Is it possible to calculate the impact force in real life if we dont know the 'time in contact' value. For example to compare two forces, one of static brick on some surface as F=mg and a brick free falling from some distance. We know that the impact force has to be higher then the static force for the bricks with the same mass but how to compare it or calculate it? Thanks for your answer!
Hello Michel: Is the force exerted by the wall in the x-direction related to the normal force (reactionary force) pushing back at impact and the force in the y-direction (parallel to the wall) the frictional force? How do these forces affect the ball upon impact? Moreover, if we assume there is no friction when the ball hits the wall, how would I find the average force that is on the ball from the wall? Many thanks!
Sir i have a question, does the force in the Y direction include the weight of the body, so if we want to calculate the force done by the wall to the body in Y direction would it be 5000-5x10=4950 N ?
what would happen if you did not have the mass of the ball and it was a linear collision no angles how would one find out the mass? I need the find the F and all i have is the velocity. Please help if you can thanks
Think about the unit square in the x and y coordinate plan. The x axis represents the cosine of an angle and the y axis represents the sine of an angle(relative to the x axis). Then you draw the radius of the circle which has an angle to the x axis of approximately 36.9 degrees. Now you can divide the radius into its components. You will see that the horizontal component is larger, which means that the cosine of the angle is the larger value( in this case between 0.6 and 0.8, the greater is 0.8)
Thank god! After watching I think I calc'd the Baltimore Bridge problem right roughly as a rough est.. It's been a yr or so since College Physics. Unfortunately, I destroyed my precooked crab dinner by boiling it. Gotta work on that. But I got the Physics right.
Thank you so much for all the examples regarding this topic! I'm working hard on understanding all the situations but I'm a bit confused about when the conservation of momentum is applicable or not. In this case, in which friction is involved, we can't use momentum conservation for reasons of simplicity in calculations, but the momentum is in fact conserved? Also, I understood that momentum conservation is applicable even for objects in free fall, is that right? Than you very much again for all your great work and invaluable help!
Momentum is ALWAYS conserved, regardless of the situation. WIth friction, without friction, etc. Whenever there is a collision of any kind, we can be guaranteed that momentum is conserved.
My values for Sin(53.1)x20, Cos(53.1)x20, Sin(36.9)x10 & Cos(36.9)x10 are completely different. They are as follows; 6.04, -19.06, -7.17 & 6.97. What am I missing here? Is there another calculation to convert these values to m/s?
I'm slightly confused. When a ball is thrown at wall at an angle, shouldn't its rebound angle be the same? Theta initial = -1 * theta final in this case?
That depends. If there is no friction between the wall and the ball, the incident angle will be the same as the "reflected" angle. But when there is friction, that will not be the case since there will be an impulse in the vertival direction.
initial momentum in the x-direction is not equal to the final momentum in the x-direction 5*20*cos(53.1), not equal 5*10*cos(36.9) how this possible if momentum is ALWAYS conserved??
Yes, this is something that often confuses students. Note that the wall is connected to the Earth and thus the remainder of the momentum is absorbed by the Earth.
Sir if the mass hits the wall exactly at 90degrees with a certain speed and bounces off again at 90degrees, so how can we calculate the impulse exerted by the wall on the object? Are we using pythagoras Therom to solve for the impulse?
Professor I have a question why after hitting the wall , the ball is not come with the same direction as the action and reaction pair is acting in equal but opposite direction ?
It depends on the amount of friction between the ball and the wall. And the friction only acts in the y-direction, thus energy is lost in the y direction but not in the x direction.
An kid is standing 15m away from the brick wall.The kid throws a ball toward the brick wall with 60 degree.Vf=Vo Vf=20m/s final theta= initial theta.In order to catch up the ball how far does the kid have to stand away from the brick wall?I tried to translate from Mongolian,Sir
Wow that is amazing. I have always been fascinated with Mongolia and I have a huge respect for its people. It is a beautiful but rugged country and the people thrive by working very hard.
That problem takes a lot of calculations, too complex to try and explain it in a comment, , but it is a very interesting problem. We'll prepare to do a video on it. The answer is 20.3 m
Assuming that the "wall" is the horizontal floor and the object is moving from left to right. Then in the x-direction there is no change of momentum. In the y direction impulse = I = change in momentum = I final = I initial = m V final - m V initial = (0.1 kg) (20 m/sec) (sin (30)) - (0.1 kg) (- 20 m/sec) ( sin (30))
awesome Gando When 2 objects collide the sum of their momentum before the collision is equal to the sum of their momentum after the collision When you look at one of the 2 objects in isolation, the change of its momentum is equal to the impulse = F * delta t
Thank you for all your videos! I would not survive Physics without you
excatly
if we find arctan(5000/10000) have any meaning of this angle?
Loved this lesson!
What about the final force F? You've calculated only the force in x and y direction.
you are an AMAZING teacher
My final is in an hour and this literally saved my life lol
Really helpful ....going to write my engineering science tomorrow with confidence 🙏😊😊
Is it possible to calculate the impact force in real life if we dont know the 'time in contact' value. For example to compare two forces, one of static brick on some surface as F=mg and a brick free falling from some distance. We know that the impact force has to be higher then the static force for the bricks with the same mass but how to compare it or calculate it? Thanks for your answer!
