Respuesta :
Answer:
1. 13 m/s
2. [tex]45^{\circ}[/tex] north of east
Solution:
As per the question:
Velocity of the two pieces with equal masses, 'm', v = 18 m/s
Mass of the third particle, M = 2m
Now,
To calculate the speed of the third piece
We know that:
Mass 1 flies off to South and mass 2 to West
Now, by the conservation of momentum in the x and y direction:
[tex]Mv'_{y} = mv[/tex]
[tex]2mv'_{y} = 18m[/tex]
[tex]v'_{y} = 9\ m/s[/tex]
Similarly,
[tex]Mv'_{x} = 18m[/tex]
[tex]v'_{x} = 9\ m/s[/tex]
The resultant velocity of the third piece:
[tex]v' = \sqrt{v'_{x}^{2} + v'_{y}^{2}} = \sqrt{9^{2} + 9^{2}} = 12.73\ m/s[/tex] ≈ 13 m/s
Now,
The direction of the third piece can be calculated as:
[tex]tan\theta = \frac{v'_{y}}{v'_{x}}[/tex]
[tex]\theta = tan^{- 1}(\frac{9}{9}) = 45^{\circ}[/tex] in the north of east direction.
