A 2kg block initially hangs at rest at the end of two 1 meter strings of negligible mass. A 0.003 kg bullet, moving horizontally with a speed of 1000 m/s, strikes the block and becomes embedded in it. After the collision, the bullet/block combination swings upward, but does not rotate.

a. Calculate the speed v of the bullet/block combination just after the collision.
b. Calculate the ratio of the initial kinetic energy of the bullet to the kinetic energy of the bullet/block combination immediately after the collision.
c. Calculate the maximum vertical height above the initial rest position reached by the bullet/block combination.

Hello, everyone. This is Tom for everystepphysics.com and everystepcalculus.com. I’m going to do a physics problem with regard to average force and I’ll show you how that works in my programs. Physics to get to my main menu. We’re going to scroll down to average force. I just did a video on average acceleration which is a change in velocity over the change in time. And the difference of force and acceleration is that mass is included. It’s interesting to note that mass in weightlessness is not weight but it is size. And for instance the space shuttle etc., in space has no weight however still has size and that makes a difference in the calculations. So we have average force on mass on mass in collision, okay? You have to press alpha before you enter anything into these entry lines, here. And the mass is given as alpha .045 choose kilograms. Initial velocity is alpha 44. Number four meters per second. Notice I did all the conversions for you. That’s one of the tricks in Physics. And then the final velocity is 0. There’s always a in every physics problem, there’s always a final velocity and an initial velocity. There’s also always a final time and a beginning time. Even though they don’t state it in a lot of problems, you have to think about that and and I help you with that as much as I can in physics. So the initial time given as a alpha 4.9 times E times to the minus 3. The final time, if not given, enter 0. It’s not given so we’re going to enter alpha 0. I always show you what you’ve entered, you can change it if you want. I say it’s okay. Here’s the calculations. Mass times the change in velocity divided by change of time It’s always this. The final velocity minus the initial velocity. Always the final time minus that initial time. We get 404.08 Newtons.

A greased pig slides from rest down a frictionless ramp into frictionless circular track of a radius of four meters. What is the magnitude of this centripetal acceleration is it reaches the lowest point? The diagram has the pig starting from a height that eight meters.

Raw Transcript

Hello, everyone. Tom from everystephysics.com and everystepcalculus.com. This is a physics problem. A greased pig slides from rest down a frictionless ramp into frictionless circular track of a radius of four meters. What is the magnitude of this centripetal acceleration is it reaches the lowest point? The diagram has the pig starting from a height that eight meters. So since this asks for centripetal acceleration, we’re gonna go to that in the menu. Physics to get to the main menu. Scroll down to the C section, all alphabetical. I Try to do that in of physics program problems is to search what it has asked for. That would be intelligent way to do it as far as I’m concerned. So we got centripetal acceleration. And you know the formula for that is V squared over R. But there is no V given. No velocity given in the problem. So, because we’re students of engineering or something like that, we have to think a little bit in our own minds. Everything can’t be just handed to us. And so we we need to say what is V and where can we get V. Well, we are going to practice on my program and we would then go. You would get out of the program. You do that by pressing on escape and back home to physics and you choose physics again. And now we choose centripetal velocity. And what we’re given here radius and height in the problem. Choose that. However when height is given, radius is not in the formula. So, we enter the height that’s given. Which is alpha, you have to press alpha before we enter anything into these entry lines, here. Alpha 8 meters. I always show you what you’ve entered, you can change it if you want. I say it’s okay. And the velocity is 12.5 meters per second. The formula is the square root of 2 times gravity times height. And we enter these in here and come up with these calculations. So now we go back to the main menu. Remember that. Write it down, of course. And we’re going to scroll down to centripetal, you can use second to go down screen by screen. Second to get to the menu quicker. And we’re going to choose centripetal acceleration again. We going to enter the initial velocity. Alpha 12.5. That’s meters per second, number 4. We’re going to enter the radius. Alpha 4 meters, number 3. I ask you if it’s okay. And here’s the formula’s. Velocity squared times 4 which is these calculations which is 39.06. And the answer to the problem is 39.1. In a multiple choice problem. So you circle that and you got it perfectly correct. Have a good one.