If you don't know the time duration of the impact you will not be able to determine the magnitude of the force.
TYSM Proffesor!!! Cool bow tie too :3
many thanks for effective and methodical teaching
Hello Michel: Is the force exerted by the wall in the x-direction related to the normal force (reactionary force) pushing back at impact and the force in the y-direction (parallel to the wall) the frictional force? How do these forces affect the ball upon impact? Moreover, if we assume there is no friction when the ball hits the wall, how would I find the average force that is on the ball from the wall? Many thanks!
Hi maria! I was really hopping someone got you an answer, because right now I'm having the same questions
How does one do cos(36.9 degrees), in one's head?
Great video.
you cant
Sir you are a lot helpful for students
sir,can we apply the impulse formula if two object of mass m and 2m are stricking,for finding the force with which they strike??
Only if you know 1) the time of contact, and 2) the force as a function of time. The other method would be to use: impulse = change in momentum
Sir i have a question, does the force in the Y direction include the weight of the body, so if we want to calculate the force done by the wall to the body in Y direction would it be 5000-5x10=4950 N ?
No, the force of the wall on the body is only the friction force. But the total force on the body is the friction force (up) - weight (down)
what would happen if you did not have the mass of the ball and it was a linear collision no angles how would one find out the mass? I need the find the F and all i have is the velocity. Please help if you can
thanks
Start with the conservation of momentum equation and determine what the unknowns in the equation are.
What is the trick you are using when you are finding the sin and cos of those angles in your head?
No trick. I just memorize a few of the angles.
Think about the unit square in the x and y coordinate plan. The x axis represents the cosine of an angle and the y axis represents the sine of an angle(relative to the x axis). Then you draw the radius of the circle which has an angle to the x axis of approximately 36.9 degrees. Now you can divide the radius into its components. You will see that the horizontal component is larger, which means that the cosine of the angle is the larger value( in this case between 0.6 and 0.8, the greater is 0.8)
Impulse and movementum in terms of energy???
Can't you use the pythagorean triple to come up with the module of the total force exerted on the ball?
The best way to find out is to try it and see if you get the same answer.
Fabulous video sir
How much time⌚ ⚽⚽🏈🏀 balls swing before it's going to loose it's energy
Completely?????
Should the two angles created by the horizontal line be equal to each other? Just like in the way of light reflection angle.
That depends. If there is no friction between the ball and the wall, then yes. But if there is friction, then no.
Thank god! After watching I think I calc'd the Baltimore Bridge problem right roughly as a rough est.. It's been a yr or so since College Physics.
Unfortunately, I destroyed my precooked crab dinner by boiling it. Gotta work on that. But I got the Physics right.
Sorry, no cookig classes on this channel. 🙂
mean the quadrant
Thank you so much for all the examples regarding this topic! I'm working hard on understanding all the situations but I'm a bit confused about when the conservation of momentum is applicable or not. In this case, in which friction is involved, we can't use momentum conservation for reasons of simplicity in calculations, but the momentum is in fact conserved? Also, I understood that momentum conservation is applicable even for objects in free fall, is that right? Than you very much again for all your great work and invaluable help!
Momentum is ALWAYS conserved, regardless of the situation. WIth friction, without friction, etc. Whenever there is a collision of any kind, we can be guaranteed that momentum is conserved.
@@MichelvanBiezen perfect!!! thank you very much!!
But sir, if we have to find the impulse how will be doing it with the forces that we obtained ?
Impulse is defined in two ways: 1) I = F (delta T) 2) I = change in momentum = m (delta v)
Helpful!👊
My values for Sin(53.1)x20, Cos(53.1)x20, Sin(36.9)x10 & Cos(36.9)x10 are completely different. They are as follows; 6.04, -19.06, -7.17 & 6.97. What am I missing here? Is there another calculation to convert these values to m/s?
Your calculator is in radian mode. You need to reset your calculator into degree mode.
@@MichelvanBiezen thanks!
I'm slightly confused. When a ball is thrown at wall at an angle, shouldn't its rebound angle be the same? Theta initial = -1 * theta final in this case?
That depends. If there is no friction between the wall and the ball, the incident angle will be the same as the "reflected" angle. But when there is friction, that will not be the case since there will be an impulse in the vertival direction.
There is energy lost during the impact. So the KE after the impact is less than before. That’s why the angle will change and not be the same.
How do we get Viy tho? like do we do the same Viy = Vi (sin53.1) or? NVM everything ok found it
Great. Keep it going.
@@MichelvanBiezen thank you 👍
initial momentum in the x-direction is not equal to the final momentum in the x-direction 5*20*cos(53.1), not equal 5*10*cos(36.9) how this possible if momentum is ALWAYS conserved??
Yes, this is something that often confuses students. Note that the wall is connected to the Earth and thus the remainder of the momentum is absorbed by the Earth.
