Archive for the ‘Physics Test Questions’ Category
Atwood Machine Test Question
Raw Transcript
Hello, Everyone. This is Tom from everystepphysics.com. And I’m going to do an Atwood Machine problem. Show you how simple it is with my programs in the Titanium. And remember these are test questions. In other words, you’re going to get a test question like this and given certain masses and what are you going to put on your test? Are you going to put in the first diagram here? They want you to make a free body diagram that you’re gonna have to do when your own brain or own head. When it comes time to find the acceleration for this system,my programs as where it’s at. So let me do that for you right now. Physics to get to my menu. Physics menu. And we’re going to scroll down to at Atwood machine. And you notice these arrows go up and these go down so that means there’s more things in the menu. We’re gonna go up here to Atwood Machine. And you have to press Alpha before you enter anything in here. Notice that they give you the mass of one is 15 kilograms so we’re gonna press Alpha 15. Choose kilograms. Second mass is 28 kilograms. Alpha 28 kilograms. I always show you what you’ve entered so you can change if if you want. I say it’s okay. Here’s the little blurb on it, what it is. 2 masses hanging by a pulley. I show you a picture of it. And here’s the formula. Acceleration equals G times m2 minus G times m1. And summing the massesI do the calculations for you. Add it up, all step-by-step and here’s the answer. Acceleration equals 2.96. He has 2.977 meters per second squared. In the next problem, he solves for tension and I do that also in my Atwood machine. Notice we already have the answer here. Here’s the formulas and the step-by-step solutions 191.64 Newtons. He’s got 190. So you get an A on this problem on tests. You’re done within one minute and you’re onto the next problem. Pretty neat, huh? everystepphysics.com Go to my site, buy my programs. Pass your physics or calculus class. And subscribe to me see more movies like this. Thank you.
Question:
A block is pushed up a frictionless 30 Degree incline by an applied force as shown. If F=25 N and M=3.0kg, what is the magnitude of the resulting acceleration of the block?
Raw Transcript
Hello, Tom from everystepphysics.com and everystepcalculus.com. A mass on a ramp problem, test problem. Let’s do it. You have to put physics in the entry line of the calculator to get to my menu. And because the problem ask for acceleration, I always recommend you circle what’s asked for and then on the side, put down what’s given. In this case force Newtons and theta and then mass. So right now we’re gonna go to, it asks for acceleration. Press number two here. And then we’re going to scroll down to what’s given because there’s many of those in here. It’s not mass, and theta, and friction force. It’s mass, theta, and newton. There we go,letter E. You have to press Alpha before you enter anything in these entry lines. Alpha 3 kilograms. And then the force is alpha 25. And the angle given is 30. Alpha 30. I always show you what you’ve entered, you can change it if you want. I say it’s okay. Here’s the formula. Write this down on your paper right away. I like to use A equals acceleration because a lot of times and most the time in Yahoo or wherever you search, you don’t know what they’re asking for you don’t know what this letter means. Professors and people do not put that down. I like to put that down first in my problem so we all know what we’re talking about, exactly not almost. And here’s the to answer here: 25 times cosine at thirty minus 3, which is the mass times gravity, etc. divided by three, all the quantity, divided by three. Here’s the step-by-step 2.3119. You got an A on this problem. Go to my site, buy my programs everystepphysics.com. Pass your classes and subscribe if you wanna see more movies. Thank you.
A cart is given an initial velocity of 5 meters per second and experiences a constant acceleration of 2 meters per second squared. What is the magnitude of the carts displacement during the first six seconds of it’s motion?
Raw Transcript
Hello, Everyone. Tom from everystephysics.com. Here’s a test problem. I’m gonna read for you. A cart is given an initial velocity of 5 meters per second and experiences a constant acceleration of 2 meters per second squared. What is the magnitude of the carts displacement during the first six seconds of it’s motion? I always want you to in these problems to circle what’s asked for. In this case displacement and I want you to write on the side of the program what’s given. Which is in this case initial velocity, acceleration, and time. So, let’s the do it. You have to put physics in here with the open/close parentheses to get my menu. We’re gonna scroll down to displacement because that’s what it’s asked for. And we’re gonna choose what’s given which is velocity, acceleration, and time. There’s a formula for it. You have to press Alpha before you enter anything in these entry lines. So the the initial velocity is Alpha five and meters per second. I do the conversions. Is acceleration given? Yes. Acceleration alpha 2. And the time is alpha 6. I always show you what you’ve entered. You can change it if you want. I say it’s okay. Here’s the answer: letter C
A cart is given an initial velocity 5 meters per second and experience’s a constant acceleration a two meters per second square. What is the magnitude of the carts displacement during the first six seconds of its motion?
