In the two-sphere arrangement of Fig. 9-20, assume that sphere 1 has a mass of 50 g and an init

In the two-sphere arrangement of Fig. 9-20, assume that...

Key Concepts

Sequential Collisions Analysis

In problems involving more than one collision, the outcomes of the initial collision directly impact the subsequent collision(s). By applying the conservation principles sequentially, one can track the changes in velocity and energy transfer through each collision event and determine the resulting motion of each object.

Elastic Collision Equations

The formulas specifically derived for elastic collisions combine the conservation of momentum and kinetic energy to provide expressions for the final velocities of two objects after they collide. These equations are crucial for accurately solving problems involving sequential elastic collisions in a closed system.

Conservation of Kinetic Energy

In an elastic collision, not only is momentum conserved, but kinetic energy is also conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision, which, together with momentum conservation, allows for the determination of the final velocities of the colliding bodies.

Conservation of Momentum

In collision problems, particularly in elastic collisions, the total momentum of the system remains constant when no external forces act on it. This principle is used to relate the velocities and masses of the objects before and after the collision, providing one of the key equations necessary for solving such problems.

Energy Conversion between Kinetic and Potential Energy

When an object is released from a height, its gravitational potential energy is converted into kinetic energy as it falls. After a collision, the kinetic energy of a moving object is converted back into gravitational potential energy as it rises. This principle is used to calculate the maximum height an object reaches based on its velocity after the collision.

Transcript

00:01 In this question, we're told that we have two spheres that are ranged as in figure 9 -20, and we're to assume that sphere 1 has a mass m1 of 50 grams and initial height h1 of 9 .0 centimeters, and that sphere 2 has a mass of 85 grams, m2.

00:16 After sphere 1 is released and collides elastically with sphere 2, for part a, what height is reached by sphere 1, and part b, what height is reached by sphere 2? and then after the next elastic collision, what height is reached by c, sphere one, and by d sphere two, hint, do not round off values.

00:35 Okay.

00:38 So for part a here, we're going to refer to the discussion in the textbook, the sample problem that's called elastic collisions two pendulums, which uses the same notation that we use here.

00:52 And for some important details in the reasoning, we choose the right word in figure, figure 9 -20 as our positive x direction.

00:59 And we use the notation v where we refer to the velocities.

01:04 Since the algebra is fairly involved, we're going to find it convenient to introduce the notation delta m equals m2 minus m1, just to make things a little simpler.

01:15 Okay, so we'll write that up over here just so we know.

01:18 Delta m, the change in m is just m2 minus m1.

01:23 That's going to pop up a lot, and so it's just going to save us some time in our writing here.

01:27 Okay, so v1 initial then is going to be equal to the square root of 2 times gravity times h1.

01:41 Okay.

01:43 And then v1 final, so this is part a, v1 final is equal to m1 minus m2 and, or excuse me, yeah, m1 minus m2.

02:12 Divided by m1 plus m2, okay, multiplied by v1 initial, which equals minus delta m over m1 plus m2 times the square root of two times g times h1.

02:44 But since we have conservation of energy, which says that m1 times g, times h1, is equal to 1 1⁄2 m1 times v1 final squared right so if we use that we can solve for h1 final right since the m1s are going to cancel here and h1 final is going to be equal to delta m so we're just solving for h1 final in this equation here so h1 final then would be 1 v1 final squared over g but we're going to plug in v1 final up above for the v1 final squared in here so we're going to have delta m over m1 plus m2 squared and this course when you squared the root 2 g h1 that just becomes 2g h1 and the gs are going to cancel and the twos are going to cancel so you're just left with this being multiplied by h1 so you plug in all those values for m1 and h1 remember delta to m is m2 minus m1 when this comes out to be 0 .60 and the units here are centimeters and we can box that in as our solution for part a now for part b we're asked to do the same thing except for doing it for sphere 2 okay so now we have v2 final is equal to 2m1 divided by m1 plus m2 and then this ratio here is multiplied by v1 initial, which is the square root of 2g h1.

04:55 Now, as before, we're going to use conservation of energy.

04:58 So we're going to have m2 times g times h2 final is equal to 1 half m2 times v2 final squared.

05:22 Okay.

05:23 Well, once again, the mass is canceled.

05:27 And we're going to solve for h2 final, which is going to give us h2 final then is one half v2 final squared over g.

05:39 So we plug in the value for v2 final squared.

05:42 And this comes out to be 2m1 over m2 plus m1.

05:53 And this is all squared.

05:54 And then when you square root 2 g h1, the 2 and the g cancel and all you're left with is h1.

06:01 So you plug this value into the equation.

06:02 And this comes out to be 4 .9 centimeters...