Abody of mass 1.5kg, traveling along the positive x axis with speed 4.5m/s, collides with another body b of mass 3.2kg which, initially is at rest. a is deflected and moves with a speed of 2.1m/s in a direction which is 30 degrees below the x axis. b is set in motion at angle b above the x axis. calculate the velocity of b after collision.,

Data:
m₁ = 1.5kg
m₂ = 3.2kg
α = -30° (negative because it is below the x-xis)
= initial speed of object 1 = 4.5m/s
= initial speed of object 2 = 0m/s
= final speed of object 1 = 2.1m/s
= ?
β = ?

Since the motion after the collision is in 2 dimentions, it is better to write the speeds with their components along the x and the y-axis:
= initial speed of object 1 along x-axis = 4.5m/s
= initial speed of object 1 along y-axis = 0m/s
= initial speed of object 2 along x-axis = 0m/s
= initial speed of object 2 along y-axis = 0m/s

= final speed of object 1 along x-axis = 2.1 cos(-30) = 1.82m/s
= final speed of object 1 along y-axis = 2.1 sin(-30) = -1.05m/s 

In this kind of collision, we have the conservation of momentum, therefore we can write the system:


Considering the terms that are zero, it becomes:


Let's face first the y-component:
=

therefore:
= = = 5.04m/s

Now, let's face the x-component:
= =
=  = 1.26m/s

Now that we have the two components, we can find:

=  =  = 6.35m/s

Lastly, the angle can be found with trigonometry:

β = tan⁻¹() = tan⁻¹() = 14°

Answer8.0m;