What Is the Direction of Two Colliding Cars Stuck Together?

[friedpork]

Homework Statement

A 1300kg car moving east collided with a 2600 kg SUV moving north at 28 m/s. The vehicles became stuck together . If the speed of the vehicles immediately after the collision was 30 m/s, what was their direction?

A) 21 degrees E of N
B) 52 degrees E of N
C) 58 degrees E of N
D) 69 degrees E of N

Answer is suppose to be B

Homework Equations

P1 + P2 = P aka m1v1 + m2v2 = (m1 + m2) v

The Attempt at a Solution

I first drew vectors and then tried to add them together but i don't know how to draw them here so yea..

I used the formula given ... m1v1 + m2v2 = ( m1 + m2) v.

Not given v1 so i attempted to solve that with algebra... v1 = ((m1 +m2)v - (m2v2))/ m

v1 = ((1300+2600) x 30) - (2600 x 28)) / 1300
v1 = 34 m/s

I drew a vector triangle eventually got tan ((2600 x 28)/(1300 x 24) = 58.74 degrees east of north

Please help and Thank You... Sorry if this was to messy to understand first time doing this :p

 

[collinsmark]

Homework Equations

P1 + P2 = P aka m1v1 + m2v2 = (m1 + m2) v

The above equation is fine, as long as you treat the velocities as vectors.

The Attempt at a Solution

I first drew vectors and then tried to add them together but i don't know how to draw them here so yea..

I used the formula given ... m1v1 + m2v2 = ( m1 + m2) v.

Not given v1 so i attempted to solve that with algebra... v1 = ((m1 +m2)v - (m2v2))/ m

v1 = ((1300+2600) x 30) - (2600 x 28)) / 1300
v1 = 34 m/s

Sorry, but that's not going to work for you. You need to treat the velocities as vectors, and add them appropriately as vectors.

In this problem it's probably easiest to break up this problem into separate directions. If you consider the East-West direction as being the 'x' direction, and the South-North direction as being the 'y' direction, then

[tex] m_{1}v_{1x} + m_{2}v_{2x} = (m_1+m_2)v_x [/tex]

[tex] m_{1}v_{1y} + m_{2}v_{2y} = (m_1+m_2)v_y [/tex]

[tex] |v| = \sqrt{v_x^2 + v_y^2} [/tex]

See if that helps.

 

[friedpork]

I think i got it..

So for the [tex]m_{1}v_{1x} + M_{2}v_{2x} = (m_1+m_2) v_x[/tex] I wasnt able to solve it since i have two variables.

But in the vertical direction i was able to solve it and i ended up getting a velocity of 18.66667 m/s

Drew a vector additional triangle. 30 m/s was my hypotenuse and 18.66667 became my opposite side.

Sin ( 18.6666666667/ 30 ) = 38.4786 degress... I'm going to assume that this value is North of East. which is also the same as 51.5212 degrees East of North.

 

[collinsmark]

That's the way to do it.

 

[rdl]

I would approach this problem by first considering the conservation of momentum principle. This states that the total momentum of a system before and after a collision remains constant, as long as there are no external forces acting on the system. In this case, the two cars are the system and we can use the formula m1v1 + m2v2 = (m1 + m2)v to solve for the final velocity (v) of the combined cars.

Using the given values, we can plug them into the equation as follows:

1300 kg x v1 + 2600 kg x 28 m/s = (1300 kg + 2600 kg) x 30 m/s

Solving for v1, we get v1 = 34 m/s

Now, to determine the direction of the combined cars, we can use trigonometry to find the angle between the initial and final velocities. We can draw a vector diagram or use the formula tanθ = (vy/vx) to find the angle θ.

Plugging in the values, we get tanθ = (28 m/s)/(34 m/s) = 0.82

Using a calculator, we can find that θ is approximately 38.66 degrees. However, this is the angle between the combined velocity and the initial velocity of the 1300 kg car. To find the angle between the combined velocity and the initial velocity of the 2600 kg car, we need to subtract this angle from 90 degrees (since the initial velocity of the 2600 kg car is perpendicular to the initial velocity of the 1300 kg car).

Therefore, the final angle between the combined velocity and the initial velocity of the 2600 kg car is 90 degrees - 38.66 degrees = 51.34 degrees.

Since the combined velocity is in the direction of the 2600 kg car, we can say that the final direction of the combined cars is 51.34 degrees east of north, which is option B.