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Two identical masses are released from rest in a smooth hemispherical bowl of radius $R$ , from the positions shown in Fig. 8.45 . You can ignore friction between the masses and the surface of the bowl. If they stick together when they collide, how high above the bottom of the bowl will the masses go after colliding?

$R / 4$

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Simon Fraser University

Hope College

University of Sheffield

McMaster University

in this problem, We're gonna talk about conservation of lunar moment. So we need to remember that the momentum off a particle is equal to its mass times its velocity V and that the total momentum of a system in the absence of external forces is concerned. So even though the individual Momenta other particles may change the total one doesn't s so what we have in our problem is a hemisphere is shown here in a picture that has a radius R and two blocks that are placed at that are they are identical and they're placed as shown here. Eso The first one is that a height are from the bottom. The second one is at the bottom and they are released. They collide at the, uh at the bottom of the hemisphere and then stick together. And our goal is to find how high the two blocks you gather Reach What is the height? H Okay, so what we have to do first is to notice that the first block, the one on the left falls while the second one since it's at the bottom, does not eso the second block just stay still until the collision happens since the block has just released, we have to find what is the velocity with which it reaches the bottom of the hemisphere. Now from conservation off energy, we have the gravitational potential energy MGH must be able to MV squared over two eso the ems Yes, watch And then we have a V is equal to the square root of two GH. Yeah, so at the bottom, when the collision happens, the velocity off the block points to the right with the speed of the square or to you th now at the bottom the collision happened. So at the collision, the momentum is concerned right at the bottom. No external forces act on the system known at external forces. Eso we can write that the mass off the falling log times its speed is equal to the total mass its final total mass times the final speed v prime So the prime is V divided by two meaning that the prime is the square root of two GH divided by Chu. Okay, so this is the speed off the system of the two blocks and what we have to find is how high this system reaches. So Basically, we have to apply conservation off energy again. So at the bottom, the kinetic energy is the mass of the system, which is two AM times the mad, temporary times view prime squared over to, and this is equal to the mass times g times the height to which the two blocks rate rise. So the two m's canceled. The prime squared is to G H over four. Then we have two divided by two. This is able to g age. And actually, I'm sorry. This shouldn't be age. Um h should be substituted by our since the initial position off the first block is are Okay, so you're subject to the height of the first block by our So sorry for that. I'm sorry. This should be our as well. Okay, So the geese cancel out Tuesday as well. And age is able to our over four, which is the answer to your question.

Universidade de Sao Paulo