Physics 2 - Motion In One-Dimension (20 of 22) Two Objects
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- Опубликовано: 30 сен 2024
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In this video I will show you how to calculate the height when 2 objects will pass each other - one object thrown upward and the other dropped (free-fall).
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I wish I could have found this video sooner, I've been losing my mind for about two hours now.
Rick Porter same, i hate physics
Atleast it took you 2 hours, it took me days
i dont understand why y1 = y2. i would thought the equation y= y0 + v0 + 1/2gt^2 is relating displacement. By saying y1 =y2, are you saying that they both displaced the same amount?
You can try with this method:
Let both the balls meet at 'x' m from the ground. Now, Ball 1 will travel 'x' and Ball 2 has to travel '20 - x' m to cross each other. At this instant, 't' becomes equal.
For Ball 1 : y = y0 + v0*t + 1/2gt^2
==> x = 0 + 25*t - 1/2gt^2 ....(1)
For Ball 2 : y = y0 + v0*t + 1/2gt^2
==> 0 = (20 - x) + 0*t + 1/2gt^2 ....(2)
Here y0 = h is not taken from the ground level for Ball 2 because we're only interested in distance travelled by Ball 2 to cross Ball 1 which = 20 - x. This '20 - x' distance decreases as the Ball 2 reaches towards Ball 1 and becomes 0 when they meet.
Equate them you'll get the answer.
y1 and y2 here are the vertical y positions. Because the problem wants you to solve at what height they will meet, so you can assume that the two balls will arrive at the same height at a certain moment. That's why you should assume y1 = y2.
It is height from the ground when they meet
This was an outstanding way to derive the t = 0.8
I know you're super used to deriving answers in this way and have derived far, far more difficult problems than this using much cleverer logic but to my inexperienced mind, this was awesome lol
By looking at 25m/s for the ground ball and 20m high for the second ball, we could see immediately that at the very least the balls would pass each other in less than a second.
isn't g different for both the objects. For one it is +g and for the other it is -g.hence we can't cancel 0.5gt^2 on both the sides ,right?
Prathap,
g is the acceleration due to gravity and it is always -9.8 m/sec^2 regardless of what a projectile is doing
For the equations of kinematics to work properly, you must always consider g negative when solving projectile problems.
i got Yf2 and Yf1= 23.13m idk Why i got everything right and i used other calculators same idk why
There are often different ways in which you can solve these types of problems
@@MichelvanBiezen Thanks
That was a bit confusing because I thought x should not be equal for both. What I tried to do and failed was if ball one travelled 1 m then ball two 20-1 = 19m, if ball 1 travelled 2 m then ball two travels 20-2 or 18 but I couldnt take it from there. Basiciallly time of x(ball one) = time of (20-xball one) for the second ball. I do not know what to do next.
He means the y vertical position, not the displacement.
Sir, why we don't use the second ball for the equation to see when they will meet? and how we can know if the balls will meet when y=y2 ? maybe because of the vi ball number one will reach the second one with in bigger distance . i hope you get it because my English language is weak :( thanks
What about the second T at the Y2 side. Please can anyone help me?
For part b, the first ball will be higher than 20 m after 1 second, so they will never be at the same height, after you drop the second ball.
I love it when terms cancel! ❤ 😂
Sure do.
Sir, how about finding the time?
Given:
1st object speed thrown downwards = 20m/s
2nd object speed thrown upwards = 30m/s
At a building with a height of 230m
Time =?
I think you'll need to know the Y for that to work
Sir when two cars, or, say a truck and a car are moving in the same direction at different velocities and the driver suddenly sees the truck, how do you calculate the minimum acceleration required to avoid collision?
Look at # 6 in this playlist to see how that works: PHYSICS 2.5 - 1D MOTION : GRAPHIC SOLUTIONS (you just have to add the motion of both).
Ramneet Singh Kumar could u pls give some more information regarding your question???
At the beginning, how do you say that the distance traveled by both the balls would be the same when they meet each other. The first ball is thrown at a velocity of 25m/s but the second ball is just dropped at 0m/s ?
Not the distance, but the vertical POSITION y.
