Collision accident/involves momemntum

[Acestein]

Envision a minor motor-vehicle accident. Car X backs out of a parking space at 1.000m/s toward the east. Car Y, whose driver searches for a place to park, travels north at 1.000m/s. Neither driver sees the other car, and the cars collide. Suppose that each car (including its driver) has a mass of 1000kg. The total system momentum vector before the collision is approximately?

a) 1000kg x m/s toward the northeast.
b) 1414 kg x m/s toward the northeast.
c) 2000kg x m/s toward the northeast.
d) zero, because the vehicles haven't hit each other yet!

My textbook has been teaching me with formulas of momentum. However, when i attempted it by finding the momenutm of each object and then adding them together i got 2000kg x m/s toward the northeast. What am i doing wrong? some people have told me to use the triangle method, but i haven't been taught that way yet.

 

[PhanthomJay]

This is a vector addition problem...how did you get the direction? The magnitude of the resultant initial momentum vector is the magnitude of longest side of the triangle formed by representing the legs as the magnitudes and directions of the individual initial momentum vectors.

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

 

[Acestein]

i used a formula involving:

mx = 1000kg
vx = 1.000m/s going east.

my = 1000kg
vy = 1.000m/s going north.

the momentum of Px = 1000kg x 1.000m/s going east
= 1000kg x m/s going east

the momentum of Py = 1000kg x 1.000m/s going north
= 1000kg x m/s going north


and their sum of momentums is P=Px+Py

= 2000kg x m/s going north east

what am i doing wrong?

 

[PhanthomJay]

i used a formula involving:

mx = 1000kg
vx = 1.000m/s going east.

my = 1000kg
vy = 1.000m/s going north.

the momentum of Px = 1000kg x 1.000m/s going east
= 1000kg x m/s going east

the momentum of Py = 1000kg x 1.000m/s going north
= 1000kg x m/s going north


and their sum of momentums is P=Px+Py

= 2000kg x m/s going north east

what am i doing wrong?

You can't just add the numbers algebrically and take the average direction. You need to look at the horizontal and vertical directions separately, then combine them in accordance with the vector addition laws. Take one vector and sketch it on a sheet of paper. Then place the tail of the 2nd vector on the arrow of the first vector, and draw it on the paper in its proper direction. The resultant is then determined by connecting the tail of the first vector to the arrow of the 2nd vector. You will see in your problem that it is the hypotenuse of a right triangle. Use basic pythagorus and trig to determine its magnitude and direction. See the examples on the site to which I referred you.

 

[Acestein]

Thanks a lot. I finally understand. I'm just a bit confused still on what the hypotenuse of the triangle represents. Is that the momentum before the collision has occurred?

 

[PhanthomJay]

Thanks a lot. I finally understand. I'm just a bit confused still on what the hypotenuse of the triangle represents. Is that the momentum before the collision has occurred?

Yes, that's the initial momentum of the system before the collision takes place ( the total initial momentum of the system (2 cars) is the vector sum of each objects initial momentum).