You are working at the postal service center to design their box distribution sy

Question

You are working at the postal service center to design their box distribution system. The system consists of 55 kg carts which roll under chutes at 5.5 m/s while boxes drop from chutes 5.0 meters above the cart. The wheels of the cart are smooth, so you can ignore friction between the cart and the ground. A 15 kg shipping box slides down the chute which is angled at 30 degrees and leaves the end of the chute traveling at 2.0 m/s. The cart is synchronized such that the package lands in the cart and they roll off together What is the speed of the package in the horizontal direction just before it lands in the cart? What is the speed of the package in the vertical direction just before it lands in the cart? What is the angle with respect to the horizontal the box makes as it lands in the cart? What is the speed of the package plus the cart after the package lands in the cart How long will it take the cart plus the package to travel from where it catches the package to be directly underneath the next chute which is 15.0 meters away from the bottom of the chute? (NOTE: You need to take the initial position of the cart into account)

Explanation / Answer

As the 15 Kg shipping box slides down the incline, it has a velocity of 2 m/s making an of 30 degrees with the horizontal. Here, it needs to be understood that the velocity of the box along the horizontal will not change as there is no acceleration along the horizontal, however, it will undergo an acceleration along the vertical direction.

Part a.) The speed of the package along the horizontal will see no change. Hence will be given as:

Vx = 2 Cos30 = 1.732 m/s

Part b.) For the vertical direction, it will undergo an acceleration g, while it falls through the height of 5 m

So we can write: V2 = u2 + 2gS

or, Vy2 = (2Sin30)2 + 2 x 9.81 x 5

or, Vy = 9.955 m/s is the required speed.

Part c.) The angle with respect to the horizontal would be given as:

tan = Vy/Vx

or, = 80.13 degrees. with the horizontal is the required angle.

Part d.) Now the 55 kg cart is moving at 5.5 m/s while the package of mass 15 kg lands on it with horizontal speed of 1.732 m/s

Using the principle of conservation of linear momentum, we get:

70 V = 55 x 5.5 - 15 x 1.732 = 276.52

or, V = 3.95 m/s is the required speed of the cart plus package.

Part E.) We need to first determine the location of the cart with respect to the incline.

So the time needed for it fall through the 5 m height would be same as that needed for it to accelerate the vertical velocity from 1 m/s to 9.955 m/s

That is. t = (9.955 - 1 ) / 9.81 = 0.913 seconds.

So in this time it will cover a horizontal distance of 2 cos 30 x 0.913 = 1.58 m

Therefore the horizontal distance of the cart from the bottom of the incline = 1.58 m

That is, the distance from the next chute = 1.58 + 15 = 16.58

Hence the time needed for it to travel to the next chute = Distance/ speed = 16.58 / 3.95 = 4.197 seconds

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