You set up a ballistic pendulum in order to conduct the experiment. The pendulum is made of a block of wood suspended from a set of strings. You fire the bullet into the stationary block and measure the speed at the combined bullet and block since the bullet lodges inside the block. I certainly would like to see these kids shoot a gun into the block, here for the experiments. I don’t know. You measure the mass of the bullet to be 4.2 grams, the mass of the block to be 500 grams, and the speed at the combined bullet and block to be 6.75 meters per second. Calculate the speed of the bullet (muzzle speed) and give your answer in meters per second.

Raw Transcripts

This is Tom from everystepphysics.com and everystepcalculus.com I’m going to an experimental problem in physics that they give you to experiment with in doing the lab class. And let’s get started. You set up a ballistic pendulum in order to conduct the experiment. The pendulum is made of a block of wood suspended from a set of strings. You fire the bullet into the stationary block and measure the speed at the combined bullet and block since the bullet lodges inside the block. I certainly would like to see these kids shoot a gun into the block, here for the experiments. I don’t know. You measure the mass of the bullet to be 4.2 grams, the mass of the block to be 500 grams, and the speed at the combined bullet and block to be 6.75 meters per second. Calculate the speed of the bullet (muzzle speed) and give your answer in meters per second. So, let me show you how my program would do for this lab thing in your homework a little bit, maybe. Physics to get to my menu. Scroll down to bulletin and pendulum which is the subject matter here. And we’re gonna wait for load. Notice when things are loading, you get a busy signal.here on your calculator indicating that is loading. And if it’s a long program like this one take a little bit in my but only once. After that, it loads very quickly. And they the bullet is going to embed into the block or the wood. So it’s, bullet in. We’re going to choose that one. And we want to get initialvelocity of the bullet. We’re going to choose that. And they give you instead of height, sometimes they give you height which there’s a way of calculating that and there’s also the velocity block and bulllet. So that’s what they’re giving, choose number two. You can choose the numbers before these items. And we’re gonna enter the mass. You have to press Alpha before you enter anything in here. Alpha 4.2 is given, 4.2 grams. It’s a very small bullet. And the 2 is here I just haven’t indicated it but it’s still in the calculations. And then we’re going into the mass of the block. Alpha 500 grams goes here. And the final velocity is alpha 6.75 meters per second, number three. In the positive direction. I always show what you’ve entered so you can change it in case you made a mistake and don’t like it. I say it’s okay. And here’s the calculations, here’s the formula. All the variables are put in. Notice it’s very clear. Initial velocity of one. Notice initial velocity, the used ot have a lot over professors use VO, which meant original velocity. But now lately they’ve been using initial and I like initial. And of course one is the bullet and two would be the block and bullet. I make everything clear of what in my programs of what I’m talking about. It always bugged me in physics that they throw around letters and minus signs and everything else and you don’t know what is going on. But in my programs , you do. Here’s the calculations. Here it is110 meters per second. Pretty neat, huh? EveryStepPhysics.com Thank you

A warehouse worker pushes a 10 kilogram crate along the frictionless floor, as shown in the figure,by a force of 10 Newtons that points downward at an angle of 40 degrees below the horizontal.

