Physics 12 momentum and vector question

[gdhillon]

A 40000N truck moving West at a velocity of 8.0 m/s collides with a 30000N truck
moving South at a velocity of 5.0 m/s. If these two vehicles lock together upon impact, what
is the initial velocity of the vehicles after the collision? I first found the momentum's of both trucks p=30000(5)=15306 and p=40000(8)=32653
I then added these vectors tip to tail and using Pythagoras a^2+b^2=c^2...15305^2+35653^2=c^2 ...c=36061 then divide that by the total force of both trucks and get .52m/s @ 25degrees S oW. Now that I think of it should i have divided by both vehicles momentums? which was 4759kg*m/s

 

[gneill]

Please follow the posting template.

A 40000N truck moving West at a velocity of 8.0 m/s collides with a 30000N truck
moving South at a velocity of 5.0 m/s. If these two vehicles lock together upon impact, what
is the initial velocity of the vehicles after the collision?


I first found the momentum's of both trucks p=30000(5)=15306 and p=40000(8)=32653

30000 x 5 does not give 15306. So you must have done something else first... hint: they gave you the weights of the trucks, not the masses. Momentum involves mass and velocity. If you state that something is a momentum value, it should have the appropriate units associated with it.

Also, momentum has both magnitude and direction (it's a vector quantity). What directions are associated with your momenta?

I then added these vectors tip to tail and using Pythagoras a^2+b^2=c^2...15305^2+35653^2=c^2 ...c=36061 then divide that by the total force of both trucks and get .52m/s @ 25degrees S oW.

What is "the total force of both trucks"? What is the "force" of a truck? Do you mean the weight?
You should be using masses here rather than weights (even though the conversion constants will fortunately cancel out in the math that ensues, it's proper form to use the correct quantities in the expressions).

Your result of 0.52 m/s for the magnitude of the final velocity does not look right.

Now that I think of it should i have divided by both vehicles momentums? which was 4759kg*m/s

How did you arrive at that value for "both vehicles momentums"? What direction is associated with it, since it should be a vector quantity?

 

[gdhillon]

O okay, I did convert the weights to masses I just forgot to right that down in my notes, so yes that's 3061.22KG(5m/s) which is 15306. Sorry I left that out.

I am not to sure what you mean, I put them tip to tail. The 4081KG truck is going west at a momentum of 32653N*s then the 3061KG truck is moving south at a momentum of 15305N*s. So then by Pythagoras c^2=a^2+b^2...so that would be 32653^2+15305^2=Sqrrt(c)=36061N*s. which would be the momentum of the vehicles so to find the initial velocity I would divide that number by both masses of the trucks combines so 36061/7142.963=5.0m/s Hmm that's a slightly different velocity than I had before.
Okay now to find the angle its invrsetan(3061/4081.63)=25 degrees So 5.0m/s @37 degrees Sof W...how does that sound?

 

[gneill]

O okay, I did convert the weights to masses I just forgot to right that down in my notes, so yes that's 3061.22KG(5m/s) which is 15306. Sorry I left that out.

I am not to sure what you mean, I put them tip to tail. The 4081KG truck is going west at a momentum of 32653N*s then the 3061KG truck is moving south at a momentum of 15305N*s. So then by Pythagoras c^2=a^2+b^2...so that would be 32653^2+15305^2=Sqrrt(c)=36061N*s. which would be the momentum of the vehicles so to find the initial velocity I would divide that number by both masses of the trucks combines so 36061/7142.963=5.0m/s Hmm that's a slightly different velocity than I had before.
Okay now to find the angle its invrsetan(3061/4081.63)=25 degrees So 5.0m/s @37 degrees Sof W...how does that sound?

Okay, much of that looks much better. The magnitude of the final velocity looks better. However, I'm not seeing how you're calculating the resultant's angle. Where are the numbers 3061 and 4081.63 coming from ? They look more like the masses of the trucks in kg rather than components of the momentum vector.

 

[Nickie]

Great job on using the concept of momentum and vector addition to solve this problem! Your method of adding the momentums of both trucks is correct, as momentum is a vector quantity and can be added using vector addition. However, in order to find the initial velocity after the collision, you will need to divide the total momentum by the total mass of both trucks, not just the total force. This will give you the final velocity of the combined trucks after the collision, taking into account the mass of both trucks. So, the final calculation would be (15306 + 32653)/(30000 + 40000) = 0.80 m/s @ 25 degrees S oW.

Also, it is important to note that the direction of the final velocity will depend on the direction of the initial velocities of the trucks. In this case, since the initial velocities are perpendicular to each other, the final velocity will also be at a 90 degree angle to both initial velocities. Therefore, the final velocity will be at a direction of 25 degrees S oW, as you calculated.

Overall, your approach was correct, but just remember to divide by the total mass instead of the total force to get the final velocity after the collision. Keep up the good work!