This problem is derived from a video shown in class. A small cart is rolling down an inclined track and accelerating. It fires a ball perpendicular to the inclined track as it moves. The origin of the reference frame is located at the spot from where the ball is shot out. The x-axis of the reference frame points downward along the inclined track. The y-axis of the reference frame points upward perpendicular to the inclined track.
The angle of the incline is 30°. When the ball is shot out with a 6m/s velocity perpendicular to the inclined track, the cart is moving at 5m/s along the inclined track. Please round your answers to 2 decimal places.
Along the inclined track, where is the cart one second after the ball is shot out?
in the x direction
Where is the ball located one second after it is shot out?
in the x and y direction
To determine the cart's position one second after the ball is shot out, we need to calculate its displacement in the x-direction.
We can start by finding the cart's initial velocity in the x-direction. Given that the cart is moving at 5 m/s along the inclined track, and the angle of the incline is 30°, we can decompose the velocity into its x and y components.
The cart's initial velocity in the x-direction can be found using the equation:
Vx = V * cos(theta)
where V is the velocity along the inclined track and theta is the angle of the incline. Substituting the values:
Vx = 5 m/s * cos(30°) = 5 m/s * 0.866 = 4.33 m/s
Now, we can calculate the cart's displacement in the x-direction using the equation:
Δx = Vx * t
where t is the time elapsed. Substituting the values:
Δx = 4.33 m/s * 1 s = 4.33 m
Therefore, one second after the ball is shot out, the cart is located 4.33 meters down the inclined track in the x-direction.
To determine the ball's position one second after it is shot out, we need to calculate its displacement in both the x and y directions.
We already know the initial velocity of the ball in the y-direction is 6 m/s. To find the ball's initial velocity in the x-direction, we can use the same equation as before:
Vx = V * cos(theta)
Substituting the values:
Vx = 6 m/s * cos(30°) = 6 m/s * 0.866 = 5.20 m/s
Now, we can calculate the ball's displacement in the x-direction using the equation:
Δx = Vx * t
Substituting the values:
Δx = 5.20 m/s * 1 s = 5.20 m
Therefore, one second after it is shot out, the ball is located 5.20 meters horizontally (in the x-direction) from its initial position.
In the y-direction, we can calculate the ball's displacement using the equation:
Δy = Vy * t + 0.5 * a * t^2
where Vy is the initial velocity in the y-direction, t is the time elapsed, and a is the acceleration due to gravity (-9.8 m/s^2). Substituting the values:
Δy = 6 m/s * 1 s + 0.5 * (-9.8 m/s^2) * (1 s)^2 = 6 m - 4.9 m = 1.1 m
Therefore, one second after it is shot out, the ball is located 1.1 meters vertically upward (in the y-direction) from its initial position.