A 2 -kg sphere moving to the right with a velocity of 5 m /...

Key Concepts

Conservation of Linear Momentum

In collision problems, the principle of conservation of linear momentum states that the total momentum of a system remains constant if no external forces act on it. This principle is crucial for analyzing impacts between bodies, as it allows us to relate the velocities and masses of the colliding bodies before and after impact, even when other forms of energy (like rotational energy) might be involved.

Coefficient of Restitution

The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the relative speed after impact to that before impact along the line of contact. This concept helps determine how much kinetic energy is conserved in the normal direction during the collision, which is essential for calculating post-impact velocities.

Energy Conservation and Elastic Potential Energy

When a moving object interacts with a spring, energy conservation principles are used to relate the kinetic energy of the moving body to the elastic potential energy stored in the spring. This concept is employed to determine the maximum compression of the spring and the corresponding force, as the work done on the spring is stored as potential energy.

Rotational Dynamics in Collisions

In problems involving collisions with bodies that can rotate, such as a quarter cylinder, it is important to consider rotational dynamics. This entails analyzing both linear and angular momentum, and understanding how the hit or impact point (relative to the body's center of mass) influences the rotational motion. Considering the moment of inertia and torque generated helps in accurately describing the system’s post-impact behavior.

Transcript

00:01 Question number 187 wants us to find the velocity of a sphere after it strikes a point on a cylinder at 5 meters per second.

00:09 Our problem is illustrated in figure 13 .187, which i've recreated here, our sphere, traveling at 5 meters per second, strikes the point a on our quarter cylinder, which is oriented at an angle of 45 degrees from the horizontal.

00:29 In addition, there's a spring attached to the back of the quarter cylinder with two cables at either side of it.

00:36 So when the sphere hits the quarter cylinder, the spring becomes compressed a bit.

00:41 We're given a few constants in relation to this problem.

00:45 Namely, we're given the mass of our sphere, m sub s, which is equal to two kilograms.

00:53 The mass of the quarter cylinder, m sub c, is equal to nine kilograms.

00:59 And the spring constant at k for the spring behind the quarter cylinder is valued at 20 kilonutons per meter.

01:07 Now there are going to be two different conditions for this problem.

01:11 The first is the initial condition before the sphere impacts with the quarter cylinder.

01:18 Here is the curve of the cylinder, and here is the velocity vector v .s of the sphere before it impacts.

01:29 The cylinder's surface.

01:31 Now there are two different axes we can draw in relation to the surface of the cylinder.

01:37 The first is the axis t, which is tangential to the cylinder's surface, and the second is the axis n, which is perpendicular to it.

01:48 Now our velocity vector is going to be oriented at a 45 degree angle from the normal axis, that axis ends.

01:59 There.

01:59 After our sphere's impact with the surface of the cylinder, the orientation of the sphere's velocity vector will be different with respect to these two axes.

02:11 What we're looking at now is our final condition.

02:14 Once the sphere impacts the surface of the cylinder, it imparts some kinetic energy to it and gives the quarter cylinder a bit of velocity.

02:24 Now we get a velocity vector v.

02:27 Sub c, the velocity of the cylinder, going at an angle of 45 degrees from the t axis.

02:34 Our sphere also keeps moving after the impact with the cylinder, but the direction of its velocity will change.

02:42 Now our velocity vector v .s will be oriented at some unknown angle theta from the t axis.

02:50 Now along both the t and the n axis in this problem, most of the momentum will be conserved.

02:58 First, let's look at the momentum conservation along the t axis.

03:03 We're only going to be looking at the sphere's momentum conservation.

03:08 So what we need to do is take the mass of the sphere, m sub s, multiply by the sphere's initial velocity, v.

03:18 Sub s, i, and set it equal to the mass of the sphere times the sphere's final velocity v, sub s, f.

03:29 We know the values of everything in this expression, except for the final velocity of the sphere, so let's go ahead and plug those in.

03:37 What we all have on the initial side is two kilograms times five meters per second, times the sign of 45 degrees, which we need to include, since the sphere is traveling at that angle with respect to the t.

03:56 Axis and set that equal to 2 kilograms times v sub s f.

04:04 If we just cancel out the 2 kilograms on either side of this expression, we find that the final velocity of the sphere after it's struck the quarter cylinder is the equivalent of 5 meters per second times the sign of 45 degrees, which multiplies out to give us a value of 3 .54 meters per second for our sphere's final velocity along the t axis.

04:34 Now, since our sphere's final velocity is at an angle from the horizontal, there are going to be two components for it.

04:42 We've just found the t component for it.

04:45 Now what we need to do is look for the velocity of the sphere along the, the n axis.

04:53 This time, in addition to the initial and final momentum for our sphere, we'll also have to look for the initial and final momentum of the quarter cylinder that the sphere impacts.

05:06 So our momentum conservation expression will look like this.

05:11 The mass of the sphere times its initial velocity, plus the mass of the cylinder times the cylinder's initial velocity, is equal to the mass of the sphere times the sphere's final velocity, plus the mass of the cylinder times the final velocity of the cylinder.

05:33 Let's plug in all our known values here.

05:37 We will have two kilograms times five meters per second, times the cosine of 45 degrees this time, plus 9 kilograms, the mass of the quarter cylinder, times 0 meters per second, because the quarter cylinder is not moving initially, is equal to 2 kilograms times v .s .f.

06:02 Our sphere's velocity component along the n axis will be different from what it was along the t axis, plus 9 kilograms times v.

06:14 Sub c, f, times the x, f, times the x, the sign of 45 degrees, the angular orientation of our cylinder's velocity component from the n -axis.

06:27 Working out our calculations, we end up with 7 .07 kilograms times meters per second is equal to 2 kilograms times v sub s f plus 6 .36 kilograms times v.

06:47 Subcf.

06:48 Let's first try to find an expression for v.

06:52 Sub s f, and then we'll worry about v sub c f later.

06:57 So let's rearrange this expression and put v.

07:01 Sub s f on one side.

07:03 We'll have two kilograms times v.

07:06 Sub s f is equal to 7 .07 kilograms times meters per second, minus 6 .07 kilograms times meters per second, 36 kilograms times v sub cf.

07:21 Now let's divide both sides by two kilograms.

07:25 That allows us to get rid of that constant on the left -hand side of this expression.

07:32 And now we get that v -sum -s -f is equal to 3 .54 meters per second minus 3 .18 times v -sub -ccc.

07:46 Okay, well, we've got an expression for the n component of the final velocity of our sphere here, but we have two unknowns, the other one being the final velocity of the cylinder after the impact.

08:02 What we can use to help us solve for both of these unknowns is the coefficient of restitution.

08:10 The coefficient of restitution, which is denoted by the letter e, is the ratio of the final and initial velocities of the sphere and the quarter cylinder.

08:24 The value of e will be equal to the cylinder's final velocity minus the sphere's final velocity, divided by the sphere's initial velocity, minus the initial velocity of the cylinder.

08:38 We're actually given a value of e...

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