High school physics - NGSS
When an object with known velocity collides with an object at rest, and the velocity of one of the objects after the collision is known, conservation of momentum can be used to determine the velocity of the other object after the collision. It is important to consider the directions of velocity and momentum when interpreting signs in the calculation. Created by David SantoPietro.
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- I guess the apple overcame the string pulling it after it was hit by the orange .. doesn't that mean that a part of the momentum of the orange was used to overcome the string tension force .. so that less momentum will transfer to the movement of the apple .. !!
is that right ?(35 votes)
- We're assuming the stem was very weak and that the apple was not held tightly to the tree, otherwise yes, we would have to take into account the force from tension.(45 votes)
- If the orange hits the apple at the apex of its trajectory, then shouldn't its velocity be 0 m/s (since that's the point at which it changes direction)??(11 votes)
- What would happen if the two stuck together?(13 votes)
- Does the conservation of momentum apply in the dodge ball example discussed in last video ?(6 votes)
- Yes momentum is conserved for the dodgeball example, as Andrew says, but since your head is connected to your body, and your feet are stuck to the floor, some of the momentum gets transferred into the ground/Earth which is hard to account for, so conservation of momentum is not as useful in the dodgeball example.(5 votes)
- How do you find the velocity of an object after a collision if they don't give you the mass of the object?(2 votes)
- For a unidimensional collision between two bodies of masses M and m, each with a initial speed of V1 and v1, respectively. The speeds after the collision become V2 and v2.
Conservation of energy:
M(V1)² + m(v1)² = M(V2)² + m(v2)²
M[(V1)² - (V2)²] = m[(v2)² - (v1)²] 1
Conservation of linear momentum:
M(V1) + m(v1) = M(V2) + m(v2)
M[(V1) - (V2)] = m[(v2) - (v1)] 2
Since a² - b² = (a + b)(a - b) , dividing equation 1 by equation 2, we have
(V2) + (V1) = (v2) + (v1)
Which should probably be of some help(9 votes)
- How is momentum conserved when a ball hits the ground?(5 votes)
- Momentum is conserved in a system when there are no external forces acting on the system. Keeping this in mind, we can analyze the problem from the perspectives of different systems.
First, how about just the ball. When the ball hits the wall, the wall provides a normal force that is external to the system, so its correct to say that momentum isn't conserved.
How about the ball plus the wall? Now the normal force from the wall to the ball is an internal force, and momentum should be conserved if there are no other forces. If you just had a huge wall floating in space, the ball would transfer some of its momentum to the wall (which would correspond to an almost nonexistent change in velocity because the wall is so massive). But if the wall is like a normal wall and is fixed to the ground, the ground will exert an external force on the wall against the wall's minuscule motion, making momentum again not conserved.
But if you really expand your system, then the ball transfers some of its momentum to the wall, which transfers it to the ground, which is still in the system.(3 votes)
- If I throw a 5 kg ball at a wall with a velocity of 20m/s , the momentum of the ball is 100kgm/s. If the ball comes back with the same speed but in the opposite direction with a momentum of -100 kgm/s, isn't the law of conservation of momentum defied, since the wall is stationary (has a momentum of 0) and the ball lost 200 kgm/s?(4 votes)
- Actually the wall is not stationary. The wall has been deformed slightly and the motion of the wall (and the earth by extension) has been altered. It is just that the earth+wall is so massive compared to the ball that this change in velocity is so small as to be considered negligible. Hope this helps!(5 votes)
- Why can we not use the conservation formula if there is external impulse?(2 votes)
- Momentum is always conserved for any closed system.
Example if you have two billiard balls colliding, the momentum between the two balls remains unchanged before and after collision. If you push one of the balls, then the momentum of the two ball system will change since your hand is external to that system. Now if you change the system to include your body in addition to the two balls, the total momentum of the two balls and yourself will remain unchanged when you push a ball (assuming there are no other external forces such as friction). This example works best when you think about the two balls and yourself floating in space with no friction.(6 votes)
- How did an orange with total momentum of 2.0kgf managed to pass all of its momentum and then some more to move the apple with 2.1kgf?
And then even generate some more energy out of nowhere to also accelerate itself too; generating more work (10% more than all energy it had, 2.2kgf total from its 2.0).
If it could accelerate the apple as much as it like, as long as it also accelerate itself proportionally on the opposite direction, what is preventing it from accelerating the apple to 100kgf of momentum and itself to -98kgf? Or infinitely at opposite directions like those glitched videogames.(2 votes)
- (Just a note, we generally use kg*m/s as the unit of momentum and kg*m^2/s^2 or Joules as the unit of energy.)
