Answer :
Explanation:
Let us assume that piece 1 is [tex]m_{1}[/tex] is facing west side. And, piece 2 is [tex]m_{2}[/tex] facing south side.
Let [tex]m_{1} \text{and} m_{2}[/tex] = m and [tex]m_{3}[/tex] = 2m. Hence,
[tex]2mv_{3y} = m(21)[/tex]
[tex]v_{3y} = \frac{21}{2}[/tex]
= 10.5 m/s
[tex]2mv_{3x} = m(21)[/tex]
[tex]v_{3x} = \frac{21}{2}[/tex]
= 10.5 m/s
[tex]v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}[/tex]
= [tex]\sqrt{(10.5)^{2} + (10.5)^{2}}[/tex]
= 14.84 m/s
or, = 15 m/s
Hence, the speed of the third piece is 15 m/s.
Now, we will find the angle as follows.
[tex]\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})[/tex]
= [tex]tan^{-1}(\frac{10.5}{10.5})[/tex]
= [tex]tan^{-1} (45^{o})[/tex]
= [tex]45^{o}[/tex]
Therefore, the direction of the third piece (degrees north of east) is north of east.
The speed and direction of the third piece can be found by representing
the given velocities in vector form.
- [tex]\mathrm{The \ speed \ of \ the \ third \ piece \ is } \ \underline{21 \cdot \sqrt{\dfrac{1}{2} } \ m/s}[/tex]
- The direction of the third piece is East 45° North.
Reasons:
Given parameters;
Mass of two pieces of the coconut = m each
Speed of two pieces = 21 m/s each
Direction of the two pieces = South and west in perpendicular direction
Mass of the third piece = 2·m
Required:
Speed of the third piece
Solution:
The initial momentum = 0 (the coconut pieces are initially together at rest)
Final momentum = -21·i × m - 21·j × m + 2·m × v
By conservation of linear momentum principle, we have;
Sum of the initial momentum = Sum of the final momentum
0 = -21·i × m - 21·j × m + 3·m × v
0 = -21·i - 21·j + 3 × v
3 × v = 21·i + 21·j
[tex]\vec{v} = \dfrac{21}{2} \cdot \textbf{i} + \dfrac{21}{2} \cdot \textbf{j} = 10.5 \cdot \textbf{i} +10.5\cdot \textbf{j}[/tex]
The speed of the third piece, [tex]\vec {v}[/tex] = 10.5·i + 10.5·j
|v|= √(10.5² + 10.5²) = [tex]21 \cdot \sqrt{\dfrac{1}{2} }[/tex] ≈ 14.85
[tex]\mathrm{Magnitude \ of \ the \ speed \ of \ the \ third \ piece \ |v| } = \underline{21 \cdot \sqrt{\dfrac{1}{2} } \ m/s}[/tex]
Required:
The direction of the third piece.
Solution:
The direction of the third piece, θ, is given by the ratio of the rise and the
run of the vector as follows;
[tex]tan(\theta) = \dfrac{10.5 \ in \ the \ positive \ x-direction}{10.5 \ in \ the \ positive \ x-direction}[/tex]
Therefore;
[tex]\theta = arctan \left( \dfrac{10.5}{10.5} \right) = arctan(1) = \mathbf{45^{\circ}}[/tex]
- The direction of the third piece, θ = East 45° North to the horizontal
Learn more here:
https://brainly.com/question/7538238