I know you’ve all been on the edge of your seats, couches and bikes in the past week wondering if both mechanical energy and momentum are conserved in a collision.

Well, the answer is…sometimes. Let’s delve.

A collision in which the internal forces are conservative is an elastic collision. In such a case both total momentum and total energy are constant. So, during a collision of this ilk, all the initial KE not required to conserve momentum is converted to U (potential energy) and, after the collision, all the U is converted back to KE. Think about our mini-billiard balls colliding!

Now, let’s think about a “fruit fight”. If two apples are traveling horizontally through the air and they collide, they will explode. Our flying fruit have no chance to return to their original form, in some cases they might even stick together. This is an example of an inelastic collision and it is here where mechanical energy is not conserved and we say KE has been dissipated. And we can quantify the amount of KE that was lost by comparing the KE before the interaction with the KE after the collision. If the objects stick together, it is referred to as a completely inelastic collision and we use the combined mass in the kinetic energy equation.

So let’s take a look at the interaction between Sydney and Jaromir again here, along with another example of a head-on collision during a stunt scene in Mission: Impossible 13; Zero Velocity, to see if we can grasp the concept.

In filming Mission Impossible 13: Zero Velocity, a 1500 kg car moving north at 35.0 m/s collides head on with a 7500 kg truck moving south at 25.0 m/s. The car and the truck stick together.

(a) How fast and in what direction is the wreckage traveling just after the collision?

(b) How much mechanical energy is dissipated in the collision?

Have a go…A large bass with a mass of 25.0 kg swims at 1.00 m/s towards and swallows a small fish that was stationary. If the smaller fish has a mass of 1.00 kg, what is the speed of the bass immediately after it dines? What type of collision is this? What is the percentage of kinetic energy dissipated, if any? Answer

And finally, I wanted to end with a look at how collisions in two dimensions can be quite daunting. I would love for you to appreciate this complexity qualitatively more than quantitatively for the moment!!

Thanks for being such amazing students!!

This video depicts was one of my best Halloween costumes. I even had the squash racquet!! I lost the contest to a Banana but I love airport walkways! The feeling of being fast or the accomplishment of treading against the moving earth are exhilarating!!

If you recall, in linear motion with constant acceleration, we were able to use a set of equations that related position, velocity and acceleration for analysis. Similarly, in rotational motion with constant angular acceleration, the same relationship appears for the rotational equivalents of the corresponding quantities in straight-line motion!! Therefore, we can define angular acceleration as the rate at which the angular velocity changes with respect to time in much the same way that we defined acceleration of an object moving in a straight line.

Now just as in straight line motion, a positive value of angular acceleration does not necessarily mean the extended object’s rotation is speeding up and a negative value doesn’t necessarily correspond to slowing down. If angular velocity (w) and angular acceleration (alpha) have the same algebraic sign, the extended object is speeding up. Conversely, if w and alpha the opposite sign, the object is slowing down. Now, let’s have go at the spinning top example!!

OK!! Let’s think practically and qualitatively about what’s happening here (also a bit old school!).

Optical disk drives, such as CDs and DVDs, have information stored in small depressions “burned” onto the disk surface. Hence the term, “burning a CD” which is what we were doing when we downloaded songs off Napster at the turn of the century! Anyway, a laser is used to detect changes in the depth on the surface as the disk spins. A change is interpreted as a 0 and no change in depth is interpreted as a 1. There are millions of these burns on a disk’s surface and the DVD player retrieves this stored information. Now, these optical technologies are designed to maintain a constant rate of information retrieval. If the laser samples depth on the surface at constant intervals of time, why do you think the angular velocity of the disk cannot be constant?

Let’s see if we can answer these questions!!

(1) Orla pushes a merry-go-round that has a diameter of 4.00 m and goes rest to an angular speed of 18.0 rpm in a time of 43 s.

(a) Calculate the angular acceleration of the merry-go-round in rad/s^2.

(b) Calculate the angular displacement (in radians) of the merry-go-round during this time interval.

(c) What is the maximum linear speed of Orla if she rides on the edge of the platform?

(2) Interpreting Data!!

I have provided the answers for the Pure Roll worksheet and the notes for the Hoop and Disk race scenario about rotational inertia. I have also written out a couple of problems that might put a bow on these moments (of inertia).

Thanks so much!! Y’all are wonderful!!

Please enjoy this feature of Mr. Burns driving and colliding with everything!!

“Once a song gets momentum and gets away from you, that’s a good sign” – Dave Matthews.

“What?” – Darren

So let’s begin our discussion about momentum by defining a collision as any interaction where the internal forces dominate. If a hockey player checks another, the internal forces are strong enough to seriously injure the other player. The external forces of friction very minimal. Also evident in the tennis serve, the ball and the racquet forces are so large that the ball distorts noticeably. However, by comparison, the external forces of gravity on the ball and your hand on the racquet are very insignificant.

In summary, if the internal forces during a collision are much greater in magnitude than the external forces, the total momentum of the interacting objects has the same value just before as just after the collision. This means that momentum is conserved and leads us to one of the incredible fundamental laws that governs the interactions of all objects in the universe, the Law of Conservation of Momentum.