Momentum is conserved in this situation because right after the collision, the apple's momentum is +2.1 kg*m/s while the orange's momentum at that instant is -0.1 kg*m/s. We add these together to get +2.1 kg*m/s+(-0.1 kg*m/s)= +2.0 kg*m/s, which is the same as the system's momentum right before the collision. Basically the apple can have a greater momentum than the orange initially did because the orange has a lesser velocity after the collision than it did before.
If we use the velocities found with conservation of momentum in the video, we find that the system actually loses kinetic energy during the collision. Kinetic energy = 1/2 mv^2, so the orange's initial kinetic energy is (1/2)(0.4)(5)^2 = 5 Joules. Since the apple begins at rest, its initial kinetic energy is 0 J, so the entire system's initial kinetic energy is 5 J. Substituting the fruits' final velocities into the kinetic energy equation tells us that the orange's final kinetic energy is 0.0125 J, and the apple's final kinetic energy is 3.15 J. Adding these together, we get the system's final kinetic energy: 3.1625 J.
As long as the total final kinetic energy was less than or equal to the total initial kinetic energy, the fruits could have any velocities. If the collision released some stored energy (like in a spring maybe), the total final kinetic energy could even be greater than the initial.(3 votes)
- One thing i have never understood in physics is the difference between internal and external force and what a system of particles is.Does this change depending on the system we consider?
Suppose there are two blocks attached to either ends of a pulley,are gravity and tension internal or external?
This thing has been bugging me since centre of mass and conservation of momentum.Please help..(2 votes)
- Gravity is external and tension is internal, given the description of the system as you gave it.
If the system is just one of the blocks, then both forces are coming from outside the system.
If you include earth in the system along with the two blocks, then gravity becomes an internal force in that system.(3 votes)
- [Instructor] There's a big fat, juicy apple hanging from a tree branch, and you want this apple, but you can't climb the tree? Luckily, you've got an orange in your pocket. So you take this orange and you chuck it at the apple, and it strikes it at the apex of its trajectory, causing the apple to fly off, and now you've got an orange and an apple. Now this is technically fruit vandalism if it's not your apple tree. So make sure you're only picking your own apples or you've paid someone to do this and everything's legit. But this is also a collision problem. And in physics, you could solve for the velocities involved and the mass and the momentums involved by using conservation momentum, if we had some numbers. So let's give ourselves some numbers. Let's see if we can solve for some quantities here. So let's say this apple, I told you it was big and fat, let's say this apple was 0.7 kilograms. Let's say the orange is probably not as big, so 0.4 kilograms. And let's give some numbers, let's say the speed of the orange right before the collision was five meters per second. So since this orange was at its apex, right? It was just heading in this way and it was going horizontally at that moment, five meters per second, right before it struck the apple. And let's say the apple was moving three meters per second right after the collision, so right after the orange hit the apple, the apple starts flying at three meters per second. One question we could ask, one obvious question, is, if this is the speed of the apple after the collision, what was the speed of the orange after the collision? What was the velocity of the orange? And which way was it going? So we'll call it VO for V orange, and was that orange going left or right immediately after the collision took place? Sometimes this isn't so obvious, so let's see if we can solve for this now. We've got enough to solve. We can do this using conservation momentum, and conservation momentum says that if there's no external impulse on a system, and our system here is the orange and apple, if there's no external impulse on these fruit, that means the total momentum before the collision took place, so right before the collision took place, has got to equal the total momentum right after the collision took place, and it's important that we denote right before and right after. We're not talking, like, right as someone threw this fruit before it got up here, and we're not talkin' finally, like after the apple gets back down to the ground. You can't do that. For most collision problems, you're gonna want to consider right before the collision and right after, and the reason is, remember, this formula here is only true if there's no external impulse. So only if external impulse is zero. And you might be like, well, isn't it always zero? Shouldn't it be zero in this case? It's not so obvious. If you're clever, you might be like, wait a minute, there's a force of gravity on this apple. There's a force of gravity on the orange. So doesn't that mean there's an external force? And if there's an external force, doesn't that mean there's an external impulse? And doesn't that mean that the momentum shouldn't be conserved? Well, not really. And the reason is, for one, this force of gravity is directed downward, so it's only gonna effect the vertical momentum. And we're just talking about the horizontal momentum here. I wanna know what happens to the horizontal momentum of this orange, but secondly, the definition of impulse is that it's the force that acts multiplied by the time duration. We're