Data:
m₁ = 1.5kg
m₂ = 3.2kg
α = -30° (negative because it is below the x-xis)
[tex] v_{1i} [/tex] = initial speed of object 1 = 4.5m/s
[tex] v_{2i} [/tex] = initial speed of object 2 = 0m/s
[tex] v_{1f} [/tex] = final speed of object 1 = 2.1m/s
[tex] v_{2f} [/tex] = ?
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:
[tex] v_{1ix} [/tex] = initial speed of object 1 along x-axis = 4.5m/s
[tex] v_{1iy} [/tex] = initial speed of object 1 along y-axis = 0m/s
[tex] v_{2ix} [/tex] = initial speed of object 2 along x-axis = 0m/s
[tex] v_{2iy} [/tex] = initial speed of object 2 along y-axis = 0m/s
[tex] v_{1fx} [/tex] = final speed of object 1 along x-axis = 2.1 cos(-30) = 1.82m/s
[tex] v_{1iy} [/tex] = 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:
[tex] \left \{ {{m_{1} v_{1ix} + m_{2} v_{2ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop { m_{1} v_{1iy} + m_{2} v_{2iy} = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right. [/tex]
Considering the terms that are zero, it becomes:
[tex]\left \{ {{m_{1} v_{1ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop {0 = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.[/tex]
Let's face first the y-component:
[tex]m_{2} v_{2fy}[/tex] = [tex]-m_{1} v_{1fy}[/tex]
therefore:
[tex]v_{2fy}[/tex] = [tex]\frac{-m_{1} v_{1fy}}{m_{2}}[/tex]=[tex]\frac{-(1.5)(-1.05)}{3.2}[/tex] = 5.04m/s
Now, let's face the x-component:
[tex]v_{2fx}[/tex]=[tex]\frac{m_{1} v_{1ix} - m_{1} v_{1fx}}{m_{2}}[/tex] =
[tex]\frac{m_{1} (v_{1ix} - v_{1fx})}{m_{2}}[/tex] = [tex]\frac{(1.5)(4.5-1.82)}{3.2}[/tex] = 1.26m/s
Now that we have the two components, we can find:
[tex]v_{2f} [/tex] = [tex] \sqrt{ v_{2fx}^2 + v_{2fy}^2 } [/tex] = [tex] \sqrt{5.04^{2} + 1.26^{2} } [/tex] = 6.35m/s
Lastly, the angle can be found with trigonometry:
β = tan⁻¹([tex] \frac{ v_{2fy} }{ v_{2fx} } [/tex]) = tan⁻¹([tex] \frac{ 1.26} }{ 5.04} } [/tex]) = 14°
