 / 1500 = 6.6 m/s

Thanks.

2. Apr 9, 2013

### rude man

Momentum has sign as well as magnitude!
Both answers are wrong.

3. Apr 9, 2013

### climb515c

Draw a picture of what the problem states, and assign a direction for +x. You'll see what you missed the first time around.

4. Apr 9, 2013

### chestycougth

I don't understand what you mean by +x. Do you mean I need to give a direction to the velocity?

5. Apr 9, 2013

### climb515c

Yes, you do have to.

If truck A is going 10m/s to the right (positive), Truck B is going 3 m/s to the left (negative), or -3m/s. Since the question stated they are going towards each other.

Its best to draw a picture, then label the diagram with a direction you've decided to call "positive". Sometimes the problem already tells you which way is positive, but in this case it didn't.

Last edited: Apr 9, 2013
6. Apr 9, 2013

### chestycougth

Will either method give me a correct answer if I make sure to take the direction into account?

I got an answer of 3 m/s for method 1 and -6.6 m/s for method 2. I don't see how it can be method 2 since truck A has a positive velocity and truck B would have had to go through truck A in order to still have a negative velocity after the collision. However, method 2 is the way I was shown to use, so now i'm a little confused.

7. Apr 9, 2013

### climb515c

Method 1 of yours looks more right. Pi = Pf

therefore, (miA*viA) + (miB*viB) = (mfA*vfA)+(mfB*vfB)

Your only unknown is vfB

Just make sure your initial signs are all correct, thats when drawing a diagram helps.

Last edited: Apr 9, 2013