A. (10 points) Draw a free-body diagram for the crate.

B. (10 points) Find the acceleration of the crate.

Raw Transcript

Hello, Tom from everystepphysics.com and everystepcalculus.com. Physics problem off of a test. A warehouse worker pushes a 10 kilogram crate along the frictionless floor, as shown in the figure,by a force of 10 Newtons that points downward at an angle of 40 degrees below the horizontal.A for 10 points, draw a free-body diagram for the crate. B for 10 points, find the accelerationof the crate. So let’s do what my programs can do. Physics to get to my menu. Choose number 2, acceleration, because that’s what the problem asks for. We’re going to scroll down, you have to get a little practice of it because there’s many menus down here for choices and stuff. So you have to practice with my programs more than you practice with the sample tests given by the professor, or the book itself, or anything. So what’s given? What’s given is mass, the angle, and the force. And newtons. Not the friction force but the, the force, okay. We’re going to choose that. Press alpha before you enter anything into the entry lines, here. Alpha mass is given at 10 kilograms. 10 kilograms. The force is given too. Alpha 10 newtons. Angle is given at alpha 40 degrees. I always show you what you’ve entered, you can change it if you want. I say it’s okay. Here’s the formula for acceleration. The force times cosine of theta minus mg times the sine of theta, the whole thing divided by mass. Here’s the calculations all entered into that formula. Here’s the answer minus 5.5397 meters per second squared. With regard to the free body diagram. I’ll just say a couple things about that or a few things. The arrow points straight down always perpendicular to the horizontal is called the gravity force. The horizontal force is the arrow to the exact right along the horizontal. It’s called the horizontal force. The arrow pointing left which you would draw on this box and that is, that is the friction force. And the arrow pointing straight up is called a normal force. The arrow with the ten newtons is called the applied force. So you write that down, you get 10 points on your problem, on your test. And you’re going to pass that class or it’ll certainly help to pass the class. Have a good one.

Hello, everyone. This is Tom from everystepphysics.com and everystepcalculus.com. I’m going to do a Modified Atwood Machine but this time with an angle on the ramp. And so let’s get started. Physics to get into my menu. All Atwood Machines are in the menu under Atwood Machines. And we go from there. You notice the angle was not on the horizontal but up by the pulley because a modified, a true modified atwood machine has a 90 degree angle there. And any other angle all the way down to where just a plain atwood machine where the cord is just hanging around the pulley is between that position and the 90 degree position so any angle in between is what I’ve indicated here. This is 5 kilograms for, I’ll show you in a second here, we’re going to go up to this one would be modified atwood with angle. And I show you that, here’s the angle, here’s m1 m2 over the pulley. And we have to use Alpha before I put anything into these entry lines here. Alpha 5 kilograms. And Alpha for m2 is 7 kilograms and the angle is Alpha 40 degrees. I always show you what you’ve entered so you can change it in case you’ve made a mistake. You notice the formula here g times m2 minus g which is 9.81 times m1 times the cosine of theta. All over m1 plus m2. And you add your variables here. And here’s the acceleration 2.5913 meters per second squared. I also give you tension. Tension uses the acceleration that you found above here. All with the mass of m2. And here’s the tension 50.531. Pretty neat, huh? everystepphysics.com.
Thank you

A .0541 kg bullet is fired into a 3.25 kgblock on a ballistic pendulum. The bullet goes straight through the block (without changing the mass of either object) and exits with the speed over 183 m/s. The block rises to a maximum height to .177 meters. What was the initial speed of the bullet?

Raw Transcript

Hello, everyone. Tom from everystepphysics.com and everystepcalculus.com. Physics test problem. A .0541 kilogram bullet is fired into a 3.25 kilogram block on a ballistic pendulum. The bullet goes straight through the block (without changing the mass of either object) and exits with the speed over 183 meters per second. The block rises to a maximum height to .177 meters. What was the initial speed of the bullet? Let’s do it in my program see how that works. Physics to get to my menu. We’re going to scroll down to all alphabetical, down to bullet pendulum. And this is a bullet going through the block so we choose number two. You can use the numbers before this . And we want to find the initial velocity at the bullet, number two. It asks you if the height is given. Yes, it is. We’re going to choose number one. We’re gonna enter the mass of the bullet. You have to press Alpha before you enter anything in these lines here. Alpha .0541 kilograms. Mass of the block is Alpha 3.25 kilograms. The height is Alpha .177 meters. Choose number two. And the final velocity is Alpha 183 meters per second, number four. It’s in the easterly direction so we’re going to use that one. I always show you what you’ve entered. You can change it if you want if there’s been a mistake. I say it’s okay. We need to find the the final velocity of the block first which is 1.8635. Do this formula, here. Square root of 2 GH. Enter the variables. Here it is. And here’s the formula for finding the initial velocity of the bullet. Here’s the added the variables and here’s the answer 294.95 meters per second. Pretty neat, huh? everystepphysics.com. Go to my site, buy my programs. Thank you.

