Question

a body 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.,

Answer 1

Data:
m₁ = 1.5kg
m₂ = 3.2kg
α = -30° (negative because it is below the x-xis)
$v_{1i}$ = initial speed of object 1 = 4.5m/s
$v_{2i}$ = initial speed of object 2 = 0m/s
$v_{1f}$ = final speed of object 1 = 2.1m/s
$v_{2f}$ = ?
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:
$v_{1ix}$ = initial speed of object 1 along x-axis = 4.5m/s
$v_{1iy}$ = initial speed of object 1 along y-axis = 0m/s
$v_{2ix}$ = initial speed of object 2 along x-axis = 0m/s
$v_{2iy}$ = initial speed of object 2 along y-axis = 0m/s

$v_{1fx}$ = final speed of object 1 along x-axis = 2.1 cos(-30) = 1.82m/s
$v_{1iy}$ = 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:
$\left \{ {{m_{1} v_{1ix} + m_{2} v_{2ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop { m_{1} v_{1iy} + m_{2} v_{2iy} = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.$

Considering the terms that are zero, it becomes:
$\left \{ {{m_{1} v_{1ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop {0 = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.$

Let's face first the y-component:
$m_{2} v_{2fy}$ = $-m_{1} v_{1fy}$

therefore:
$v_{2fy}$ = $\frac{-m_{1} v_{1fy}}{m_{2}}$=$\frac{-(1.5)(-1.05)}{3.2}$ = 5.04m/s

Now, let's face the x-component:
$v_{2fx}$=$\frac{m_{1} v_{1ix} - m_{1} v_{1fx}}{m_{2}}$ =
$\frac{m_{1} (v_{1ix} - v_{1fx})}{m_{2}}$ = $\frac{(1.5)(4.5-1.82)}{3.2}$ = 1.26m/s

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

$v_{2f}$ = $\sqrt{ v_{2fx}^2 + v_{2fy}^2 }$ = $\sqrt{5.04^{2} + 1.26^{2} }$ = 6.35m/s

Lastly, the angle can be found with trigonometry:

β = tan⁻¹($\frac{ v_{2fy} }{ v_{2fx} }$) = tan⁻¹($\frac{ 1.26} }{ 5.04} }$) = 14°