In a shipping company distribution center, an open cart of mass 46.0 kg is rolli

Question

In a shipping company distribution center, an open cart of mass 46.0 kg is rolling to the left at a speed of 4.50 m/s(see the figure (Figure 1) ). You can ignore friction between the cart and the floor. A 12.0 kg package slides down a chute that is inclined at 37 from the horizontal and leaves the end of the chute with a speed of 2.90 m/s . The package lands in the cart and they roll off together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what is the speed of the package just before it lands in the cart? What if the final speed of the cart?

Explanation / Answer

The speed of the package when it leaves the chute is 2.90*sin(37)=1.745 m/s in the y-direction (down), and 2.90*cos(37)=2.316 m/s in the x-direction.

Its final speed in the y direction is given by vy^2=v0y^2+2*a*y
vy^2 = 1.745^2 + 2*9.8*4 = 3.046 + 78.4 = 81.446
vy = sqrt(81.446) = 9.02474 m/s

So, vx = 2.316, and vy= 9.025. The total velocity of the package is given by the Pythagorean theorem:
v = Sqrt(vx^2+vy^2) = Sqrt(5.364+81.446) = Sqrt(86.810) =
9.317 m/s

Then, the final speed of the cart depends on whether the package is moving to the left or right. If they are both moving to the left, then the final velocity is:
(m1*vcart+m2*vx)/(m1+m2) = (46*4.5+12*2.316)/(46+12) = 4.04 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
(46*4.5-12*2.316)/(46+12) = 3.08 m/s to the left.

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