A mass m is placed at the rim of a frictionless...

A mass $m$ is placed at the rim of a frictionless hemispherical bowl with a radius $R$ and released from rest (Figure 8.49 ). It then slides down and undergoes a perfectly elastic collision with a second mass $3 m$ sitting at rest at the bottom of the bowl. (a) What are the direction and speed of each mass just after the collision? (b) In terms of $R$, to what maximum height will each mass travel after the collision?

Key Concepts

Conservation of Energy

This concept explains that in the absence of non-conservative forces (here, friction is absent), the total mechanical energy of the system (the sum of potential and kinetic energy) remains constant. In problems like this, it is used to determine the speed of an object as it moves along a curved path by converting gravitational potential energy into kinetic energy or vice versa.

Conservation of Momentum

In collision problems, especially when no external forces act in the horizontal direction, the total linear momentum of the system is conserved. This principle is essential in calculating the speeds of the colliding bodies immediately after impact by equating the total momentum before and after the collision.

Elastic Collision

An elastic collision is one in which both momentum and kinetic energy are conserved. This concept is crucial in resolving the post-collision velocities since it allows the use of both momentum and energy conservation equations simultaneously, making it possible to uniquely determine the outcome for each object involved.

Gravitational Potential Energy Conversion

After collisions, moving masses in a gravitational field convert kinetic energy back into gravitational potential energy as they rise. This concept provides a way to compute the maximum height reached by a mass by equating its kinetic energy immediately after collision to the gravitational potential energy at the peak of its ascent.

Transcript

00:01 So in this question, we have an object of mass m that slides down on frictionless hemispherical ball with radius r, and then it collides into a mass of 3m and a perfectly elastic collision.

00:11 We want to know a few things.

00:13 The direction and speed of each mass after the collision, and then what height do each of these masses travel back to it? well, the first thing before we do anything else is we got to figure out what amount of velocity the mass m is going to have when it reaches the bottom of the ram.

00:31 And the way that we can do this is conservation of energy.

00:33 At the highest point on the ramp, that mass has gravitational potential.

00:38 And at the bottom, it's going to have kinetic.

00:41 And let's set those things equal to each other.

00:43 So that's m .g.

00:45 And then the height that it's at is one radius and one half mv squared.

00:53 So this gives me v is equal to square root of 2gr.

01:00 So now we have to turn this into a momentum question.

01:07 Now since it's an elastic collision, two things are true.

01:10 The initial momentum is equal to the final momentum, and the initial kinetic energy is equal to the final kinetic energy, and that's going to be very important.

01:21 So let's set up our two equations here.

01:25 So we have, for our momentum equation, i'll put in red, as m times the square root of 2gr, equals m times v1 plus 3m times v2.

01:43 So i'm using v1 for the velocity of the block of just mass m, and v2 for the velocity of the 3m block.

01:54 Now for our kinetic energy equation, i'm going to have 1 half m times square of 2gr squared, which is just 2gr equals 2gr equals, 1 1� m v1 squared plus 1 1 1⁄2 squared.

02:25 So the first thing that i'm going to do is i'm going to solve this red equation for v1.

02:31 So my ms cancel out, subtract of 3b2, so i have v1 equals square root of 2 gr minus 3v2.

02:50 Now what i'm going to go ahead and do is plug that in for v1 in my second equation.

02:56 And i'm going to simplify as we go.

03:00 So i'm going to get rid of all the ones that have in the ends.

03:02 So i have 2vr equals square root of 2vr minus 3b2 squared plus 3b2...

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