Sir if the mass hits the wall exactly at 90degrees with a certain speed and bounces off again at 90degrees, so how can we calculate the impulse exerted by the wall on the object? Are we using pythagoras Therom to solve for the impulse?
If the angle is 90 degrees (from the horizontal), technically, the object cannot collide with the wall.
@@MichelvanBiezen so which formula am I going to use sir?
if the ball doesn't collide with the wall then we don't have a collision, so there is no impulse.
@@MichelvanBiezen okay but what about bounces, isn't it the same as collision?
Sir, what is the actual change in velocity of the ball
Vf = SQRT ( Vxf^2 + Vyf^2)
We found Fx and Fy but now wouldn’t we use Pythagorean’s theorm to find F by doing the square root of Fx squared and Fy squared?
You can do that or you can simply express the force (which is a vector) as the sum of its components.
Michel van Biezen thank you very much:)
@@MichelvanBiezen Adding them as vectors won't give the same result .Or Will it ?
Arent we supposed to find the resultant force? please help on the question # 11
We are supposed to find the force required to change the momentum of the ball. (Change in momentum = force x time)
@@MichelvanBiezen I'm cleared 👍. Thanks for clarification
Why is the Force in y direction pointed up?
The friction force opposes the motion of the ball.
Professor I have a question why after hitting the wall , the ball is not come with the same direction as the action and reaction pair is acting in equal but opposite direction ?
It depends on the amount of friction between the ball and the wall. And the friction only acts in the y-direction, thus energy is lost in the y direction but not in the x direction.
@@MichelvanBiezen is the total energy lost for friction in y direction or some energy lost because after hitting it also moves in y direction.
Energy is a scalar, and therefore is not associated with direction. But the friction force in this case acts in the y-direction only.
Sir what would happen if a gaseous molecule strike wall inspite of a ball ?? Please
To a molecule, the wall isn't perfectly smooth, but essentially the mechanics are similar.
Sir, i don't understand why Fy isn't negative, because if the ball is going downwards shouldn't the force in the y-axis be negative?
Force is a vector and the magnitude of a vector cannot be negative. Only if we write it in vector format do we need to worry about the sign.
because the Force_y changes the y velocity postively up, even though at the end it is still negative.
Goes from -16 to -6.
thanks professor , please keep teaching , we need you
sir u did a mistake by taking both the components of Vy as Vysino
+Rahul Tiwari I tried it myself and i don't see how, could you elaborate ? Because i'm not really good at trig so i might be making a mistake as well.
can anyone help me with collisions on linkage systems?
no
do the signs matter there ¿¿¿¿
When working on momentum problems, the direction (and therefore the sign) does matter.
An kid is standing 15m away from the brick wall.The kid throws a ball toward the brick wall with 60 degree.Vf=Vo Vf=20m/s final theta= initial theta.In order to catch up the ball how far does the kid have to stand away from the brick wall?I tried to translate from Mongolian,Sir
Do you live in Mongolia?
@@MichelvanBiezen Yeah Sir,
Wow that is amazing. I have always been fascinated with Mongolia and I have a huge respect for its people. It is a beautiful but rugged country and the people thrive by working very hard.
How about this problem.Do you understand it?
That problem takes a lot of calculations, too complex to try and explain it in a comment, , but it is a very interesting problem. We'll prepare to do a video on it. The answer is 20.3 m
what if you don't know the time
If you don't know the time of contact, then you need to know the change in momentum to find the force.
Im over here thinking this dude is doing an average value of a function with integrals wth
sir. i think that you accidentally made a mistake in the value of 20sin(53.1).
Yes, but I caught my mistake.
ohh. haha good for you sir..tnx for the help youve been giving me anyway.
What am I doing at 2am in the morning
If you're looking a our videos then you're doing a "magnificant" thing.
😘
theta initial=final theta,Vf=Vo.
theta initial is 30 degree,velocity initial is 20m/s.Ball mass is 100g.what will change in impulse be?
Assuming that the "wall" is the horizontal floor and the object is moving from left to right. Then in the x-direction there is no change of momentum. In the y direction impulse = I = change in momentum = I final = I initial = m V final - m V initial = (0.1 kg) (20 m/sec) (sin (30)) - (0.1 kg) (- 20 m/sec) ( sin (30))
Make sure you pay special attention to the signs since momentum is a vector quantity.
Wait, i thought momentum is always conserved , no matter what. Then , why the concept of impulse is the change in momentum??
awesome Gando
When 2 objects collide the sum of their momentum before the collision is equal to the sum of their momentum after the collision
When you look at one of the 2 objects in isolation, the change of its momentum is equal to the impulse = F * delta t
+awesome Gando The conservation of momentum just happens if the net force = 0 , that means Impulse = 0.
I hope it helps you.!
Only when there is no external force
Adatuluka mudala
Fibi ifi......
hocam anladıysam arap olayım
Probably he is good 😁 but IIT JEE ADVANCED In INDIA has better standards of questions than this one.