A gun with a mass of 2 kg fires out a .01 kg bullet with a muzzle speed 1,000 m/s. The bullet hits a 10 kg block resting o a horizontal surface. The block and bullet (embedded in the block) slide across the surface. If the coefficient of friction is .06. How far does the block slide?

Raw Transcript

Hello, everyone. Tom from everystepphysics.com and everystepcalculus.com. I want to do a physics problem off of a test that was, with respect to you know a bullet in a block sliding on a surface and how you find the how far the block slides. So here’s the test problem. A gun with a mass of 2 kilograms fires out a .01 kilogram bullet with a muzzle speed 1,000 meters per second. The bullet hits a 10 kilogram block resting o a horizontal surface. The block and bullet (embedded in the block) slide across the surface. If the coefficient of friction is .06, how far does the block slide? It’s irrelevant how much the gun weighs. So, that was in there to trick you, of course as if physics wasn’t hard enough. So let’s get started. Physics to get to my menu. We’re going to scroll down to bullet and pendulum. And bullet is is embedded in the block after the it has launched. It’s on a surface. And there’s friction and we want to find distance, number two. We’re going to enter the mass. We have to press Alpha before we enter anything in these at these entry lines here. So the mass is Alpha point 0 one kilograms. And the second mass Alpha is .01 kilograms. Initial Velocity is Alpha 1000 meters per second. And the coefficient is Alpha point 06. I always let you look at something to see if it’s correct. I say it’s correct. You can change it if you want. Here’s the formula. As you write this on your paper and you add the variables here. I always was like to show everybody what I’m talking about. Distance sliding for D. And here’s the calculations and here’s the answer .847 meters. Pretty neat, huh. everystepphysics.com.

A ball is thrown at a 40 degree angle above the horizontal, across level ground. It is released from a height of 2 meters above the ground with the speed of 18 meters per second. What is the height above the ground that the ball reaches?

Raw Transcript

Hello, everyone. Tom from everystepphysics.com and everystepcalculus.com. We’re gonna do a physics problem of a ball being thrown. So I’m going to read it to you. It’s off a test. A ball is thrown at a 40 degree angle above the horizontal, across level ground. It is released from a height of 2 meters above the ground with the speed of 18 meters per second. What is the height above the ground that the ball reaches? It gives you 5 answers. To choose. Multiple choice. So, let’s do it. Physics to get to my menu into the entry line here. And we’re going to scroll down to You can go down the menu quick but this is a long menu so a lot of things in here so you’re going to press second and the cursor second cursor to go down page by page. I did that and went down here we’re going to choose ball thrown. An object launched with angle and initial velocity, it doesn’t make a difference whether it’s a ball or a tennis ball or a rock. Whatever. It’s the same formula’s.So you have to kinda practice with my programs, here to get used how they work. And practice them more than Physics, actually. And I give you the answer here or questions whether it’s aboveground or ground level. It makes a difference. So we’re gonna enter the initial height that they give you.You have to press Alpha before you enter anything into these entry lines, here. Alpha 2 and I always give you choices in case they try to trick you with inches or feet or conversions. We’re going to choose number 2 which is meters. And then the initial velocity is Alpha 18 meters per second, number four. Enter the angle. Alpha 40 degrees. I always show you what you’ve entered so you can change it if you made a mistake. I say it’s okay.

First off I would like to say this is a greatprogram, I went out and got a Ti-89 titanium just so I could have your program for physics. I have been playing with it doing some problems and it works great. So what are the questions I have well here is a multiple part question I was asked to do last week but I am trying to figure out how to do all the steps on your program. I have been trying to find where I can calculate the thermal energy, potential energy etc...

Thank you for your time

DD

Tom- I just wanted to let you know, it was the difference between an “A” and a “C” for me. The product worked well and I was extremely pleased to post not just an “A” on the final, but an “A” for the semester. With your help, I’ll take another stab at Calc II next spring, but for now, I’m on Cloud 9. Thanks again and Merry Christmas! -Darrin

Tom- As good as your every step calculus programs, thank you -Kevin

Tom, thanks for the quick reply and help. I can’t explain to you how much I love this program and your customer service was greatly appreciated. -Jared