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)` =
`(-m_(1) v_(1fy))/(m_(2))`=
`<span>(-(1.5)(-1.05))/(3.2)` = 5.04m/s

Now, let's face the x-component:

`v_(2fx)`=
`<span>(m_(1) v_(1ix) - m_(1) v_(1fx))/(m_(2)<span>)</span>` =

`(m_(1) (v_(1ix) - v_(1fx)))/(m_(2))` =
`<span>((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(
`( v_(2fy) )/( v_(2fx) )`) = tan⁻¹(
`\frac{ 1.26} }{ 5.04} }`) = 14°