A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 200g piece moves along the x-axis with a speed of 2.00m/s and a 235g piece moves along the the y axis with a speed of 1.50m/s. The third piece has a mass of 100g. In what direction does the third piece move?
I have done this:
tan^-1 = 235(1.50)/200(2) = 41.4
i don't think this is right please help.
5.33
To determine the direction in which the third piece moves, we can use vector addition. Since we know the motions along the x-axis and y-axis, we can treat them as components of a vector. Let's assign positive x-axis as east and positive y-axis as north.
First, we need to find the x-component and y-component of the velocity vector of the first two pieces.
The x-component (Vx1) of the first piece is given by: Vx1 = 200g * 2.00 m/s / (200g + 235g + 100g) = 0.57 m/s
The y-component (Vy2) of the second piece is given by: Vy2 = 235g * 1.50 m/s / (200g + 235g + 100g) = 0.83 m/s
Now, we can find the resultant vector by adding the x-components and y-components:
Vx = Vx1
Vy = Vy2
The magnitude of the resultant vector is given by: V = sqrt(Vx^2 + Vy^2)
The angle (θ) of the resultant vector is given by: θ = tan^(-1)(Vy / Vx)
Now let's calculate these values:
V = sqrt(0.57^2 + 0.83^2) ≈ 1.01 m/s
θ = tan^(-1)(0.83 / 0.57) ≈ 54.35 degrees
Therefore, the third piece moves in the direction approximately 54.35 degrees north of east.