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Archive for the ‘Bullets’ Category

February 22, 2015 · by physics · Bullets

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.

February 19, 2015 · by physics · Bullets, Physics Help, Physics Problem Solver

Problem:

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

February 17, 2015 · by physics · Bullets

Question:

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.

February 16, 2015 · by physics · Bullets, Help with Physics, Physics Help

Question:

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.

February 3, 2015 · by physics · Bullets, Velocity in Physics

Question:

A pendulum consists of a 2 kilogram block hanging on a 1.5 meter
length of string. A ten gram bullet moving with a horizontal velocity of nine hundred meters per second strikes passes through and emerges from the block which was initially at rest with the horizontal velocity of three hundred meters per second. To what maximum height above its initial position will the block hang?

Transcripts

Hello, everyone. I’m Tom for everystepphysics.com and everystepcalculus.com. I’m going to do physics problem here. I’m going to read it right now. A pendulum consists of a 2 kilogram block hanging on a 1.5 meter length of string. A ten gram bullet moving with a horizontal velocity of nine hundred meters per second strikes passes through and emerges from the block which was initially at rest with the horizontal velocity of three hundred meters per second. To what maximum height above its initial position will the block hang? So let’s do it and show you how my programs work on this. You enter physics here to get to my main menu. We’re going to scroll down to bullet and pendulum. Which is right. And we’re going to enter our variables. You have to press enter before you enter anything in these entry lines here. So it asks to enter mass so we’re going put Alpha 3 and grams. I show you how to convert that to meters. And second mass is Alpha 2 kilograms. Initial Velocity is Alpha 1000 meters per second. And this is at Alpha 0 because the block is a rest hanging. And then we’re going into the final velocity is the bullet passes through Alpha 300 meters per second. And given in the positive direction east. Enter length of string is Alpha 1.5. I don’t know what they gave that to you in a problem for because it’s not entered into the calculations so must be to throw you off or something which professors love to do. This is at meters. So, I show you what you’ve entered, you can change it if you want. I say it’s okay. And it’s asking for the height so we’re going to choose number 3 here. And you have to find the velocity first. Which is .75 meters per second squared. I mean per second meters per second. And then we use that velocity to find the height which is .02867 meters. Centimeters would be .029 close to the 27 centimeter answer And that’s what you’d circle and get a 100 percent in this problem. You can choose 2 also and get to the and find the actual velocity of the block. Which is .75 meters per second. Pretty neat, huh? everystepphysics.com. Thank you.

January 6, 2014 · by physics · Bullets, Momentum, Physics Test Questions

A miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun’s bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g. With the system initially at rest, the radio controlled trigger releases the bullet horizontally causing the puck and empty gun to move with a speed of 0.500 m/s. What is the bullet’s speed?

a. 4.80 m/s
b. 11.5 m/s
c. 48.0 m/s
d. 12.0 m/s

 Raw Transcript

Hello this is Tom for every step physics dot com I’m gonna do actual test problem number seven on the sheet here which I’ll show you the problem and let’s get started you go to my main menu and because you know it’s a collision problem our momentum problem you’re going to scroll down to a collision
if you didn’t know it was that you and you thought it was momentum you would go to momentum and it’d take you to the same spot but and this one here decides certain words in a problem to see whether it’s a momentum problem or a collision problem in this case number seven says is what releases scroll down to that thing that separates momentum problems from collision problems that the two masses are at rest at the beginning so they can’t collide so just a little information there
and we’re gonna find the bullets final velocities so we’re join to choose final velocity of mass 2 we’re gonna enter mass 1 which is you have to press Alpha before you enter anything in my programs Alpha 120 and it says gram so we’re gonna choose grams a show you how to convert that put that on your paper and
the bullets mass is 5 grams does a quick conversion and the releasing velocity is alpha .5 meters per second and because we’re releasing or pushing you have to you push off to the left if you were it was happening in reality and you push to the right so our velocity is gonna go to the left which is number two
becomes a minus and I check and see if tell you show you so that you can see if you’ve entered the right stuff or if you are happy with it I say it’s okay and momentum is p1 is the momentum of mass one which is the gun and the puck an that’s the mass times the velocity so it’s minus .06 kilograms per meters per second and because one of the rules in physics P one plus p2 equals zero we have the p1 is minus .06 so that becomes a a positive number for p2 because they’re equal and then we can figure the final velocity
of the bullet which is 12 meters per second pretty neat huh, every step physics dot com

October 20, 2013 · by physics · Bullets, Physics Videos

 

Bullet Hits Steel Plate on TI-89: Raw Transcript

This is a video from every step physics dot com
demonstrating how my program work on the TI-89 Titanium calculator
and other calculators in the TI system for physics and calculus problems
here’s a problem that is on many tests in physics
and ah if a bullet traveling at three hunderd fifty
meters per second strikes a steel plate
at an angle of 30 degrees as shown It recochets off
with the same angle at a speed of three hundred twenty meters
per second what is the magnitude of the impulse that
the plate gives the bullet
and in my programs ah you
find first what the problem is asking in this case it’s asking impulse
and you go to my menu and you scroll down for impulse
and then you on the side of your problem
instead of looking at it or thinking about the problem at all you just
put what’s given here which is mass
twenty grams and initial velocity
which is three hundred fifty meters per second final velocity would be
three hundred twenty meters per second and then a thirty degree angle
and here’s the picture of what’s happening and this is really vector addition
so let’s get started here ah you have to press second alpha
to put these letters in here to get to my program menu
p h y s i c s and then you have to press alpha again
to put the open and closed parenthesis and press enter and you are into my menu
I’ve scrolled down to impulse here to save some time
but you start it brings you to the top of the menu
and then you have all these other choices and uhm
we get to impulse we press enter
gives you these variables here for impulse
and right now in this problem we want the mass
the initial velocity the final velocity
and the angle we choose that
we enter your variables you have to press alpha first
before we enter any variables into the entry lines of my programs
so were going to press alpha and put down
put in twenty it also might give you kilograms or grams
in this case it gives you the bullet as grams so I’m going to put that there
it shows you you take twent and divide it by a thousand
to get point zero two kilograms which is necessary for the formula
and the initial velocity your going to press alpha
and put in three fifty that’s the way the bullet is traveling
and it’s not in centimeters it’s in meters
so were going to choose number two you can press or scroll to that choice
final velocity alpha three twenty
meters again per second and then the angle is
alpha thirty I always show you what you’ve entered
so you can change it this looks good to me
it’s been changed from grams to kilograms etcetera
and this is vector addition who would have known that
if you just saw a problem in a test ah here’s the impulse vector resultant
ah the square root of the x axis and y axis
and the I of the impulse and here’s the formula
and then we add the the all the variables
you write this on paper you don’t think about it
and the I y is six point seven newtons second
and the answer is six point seven two three one
newtons second pretty neat huh
EveryStepPhysics.com go to my site
buy my programs and pass physics