Answer:
12.02 m/s north of east
Explanation:
Let [tex]v_x, v_y[/tex] be the horizontal and vertical velocity of the 3rd piece, positive in the East and North direction. Also let m be the mass of the 2 smaller pieces, so 2m is the mass of the 3rd piece.
We can solve for the 2 velocities components using law of momentum conservation. Before the explosion, the total momentum was 0 because the coconut was at rest. After the explosion:
- In the x direction:
[tex]2mv_x + m*(-17) = 0[/tex] negative 17 because 1 piece flew west
[tex]v_x = 17/2 = 8.5 m/s[/tex]
- In the y direction
[tex]2mv_y + m(-17) = 0[/tex] negative 17 because 1 piece flew south
[tex]v_y = 17/2 = 8.5 m/s [/tex]
So the speed and direction of the 3rd piece is
[tex]v = \sqrt{v_y^2 + v_x^2} = \sqrt{8.5^2 + 8.5^2} = \sqrt{72.25 + 72.25} = \sqrt{144.5} = 12.02 m/s[/tex]
[tex]tan\theta = \frac{v_y}{v_x} = \frac{8.5}{8.5} = 1[/tex]
[tex]\theta = tan^{-1}1 = 0.79 rad \approx 45^o[/tex] north of east
Answer:
v3 = –17m/s East or north.
Explanation:
This problem involves the concept of momentum
The total momentum before the explosion of the fire cracker is zero
Let the mass of the first two particles be m1 = m2 = m, velocities v1 = v2 = 17m/s
And m3 = 2m
Taking south and west as positive directions, total momentum before collision equals total momentum after collision
Summing momentum along the north-south axis
0 = m1v1y +m2v2y+ m3v3y
0 = m×17 +m×0 + 2m×v3
0 = 17m + 2mv3y
Dividing through by m
0= 17 +2v3y
v3y = –17/2 = –8.5m/s
0 = m1v1x +m2v2x + m3v3x
0 = m×0 + m×17 + 2m×v3x
0 = 17m + 2mv3x
Dividing through by m
0= 17 +2v3x
v3y = –17/2 = – 8.5m/s
V3 = √(v3x² + v3y²)
V3 = √((-8.5)² + (-8.5)²)
V3 = √(144.5) = 12.02m/s
θ = Tan -¹(v3y/v3x) = Tan -¹(–8.5/–8.5) = 45°
θ = 45° (NE)
