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

Velocity box leaves the end of chute = 2.0 m/s
Horizontal Component = 2.0 * cos(30) = 1.732 m/s
Vertical Component = 2.0 * sin(30) = 1.0 m/s

(a)
As Horizontal acceleration = 0,
Horizontal Velocity just before it lands in the cart, v = 1.732 m/s

(b)
Vertical compnent, Vf
Vf^2 = Vi^2 + 2*a*s
Vf^2 = 1^1 + 2*9.8*5.0
Vf = 9.95 m/s
Vertical Velocity just before it lands in the cart, v =  9.95 m/s

(c)
Angle = tan^-1(9.95/1.732)
Angle = 80.1 degree

(d)
Using Momentum Conservation in Horizontal Direction,
Initial Momentum = Final Momentum
- 15 * 1.732 + 55 * 5.5 = (15+55)*vf
vf = 3.95 m/s

(e)
s = 1.0*t + 1/2*g*t^2
5 = 1.0*t + 1/2*9.8*t^2
t = 0.91 s

Horizontal distance travelled, s = 0.91 * 1.732 = 1.58 m
Total distance travelled by cart + package = 15 + 1.58 = 16.58 m
Time, t = 16.58/3.95
t = 4.2 s

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