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In a shipping company distribution center, an open cart of mass 50.0 $\mathrm{kg}$ is rolling to the left

at a speed of 5.00 $\mathrm{m} / \mathrm{s}(\text { Fig } .8 .46)$ . You can ignore friction between the cart and the floor. A 15.0 $\mathrm{kg}$ package slides down a chute that is inclined at $37^{\circ}$ from the horizontal and leaves the end of the chute with a speed of 3.00 $\mathrm{m} / \mathrm{s}$ .

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 $\mathrm{m}$ above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

(a) $v_{A 1}=9.35 \mathrm{m} / \mathrm{s}$

(b) $v_{2}=3.29 \mathrm{m} / \mathrm{s}$ to the left.

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{'transcript': "problem. 8.87. We've got a package in the shipping center rolling, sliding, gonna ramp. That's a 37 degrees and four meters above the ground. It leaves the graham with a speed of three meters per second and initially is going at this 37 degree inclination from horizontal. We have a cart as moving it five meters per second to the left and we want to know how fast is this package going right before it lands in the card? And then we want to find out what the final speed of the card is. So a is a conservation of energy question. This has some gravitational potential energy and the some of that and its kinetic energy. As it leaves the Ram, there's going to be constant as that potential energy is converted to kinetic energy. So on half and the initial squared waas M G H equals 1/2 and e squared, so it's final speed is going to be the square root of its initial speed, plus two times the speed at games. You know, this GH term has units of speeds, so this is equal to 9.35 meters per second. So now be we have a collision question and therefore looking a conservation of momentum. Now, on Lee, only the component of momentum in the X direction is going to be conserved because we have normal forces and waits here in the wider action. Because the you know the net momentum to change entity that mean it goes to zero. So m a be a initials x component us and the I am this the initial X component of bees velocity, which is the only component of it because moving horizontally is going to be equal to the combined mass of them because the package stays in the card. This is presumably some sort of weird system for getting packages and carts and getting the carts to where they need to go times what we're going to call just be X call it e f ex or something doesn't need a subsequent Really. So first order of business is going to be figuring out what this is. Now we're going to assume, but its direction of motion is still 37 degrees. Yeah, no, we don't even need to soothe. The horizontal component of the motion is constant cause there's no no forces acting on it. In the in the extractions of the fact that it doesn't matter what the actual angle is, it comes out. So this stays three meters per second throughout the entire problem, just nice. So our speed that the end is going to be and a a i x Oh, I'm sorry. This isn't going to be three meters per second. It's going to be three meters per second times co sign of the 37 degrees because it's not moving completely horizontally, it's important. It's more important to catch mistakes than it is to avoid making them in the first place because you can't always really do that. Otherwise they would really be mistakes. X, divided by the total mass and was fearless, noting that the carts ex complainer That's velocity is negative. They did three points 29 so the card is still going to the left, but it is going slower now, as you would expect"}