Conservation of Momentum, a Totally Inelastic Collision

[mvl46566]

Homework Statement

A 950kg compact car is moving with the velocity ⃗v1 = 32 ˆx + 17 ˆy m/s. It
skids on an icy, frictionless patch and collides with a 450kg hay wagon
moving with velocity ⃗v2 = 12ˆx +14ˆy m/s. If the two stay together,
what is their velocity?

Homework Equations

m1v1 (initial) + m2v2 (initial) = vf (m1+ m2)

The Attempt at a Solution

I realize this is a totally inelastic collision and I need to use the above equation to solve. However, I don't know how to change the vector components into a number that can be plugged into the formula. I have tried adding the vector components of both velocities (x+x, y+y) but the answers I am getting do not sound right. I tried doing this because I figured if I know the numbers for the masses I can factor them out and have the vectors being added to together, but I don't think this is the solution. Please help!

 

[rl.bhat]

Show your calculations. Try this one.
m1*v1x + m2*v2x = (m1+m2)*vfx.
Similarly for y components.

 

[kerimek]

I would approach this problem by first converting the given velocities into their corresponding components in the x and y directions. This can be done using basic trigonometry and the given magnitudes and angles. Once the velocities are in their component form, I would then use the conservation of momentum equation as stated in the homework to solve for the final velocity.

To start, I would label the compact car as object 1 and the hay wagon as object 2. Using the given masses and velocities, I would first calculate the momentum of each object in the x and y directions.

For object 1:
Px1 = (950kg)(32m/s) = 30400 kgm/s in the x-direction
Py1 = (950kg)(17m/s) = 16150 kgm/s in the y-direction

For object 2:
Px2 = (450kg)(12m/s) = 5400 kgm/s in the x-direction
Py2 = (450kg)(14m/s) = 6300 kgm/s in the y-direction

Next, I would add the momenta of each object in each direction to find the total momentum before the collision.

Px (initial) = Px1 + Px2 = 35800 kgm/s in the x-direction
Py (initial) = Py1 + Py2 = 22450 kgm/s in the y-direction

Since this is a totally inelastic collision, the two objects will stick together after the collision and move with a combined velocity. This means that the final momentum in each direction will be the same for both objects.

Using the conservation of momentum equation, I can set the initial momentum equal to the final momentum and solve for the final velocity.

(m1 + m2)vf = Px (initial) + Py (initial)
(950kg + 450kg)vf = 35800 kgm/s + 22450 kgm/s
1400kg vf = 58250 kgm/s

Solving for vf, we get:
vf = 41.61 m/s in the x-direction
vf = 41.61 m/s in the y-direction

Therefore, the final velocity of the combined objects after the collision is 41.61 m/s in the x-direction and 41.61 m/s in the y-direction. This can also be expressed in vector form as ⃗vf = 41.61 ˆx