Download the App!
Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.
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?
Get the answer to your homework problem.
Try Numerade free for 7 days
$R / 4$
Physics 101 Mechanics
Momentum, Impulse, and Collisions
Moment, Impulse, and Collisions
Simon Fraser University
University of Sheffield
In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.
In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.
Two identical masses are r…
Two idcntical masses are r…
two identical masses are i…
8.73 Two identical masses …
Two identical masses (each…
A mass $m$ is placed at th…
A hemispherical bowl is at…
A uniform solid sphere of …
Two identical ball bearing…
Suppose that each of the t…
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.
View More Answers From This Book
Find Another Textbook