2-D car collision, finding speed and angle

[clockworks204]

1. A 1000- kg car collides with a 1200- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 25.0 km/h in a direction of 20 o with respect to the positive x axis. The heavier car moves at 28 km/h at -44 o with respect to the positive x axis.
What was the initial speed of the lighter car (in km/h)? What was the initial direction (as measured counterclockwise from the x-axis)?



2. x-momentum= m1v1cos(angle1) + m2v2cos(angle2)
y-momentum= m1v1sin(angle1) + m2v2sin(angle2)
magnitude momentum= sqrt(Xtotal)^2 + (Ytotal)^2
= sqrt((m1v1cos(angle1) + m2v2cos(angle2))^2 + ((m1v1sin(angle1) + m2v2sin(angle2))^2




3. By plugging into this equation, I come up with 49602.6 m/s, but it's not right. I need to get the velocity correct before I can make an attempt to find the initial direction

 

[inky]

1. A 1000- kg car collides with a 1200- kg car that was initially at rest at the origin of an x-y coordinate system. After the collision, the lighter car moves at 25.0 km/h in a direction of 20 o with respect to the positive x axis. The heavier car moves at 28 km/h at -44 o with respect to the positive x axis.
What was the initial speed of the lighter car (in km/h)? What was the initial direction (as measured counterclockwise from the x-axis)?



2. x-momentum= m1v1cos(angle1) + m2v2cos(angle2)
y-momentum= m1v1sin(angle1) + m2v2sin(angle2)
magnitude momentum= sqrt(Xtotal)^2 + (Ytotal)^2
= sqrt((m1v1cos(angle1) + m2v2cos(angle2))^2 + ((m1v1sin(angle1) + m2v2sin(angle2))^2




3. By plugging into this equation, I come up with 49602.6 m/s, but it's not right. I need to get the velocity correct before I can make an attempt to find the initial direction

Firstly you need to use law of conservation of momentum for xaxis.
Then use for yaxis. Don't mix two axes. Two equations , two unknowns so you can find the correct answer.

 

[collinsmark]

2. x-momentum= m1v1cos(angle1) + m2v2cos(angle2)
y-momentum= m1v1sin(angle1) + m2v2sin(angle2)
magnitude momentum= sqrt(Xtotal)^2 + (Ytotal)^2
= sqrt((m1v1cos(angle1) + m2v2cos(angle2))^2 + ((m1v1sin(angle1) + m2v2sin(angle2))^2


3. By plugging into this equation, I come up with 49602.6 m/s, but it's not right. I need to get the velocity correct before I can make an attempt to find the initial direction

Your relevant equations seem correct, but show us your work, and maybe we can help you find out what went wrong. You're correct that 49602.9 m/s isn't right. 49602.9 m/s is greater than Earth's escape velocity.

 

[wwshr87]

I will call the car with mass of 1000Kg car A, the other one car B. Applying conservation of momentum in the x and y directions we have:
mAVAx=mAVA'x + mBVB'x for the x-direction
we know all the masses and velocities after collision, therefore we can solve for VA which is the magnitude of the velocity in the x-direction. Following the same concept for the y-direction, I obtained magnitude of VA 50Km/h with an angle of -17.25 degrees from the x-axis. Someone confirm this.

 

[clockworks204]

Thank you all for your input! I was able to get the answer by splitting into the 2 equations like everybody said to do. 49.9 km/hr and 342.8 deg were the answers I came up with. Thanks again!