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 2.0m/s. The cart is synchronized such that the package lands in the cart and they roll off together

(a) What is the speed of the package in the horizontal direction just before it lands in the cart?

(b) What is the speed of the package in the vertical direction just before it lands in the cart?

(c) What is the angle with respect to the horizontal the box makes as it lands in the cart?

(d) What is the speed of the package plus the cart after the package lands in the cart?

(e) 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

The speed of the package when it leaves the chute is 2.0*sin(30)=1.0 m/s in the y-direction (down), and 2.0*cos(30)= 1.732 m/s in the x-direction (not sure about left or right, because your question doesn't specify).

Its final speed in the y direction is given by vy^2=v0y^2+2*a*y
vy^2 = 1.0^2 + 2*9.8*5 = 99.0
vy = sqrt(99) = 9.95 m/s

So, vx = 1.732, and vy= 9.95. The total velocity of the package is given by the Pythagorean theorem:
v = Sqrt(vx^2+vy^2) = Sqrt(1.732^2 + 9.95^2) = 10.1 m/s

Then, the final speed of the cart depends on whether the package is moving to the left or right. I assume there was a diagram included in your question. If they are both moving to the left, then the final velocity is:
(m1*vcart+m2*vx)/(m1+m2) = (55*5.5 + 15*1.732)/(55+15) = 4.7 m/s to the left

If the cart is moving to the left, and the package is going to the right, then the equation gives
(55*5.5 - 15*1.732)/(55+15) = 3.95 m/s to the left.

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