A car heading north collides at an intersection with a truck heading east. if they lock together and travel 28 m. s at 46 degrees north of east just after the collision, how fast was the car initially traveling? assume that the two vehicles have the same mass.
A) 30 m/s
B) 80 m/s
C) 20 m/s
D) 40 m/s

Resolving horizontally, the car is initially traveling at the speed of 40 m/s approximately. The correct answer is option D

Given that a car heading north collides at an intersection with a truck heading east. if they lock together and travel 28 m/s at 46 degrees north of east just after the collision, it is assumed that the two vehicles have the same mass.

The degree to the horizontal will be 90 - 46 = 44 degrees

And since we are focusing on how fast was the car initially traveling, let us resolve the momentum after collision horizontally. That is,

Momentum before collision = momentum after collision

mU + 0 = (m + m) x 28 cos 44

Note: horizontally, momentum of the truck = 0

mU = 2m x 20.14

The mass m will cancel out.

U = 40.28 m/s

Therefore,  the car is initially traveling at the speed of 40 m/s approximately. The correct answer is option D

Learn more about momentum here: link

C) 20 m/s

Explanation:

The collision of the car and truck will give a single resultant velocity due to equal masses.

We'll treat this question as a vector question.

The velocity profile will form a right angle triangle.

To calculate the velocity of the car

Let the velocity be x

Sina = x/28

Sin46= x/28

0.7193*28 = x

20 = x

X = 20 m/s

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

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