The problem describes small packages moving along a conveyor belt, which drop into a 1-meter-long cart. The conveyor belt moves at a constant velocity of $v_c = 2 \, \text{m/s}$ and is inclined at 30° from the horizontal. The goal is to determine the shortest and longest distances $R$ where the edge $A$ of the cart can be placed to ensure the packages fall into the cart.

Known dataThe problem describes small packages moving along a conveyor belt, which drop into a 1-meter-long cart. The conveyor belt moves at a constant velocity of $v_c = 2 \, \text{m/s}$ and is inclined at 30° from the horizontal. The goal is to determine the shortest and longest distances $R$ where the edge $A$ of the cart can be placed to ensure the packages fall into the cart.

Known

• Speed of packages on the conveyor belt: $v_c = 2 \, \text{m/s}$
• Inclination of the conveyor: $\theta = 30^\circ$
• Height from which packages fall: $3 \, \text{m}$
• Length of the cart: $1 \, \text{m}$

Step-by-step approach:

1. Decompose the velocity: The velocity of the packages as they leave the conveyor belt can be split into horizontal and vertical components.

   • Horizontal velocity: $v_x = v_c \cos(30^\circ)$
   • Vertical velocity: $v_y = v_c \sin(30^\circ)$

2. Time of flight: The time it takes for a package to fall the vertical distance of 3 m can be found using the vertical motion equation under gravity $g = 9.81 \, \text{m/s}^2$: $y = v_y t + \frac{1}{2} g t^2$ With $y = -3 \, \text{m}$ (downward), solve for $t$.

3. Horizontal distance traveled: Once the time of flight is known, the horizontal distance $x$ the package travels before landing is given by: $x = v_x t$ This will give us the distance from the base of the conveyor where the package lands.

4. Determine the range of $R$: To find the range of $R$, we need the package to land anywhere between points $A$ and $B$ on the cart, which means the horizontal distance $x$ must be within $R$ and $R + 1$.

Let me calculate the key values.The time it takes for the packages to fall from the conveyor to the ground is approximately $t = 0.69 \, \text{seconds}$, and during this time, the horizontal distance the packages travel is about $x = 1.19 \, \text{meters}$.

Range for $R$:

For the packages to land inside the cart (which is 1 meter long), the distance $R$ from the conveyor belt to the cart's edge $A$ must satisfy: $R \leq 1.19 \leq R + 1$ This gives the following:

• The shortest distance $R$ is when the package lands at the edge of $A$, meaning $R = 0.19 \, \text{meters}$.
• The longest distance $R$ is when the package lands at the edge of $B$, so $R = 1.19 \, \text{meters} - 1 = 0.19 \, \text{meters}$.

Thus, the range for $R$ is approximately from $0.19 \, \text{m}$ to $1.19 \, \text{m}$.

Would you like more details or have any questions? Here are five related questions:

1. How would the distance change if the angle of the conveyor was steeper?
2. What happens to the range of $R$ if the cart was moving?
3. How would the time of flight change if the packages fell from a higher height?
4. Can you calculate the required conveyor speed for packages to drop directly at point $B$?
5. How would air resistance affect the problem?

Tip: Breaking velocity into horizontal and vertical components helps simplify projectile motion problems like this one.