Raw Transcript
Hello, Everyone. Tom from everystepphysics.com Here’s a test problem. I’m going to read it for you. A cart is given an initial velocity 5 meters per second and experience’s is a constant acceleration a two meters per second square. What is the magnitude of the carts displacement during the first six seconds of its motion? I always want you to in these problems to circle what’s ask for. In this case displacement and I want you to write on the side of the program what’s given. Which is in this case initial velocity acceleration in time. So let’s do it. Put physics in here with the open/close parentheses to get to my menu. We’re going to scroll down to displacement because that’s what it asks for. I’m going to choose what’s given which is velocity acceleration in time. There’s a formula for it. You have to press Alpha before we enter anything into the entry lines. So the initial velocity is Alpha five and meters per second. I do the conversions. Is acceleration given? Yes. Acceleration Alpha 2. And the time is Alpha 6. I always show you what you’ve entered. You can change it if you want. I say okay. Here’s the answer. Letter C, on the test. Pretty neat, huh? everystepphysics.com Go to my site, buy my programs and subscribe to me so you can see more movies. Thank you
A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at +3.0 m/s. Find the final velocity of the 20-kg object after the collision.
a. -1.0 m/s
b. -2.0 m/s
c. +1.5 m/s
d. +2.0 m/s
Raw Video Transcript
Okay this is Tom from every step physics dot com we’re gonna work an actual physics problem here problem number 8 on my sheet which I’ll shall show you and let’s do it get to my menu you know it’s a collision or momentum problem..ah because the words and so you know scroll down here to collision
and this decide which from other words in the problem with what they’re talking about in this case number eight they’re talking about a head-on collision so we’re gonna choose 2 which is collides and we’re going to do head on which is 7 I show you the formula for the final velocity of mass 1 and mass 2 final velocity this is the actual formula you can see CR here whic is the coefficient of restitution that’s the bounce ability of a substance and sometimes they give you that to enter into the problem or not and if they do then you’re in good shape here but they find this out by experimentation and the lists of substances about what they’re coefficient of restitution is for instance a super ball which is the most bouncy thing on earth is .78 coefficient of restitution doesn’t come up to one because if it came up to one that would mean that if you drop it from a hundred feet it would bounce right back up in your hand which gravity won’t let it do it so there is no possibility of one clay as an example the coefficient of restitution of zero and that defines whether it’s a perfect elastic collision if it has a coefficient of one or a perfectly inelastic collision which is sticking to mass is sticking and not bouncing off at all so let’s continue here um here’s a note here, if an object is at rest that’s the m2 mass to enter here so we have to press alpha first gonna press alpha and put in the mass that’s not at rest which is 10 you choose kilograms that’s what it says the second mass alpha twenty also kilograms initial velocity is coming at it at what does it say 3 meters per second we choose number three here and it’s going in the easterly direction or the plus X axis direction choose that and we have the initial velocity of the the mass 2 which is at rest if its at rest we enter 0 Alpha 0 and coefficient of restitution if it’s not given just enter one alpha 1 I show you what you’ve entered you can change it if you want looks pretty good to me were going to say okay and I give you both final velocities of both because they’re both gonna bounce off somewhat go somewhere and so the velocity final velocity mass 1 was minus point one notices its going off to the left there with the minus sign after they hit and here’s the final velocity mass 2 and 2 is positive so it’s going off to the right two meters per second so that’s the calculations notice on the test they do have plus 2 is meters per second and so you circle that one and get a hundred percent on it pretty neat huh..every step physics dot com and because you know its a collision problem our momentum problem you’re going to scroll down to a collision if you didn’t know it was that
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
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