2D Glancing Collision Problem

In summary, the problem involves two cars, one traveling due North and the other due West, colliding with each other. After the collision, the first car has a mass of 765kg and travels at 70km/h at an angle of 57 degrees North of West, while the second car has a mass of 1100kg and travels at 41km/h at an angle of 44 degrees North of West. To solve for the velocities of each car before the impact, the vectors must be resolved into x and y components and then combined to find the final momentum of each car.
  • #1
cybernerd
27
0

Homework Statement


Car two is traveling due North and collides with Car one, which is traveling due west. Car One has a mass of 765kg and after the collision it travels at 70km/h, 57 degrees North of West. Car two has a mass of 1100kg and after the collision it travels at 41km/h, 44 degrees North of West. What is the velocity of each car right before impact?

So

Car One - m1 = 765kg, v1f = 70km/h

Car Two - m2 = 1100kg, v2f = 41km/h


Homework Equations



Pyi = Pyf
Pxi = Pxf
P = mv
SINE LAW


The Attempt at a Solution



So, I started by constructing a vector triangle for each car, and using the sine law to determine the other sides.

CAR ONE TRIANGLE

Where one side is the west (x plane/ P1) momentum and one side is the north (y plane/P2) momentum. The hypotenuse (p12) is the final momentum of car one after the collision.

P12 = mv
=(765kg)(70km/h)
=53550 kgxkm/h

cos = adjacent/hypotenuse
= cos44 = P1/53550
P1 = 38520.64...kgxkm/h

sine = opposite/hypotenuse
=sine44 = P2/53550
P2 = 37198.95...kgxkm/h

CAR TWO TRIANGLE

Much the same as the first triangle, except the hypotenuse of this triangle represents the final momentum of the second car.

P12 = mv
= (1100kg)(41km/h)
45100kgxkm/h

cos57 = P1/45100
P1 = 24563.22...kgxkm/h

sine57 = P2/45100
P2 = 37824.04...kgxkm/h

Okay, up to this point, I'm fairly confident that I'm on the right track. It's right here where I'm getting confused. I know I need to add the vector components, but I'm not entirely certain how.

So...

Pyi = Pyf

Py1 = P2 + P2
=37198.95...kgxkm/h + 37824.04...kg x km/h
= 75022.99...kgxkm/h

Pyf = final momentum of car two
= 45100 kgxkm/h

...They do not match. I have accomplished nothing.
My teacher assures me I am using the correct formula, but clearly, I am using it incorrectly. What am I doing wrong?
 
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  • #2
I would rather see you resolve the final mass*velocity vectors (momentum vectors) into x and y components and then combine the vertical components of each and recognize that those components belonged to the vertically traveling car before the collision and all the horizontal momentum must have come from the horizontally traveling car.
 
  • #3


I can see that you have correctly used the Pyi = Pyf equation to determine the final momentum of car two. However, there seems to be an error in your calculation of the final momentum of car one. Instead of adding P2 + P2, you should be adding P1 + P2 to get the final momentum in the y-direction. This would give you a final momentum of 75784.27 kgxkm/h for car one, which is closer to the final momentum of car two (45100 kgxkm/h).

Additionally, in your calculation of the final momentum for car two, you used the value of 53550 kgxkm/h, which is the initial momentum of car one. The final momentum of car two should be the same as the initial momentum of car one, so you should use 53550 kgxkm/h instead of 45100 kgxkm/h.

I would also recommend using consistent units throughout your calculations. In this case, it would be best to use kilometers per hour for velocity and kilograms for mass.

Overall, it seems like you have a good understanding of the concepts and equations involved in this problem. I would suggest double-checking your calculations and units to ensure accuracy.
 

1. What is a 2D glancing collision problem?

A 2D glancing collision problem is a type of physics problem that involves two objects colliding at an angle in a two-dimensional space. It is commonly used to analyze the motion and energy transfer of objects during a collision.

2. How is momentum conserved in a 2D glancing collision?

In a 2D glancing collision, momentum is conserved in both the x and y directions. This means that the total momentum before the collision is equal to the total momentum after the collision. This is known as the law of conservation of momentum.

3. What factors affect the outcome of a 2D glancing collision?

The outcome of a 2D glancing collision is affected by factors such as the masses and velocities of the objects involved, as well as the angle of collision. The elasticity of the objects and the presence of external forces can also impact the outcome.

4. How do you calculate the final velocities of the objects after a 2D glancing collision?

The final velocities of the objects after a 2D glancing collision can be calculated using the conservation of momentum and the conservation of kinetic energy equations. These equations take into account the masses, initial velocities, and angle of collision of the objects.

5. What is the difference between a perfectly elastic and an inelastic 2D glancing collision?

In a perfectly elastic collision, the total kinetic energy of the objects before the collision is equal to the total kinetic energy after the collision. In an inelastic collision, some of the kinetic energy is lost during the collision, resulting in a decrease in the total kinetic energy. This can be seen in the final velocities of the objects, as they will be higher in a perfectly elastic collision compared to an inelastic one.

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