gonna say that if we consider right before the collision and right after the collision is our initial and final points, this time interval's gonna be so small that the force of gravity is gonna have almost no time to act. And because it has almost no time to act, it has almost no external impulse. So we're gonna ignore the impulse due to gravity, because it acts over such a small period of time and it's such a modest force, which means we get to use conservation momentum for our system. So what is this gonna look like? Well, the momentum formula is mass times velocity, so the initial momentum of the system, let's see, I'd have to add up initial momentum of the orange is 0.4 kilograms, that's the mass, times the initial velocity, that's five meters per second, plus, I'm gonna add to that the mass of the apple, 0.7 kilograms, multiplied by the initial velocity of the apple. What was the initial velocity of the apple? It wasn't three, people try to plug in three, that wasn't the final velocity of the apple. The initial velocity of the apple was just zero, 'cause it was hangin' on a tree branch and was just sittin' there. So this is gonna be zero. What that means is, this entire term is gonna be zero, 'cause zero times 7.7 is still zero. So this term just goes away. It's gonna be zero equals the final momentum. All right, so we added up the total momentum of our system initially, now we're gonna add up all the momentum of our system finally. So 0.4 kilograms is the mass of the orange, multiplied by, we don't know the final velocity of the orange, that's the thing we wanna find. So I'm gonna write that as VO, for V of the orange. This final velocity of the orange is what we wanna find. This term here represents the final momentum of the orange, but I have to, I can't stop yet. I have to add to that the final momentum of the apple. So remember, when you're writing down conservation momentum for a system, the statement isn't that the initial momentum of one object equals the final momentum of some other object. It says that the total initial momentum of the entire system equals the total final momentum of the entire system. So I'll take my .7, multiplied by my final velocity is three meters per second for the apple, and now I can solve. So we can solve this, I've only got one unknown. So .4 times five is two kilogram meters per second, plus zero, I'm not gonna write that, 'cause it would just take up space. Equals .4 times VO is the unknown, so .4 kilograms times the unknown VO, and then plus .7 times three is gonna be 2.1 kilogram meters per second. So our system started off with two kilogram meters per second of momentum to the right. That's what the orange brought in. And our system ends with 2.1 kilogram meters per second to the right, which is what the apple has, plus whatever momentum the orange has right after the collision, and you might look at this and be like, wait a minute, hold on, we screwed somethin' up. Two kilogram meters per second equals 2.1 kilogram meters per second plus something? How is this right hand side ever gonna equal two if it's got 2.1 to start with, but remember it can, momentum is a vector. And vectors can be positive or negative, depending on whether they point right or left. So this just tells us, okay, the orange is going to have to have momentum leftward after the collision, so that this whole right hand side can add up to two again, and we know we're gonna have a final velocity of the orange that's negative. But you don't have to be clever. If you just wanted to solve this equation, it'll tell you whether it's going right or left. I'll show you why. If we just do this, two minus 2.1. So if we subtract 2.1 from both sides we'll get negative 0.1 kilogram meters per second and that's gonna equal this final momentum of the orange, so equals 0.4 kilograms for the orange, times VO, the final velocity of the orange, and now if we just divide both sides by .4, we'll get negative 0.25 meters per second. That's the final velocity of the orange, and you realize, oh, I didn't have to figure out the sine beforehand, I could just solve and in the conservation of momentum formula will tell me whether it's going right or left, 'cause if I get a negative sine here, it just says, oh that velocity had to be directed in the negative direction in order to conserve momentum in this case, so this orange, right after the collision was heading leftward, that's what the negative sign means, and this .25 means it was heading leftward at a rate of .25 meters per second. So recapping, we could use conservation of momentum to solve for an unknown velocity by setting the total initial momentum of the system equal to the total final momentum of the system. We gotta be careful with negative signs. If there was an initial velocity that was negative, we would've had to plug in that velocity with a negative number, and if we find a negative velocity to end with, that means that quantity of that velocity was directed in the negative direction. Also, we can only use conservation of momentum whenever the external impulse is zero, which is why we consider points immediately before the collision and immediately after, so that this time interval is so small, gravity can't apply much of an impulse at all, and I should say, we should assume that this stem was barely hanging on by a string, 'cause if this stem was secured to the tree, then there would've been and external force that coulda caused an external impulse. So let's assume this apple was already just about to fall off, and the slightest of forces could knock it off. That way, there's no external impulse and we get to use conservation of momentum.