So, I wanted to give you a couple of problems to work on. This unit is all about practice and keeping track of your direction and your variables. Good luck!!

(1) Blythe and Bart are ice skating together. Blythe has a mass of 50 kg and Bart has a mass of 80 kg. Blythe pushes Bart in the chest when both are at rest, causing him to move away at a speed of 4 m/s. Determine Blythe’s speed after she pushes Bart?

(2) A 2 kg object is moving east at 4 m/s when it collided with a 6 kg object that is initially at rest. After the collision, the larger object moves east at 2.00 m/s. What is the final magnitude and direction of the velocity of the smaller object after the collision?

Thanks so much everyone!! You’re doing amazing!!

This is a perfect video in so many ways and extremely relevant to our lesson for the day!!

So we’ve all used the idiom “it’s just like riding a bike” in regards to an action that has become second nature to us. But our primary mode of recreational transport as young people exemplifies one of my favorite physical principles, Pure Roll.

Pure Roll is commonly referred to as “rolling without slipping.” In this situation we see a specific relationship develop between speed, v, of the CoM and the angular speed, w, of the wheel’s rotation, v = rw. The top of the bicycle wheel and the bottom are going this same speed, v = rw, but in opposite directions. Now, if we add in the velocity of the CoM, we find that, at any instant, the top of the wheel is going twice as fast as it’s center and the bottom of the wheel is at rest with respect to the ground! Whoa!! Here, the force of friction does no work because there is no relative motion!! But friction does ensure that a portion of the total kinetic energy is rotational kinetic energy. Now, please enjoy this edition of Darren and the Tiger concerning Pure Roll.

Ok, so I would like you to have a go at this scenario from the workbook in the next couple days. It is a great example of crossover concepts!! If you use Conservation of Energy, Rotational Inertia and your qualitative reasoning skills you should do great!!

Also, use your same skills to answer this question: If my Beatles record has a mass of 125g, a radius of 30cm and spins up to 33 1/3 rpm in .80s, what is the power delivered to the turntable??

Thanks so much everyone. It was so awesome to see you these past couple days. I will be doing more problems during class time using the screen share so let’s keep the momentum going!!

Mr. C

Welcome to the first episode of Physics Clas!! I truly hope you all are doing well and I wish I was going to school in the morning to see your happy faces!!

In the blog, I simply wanted to take a breath and go back to some simple concepts about momentum, its conservation and review some basic problem solving techniques with individual objects before we moved forward into collisions. So, I hope it’s helpful and sort of resets us as we move forward. I’m still working on my camera placing because iMovie shrinks the frame a wee bit but overall I’m pretty psyched about the video, except for my hair!! If you could comment on the sponsor of the episode (there is a little commercial in there) that would be great!!

So, we remember that momentum is equal to the objects mass times its velocity (p = mv) and that momentum is going to be conserved as long as there are no external forces acting upon the object. Also, since velocity is a vector, so is momentum which means it has a magnitude AND a direction. Finally, the impulse-momentum theorem states that the force applied over a time interval is equal to the change in momentum, (Ft = (delta)p, with delta implying “change in”).

So here is a question for you wrap your head around that touches on all the bits from this episode.

(1) A sudden gust of wind exerts a force of 10.0 N for 1.2 s in a direction opposite to the flight of a red-tailed hawk, whose speed was 5.00 m/s before the gust of wind. As a result, the bird ends up moving in the opposite direction at 7.00 m/s. What is the mass of the hawk??

Have fun!! Thanks again!! Y’all are amazing!!!

Welcome to the the Physics Clas blog!! I hope you’re all doing great!!

In this episode, I simply wanted to reacquaint us with Rotational Inertia or the “Moment” of Inertia (I). So I go though the bird/insect scenario again and I review the frisbee problem because many were absent and I didn’t go over it with the second group. Now, the video came out good except it shrinks a little when you put in in iMovie so the top couple lines are chopped on the problems but I read them out loud. So please forgive me as I navigate the precise placement. I hope you find it fun (there’s an outtake right at the beginning) and if you could comment on the sponsor of the episode (there’s a little commercial in there somewhere) that would be great!!

So, after I review Moment of Inertia, I talk about how how we can use the principle of the Conservation of Energy for an extended object that is rotating as its center of mass is moving through space. In such a situation, it turns out that we can write the extended object’s Total Kinetic Energy (K) as the sum of its Rotational Kinetic Energy (Kr) and its Translational Kinetic Energy (Kt). Specifically, the movement of the CoM and the rotation around the CoM!

Here are the two questions I would like you to have a go with!

(1) Three beads with masses m1<m2<m3 are each placed on a rod at distances of La<Lb<Lc from the axis of rotation of the rod with a rotational inertia of Irod (meaning the rod itself has a rotational inertia that is part of the total). Express the smallest rotational inertia of the system in terms of Irod and the masses and locations of the beads.

(2) A potter’s flywheel is a circular concrete slab, 6.5 cm thick, with a mass of 60.0 kg and a diameter or 35.0 cm. The disk rotates about an axis that passes through its center, perpendicular to its surface. Calculate the speed of the slab about its center if it’s Rotational KE is 15.0 J. The (I) for a disk is 1/2mr^2.

Thanks so much everyone!!! Y’all are simply amazing!!