What is the velocity and direction of the wreckage after the collision?

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The discussion focuses on calculating the velocity and direction of wreckage after a collision between a car and a truck. The car, weighing 2082 kg, was traveling east at 14.3 m/s, while the truck, weighing 4070 kg, was moving north at 10.8 m/s. After the collision, the combined mass of the wreckage is 6152 kg, resulting in a velocity of 8.63 m/s. The angle of the wreckage's direction is calculated to be 55.9 degrees from the east. The conversation emphasizes the need for resolving the momentum equations in two dimensions to find the correct angles.
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A 2082 kg Oldsmobile traveling east on Saginaw Street at 14.3 m/s is unable to stop on the ice covered intersection for a red light at Abbott Road. The car collides with a 4070 kg truck hauling animal feed north on Abbott at 10.8 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg)
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So, you can add the momentums like vectors to get the resultant.

Pr = sqrt {((2082)(14.3))^2 + ((4070)(10.8))^2}
Pr = 53089.9 N*s

P = P'
So, 53089.9 = (combined mass)v'

v' = (53089.9)/(6152)
v = 8.63 m/s

and the angle.

theta = tan(inverse) of (43956)/(29772.6)
= 55.9 deg

HELP ME
 
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Resolve the problem into two different equations since it's in 2D

x direction:
m1 * v1 * cos(theta) + m2 * v2 * cos(theta') = (m1+m2) * v3 * cos(theta'')

y direction:
m1 * v1 * sin(theta) + m2 * v2 * sin(theta') = (m1+m2) * v3 * sin(theta'')
 
i don't know what angles to use. can you please be more specific. i need help!
 
Someone, come on. I know basic physics. Give me a freakin bone.
 
On the LHS of the equation Nothing gave the angles are given in the problem (east and north). You'll have to solve for the angle (its the same for the x and y direction equations) on the other side.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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