In a shipping company distribution center, an open cart of mass 50 kg is rolling

Question

In a shipping company distribution center, an open cart of mass 50 kg is rolling to the left at a speed of 5 m/s. You can reglect the friction between the cart and the floor. A 15 kg package slides down a chute at an angle of 37 degrees below the horizontal. The package leaves the chute with a speed of 3 m/s. The package lands in the cart after falling for 0.75 seconds, and rools back and forth along the bottom of the cart before coming to a stop after 2 seconds. Use impluse-momentum physics to answer the following questions:

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

b) What is the final veocity of the cart?

c) What is the average scale reading during the landing of the package, If you assume that the whole bottom of the cart is a scale platform?

Explanation / Answer

The speed of the package when it leaves the chute is 3*sin(37)=1.805 m/s in the y-direction (down), and 3*cos(37)=2.395 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 = 1.805^2 + 2*9.8*4 = 81.658
vy = sqrt(81.658) = 9.036 m/s

So, vx = 2.395, and vy= 9.036. The total velocity of the package is given by the Pythagorean theorem:
v = Sqrt(vx^2+vy^2) = Sqrt(5.364+81.446) = 9.348 m/s

Then, the final speed of the cart depends on whether the package is moving to the left ,then the final velocity is:
(m1*vcart+m2*vx)/(m1+m2) = (50*5+15*2.395)/(65) = (288.6)/(68) = 4.398 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*4.7-13*2.316)/(55+13) = (228.4)/(68) = 3.36 m/s to the left.

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