Particle Collision: 3.00kg & 4.00kg Particles

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In the particle collision discussion, a 3.00 kg particle with a velocity of (-4.00 m/s, -5.00 m/s) collides with a 4.00 kg particle moving at (5.60 m/s, -1.80 m/s). To find the final velocity, the momentum formula p = mass * velocity is applied, leading to the equation vf = (p1 + p2) / (m1 + m2). For angle calculations, the tangent function is used, where tan(θ) = p2/p1, and θ = tan^-1(p2/p1). Vector notation requires adding and subtracting the respective components, while Pythagorean theorem assists in determining the resultant velocity magnitude. The discussion emphasizes the need for clarity on formulas and methods for solving the problem.
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Particle Collision Two A collision occurs between a 3.00 kg particle traveling with velocity v = (-4.00 m/s) i + (-5.00 m/s) j and a 4.00 kg particle traveling with velocity v = (5.60 m/s) i + (-1.80 m/s) j. The collision connects the two particles. What then is their velocity in each of the following notations?
(a) unit-vector notation
( wrong check mark m/s ) i + ( wrong check mark m/s ) j
(b) magnitude-angle notation
m/s angle ° (counterclockwise from the positive x axis)

I would show an attempt at the solution but i have no idea what the formulas are or where to start
 
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nobody?
 
Hey a little late, but I think for this one you have to use

p= momentum= (mass*velocity)

vf = (p1 + p2)/(m1+m2)

for angle, use tangent.

tan(o) = p2/p1

(o) = tan^-1(p2/p1).


With respect to vector notation, i think you just add and subtract, but for angle use Pythagoras, cause you will need v1, and v2 as a quantity.
 
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