You can try with this method:
Let both the balls meet at 'x' m from the ground. Now, Ball 1 will travel 'x' and Ball 2 has to travel '20 - x' m to cross each other. At this instant, 't' becomes equal.
For Ball 1 : y = y0 + v0*t + 1/2gt^2
==> x = 0 + 25*t - 1/2gt^2 ....(1)
For Ball 2 : y = y0 + v0*t + 1/2gt^2
==> 0 = (20 - x) + 0*t + 1/2gt^2 ....(2)
Here y0 = h is not taken from the ground level for Ball 2 because we're only interested in distance travelled by Ball 2 to cross Ball 1 which = 20 - x. This '20 - x' distance decreases as the Ball 2 reaches towards Ball 1 and becomes 0 when they meet.
Equate them you'll get the answer.
Sir. Did you calculate the same time based on your assumption that y1 equals y2. I get why y1 = y2 because they should be at the same height at a certain moment. If you plug in the equations of y1 and y2, then it's completely reasonable. But it's interesting that if you think backward, the two objects released at the same time has traveled an equal amount of time but they will not necessarily meet at the same height right?
The first ball has so much velocity that it can reach high height before the second ball has time to accelerate
Plus it's not exacly at 20 m but at 17 m so it's not big of a deal
Thank you so much! Very well explained!
Glad it was helpful!
If your one object is falling down and the other is thrown up, shouldnt one of the g= -9.8 and the other g=+9.8 so they cant cancel out.
No, g is always "-" when used in kinematics equations.
Never knew that thanks :D
saima arshad g is negative cis it's in the negative Y direction
In the second equation when we found the height why we took negative g while the ball is going up ?
why we are assuming the objects have the same height ?
First start with assuming that they will have the same height when they pass one another. If they don't pass one another, you will end up with an impossible answer which then implies they will not pass one another. But a quick check on the initial velocities should make it obvious that they will. (The bottom object will move upward for at least 2.5 seconds given its initial velocity)
Thank you
Physics is much more easier with you!
Thank you!
Nicol,
Thanks for the feedback. I try to show my students that physics can actually be interesting and fun once we understand it.
I disagree physics can be fun until you understand too much, anything that Is easy is no fun.
it is interesting for me now since I'm in 11th grade and did not learn almost any physics until now.
Very helpful
Sir,In physics playlist 2 (one dimensional motion) video no. 20,you assumed that the two objects will meet each other at the same height.Sir,I just want to know why you wrote y1=y2.If you help me i would be more grateful to you.
+Maheen Mahfuz
The technique to solve any physics problem is to find something that is equal to each other. In this case, since the problem wanted you to find the location where both objects are at the same height and y1 represents the height of object 1 and y2 represents the height of the second object (as a function of time) we just set those two equal to each other to find the time. (or we set the time equal to each other to find the height.
Is there any other method to solve this???
There are usually multiple methods to use any problem. We are just illustrating one of them.
Sir, I just want to understand why did you put g= -9.8 for the Ball 1, as "from my knowledge" it should be (+) for the Ball 1 and (-) for the Ball 2
In problems like this one, g represents the acceleration due to gravity. That is the same for all objects regardless of what they are doing. Therefore in projectile problems, g is always equal to - 9.8 m/sec^2
Does this work only if the balls are equal in weight? I have a problem where the ball on the ground is twice the mass of the ball being dropped
No, it works exactly the same regardless of the weights of the balls.
total height is 20m how can the two balls meet at 169m
There is a decimal point there. (16.9 m)
i got 16.8 but may be that is because i took g 10m/s^2
sorry my mistake
Sir how could we consider the Acceleration is positive in the both situations ?
as you canceled it from both equations without adding any marks
actually I'm so confused because of the direction of the Gravity sometimes we put it + others - .. :(
In projectile motion the acceleration due to gravity (g) is ALWAYS negative as is the case in this video. Note that the substitution of (1/2) g = - 4.9 m/sec^2.
+Michel van Biezen Thank you so much for your fast and useful responding i appreciate that sir :)
i would like to say thank you very much sir!!
Thanks Michael
when would negative displacement considered?
Displacement, velocity, and acceleration are vectors and therefore have direction (as well as magnitude)
@@MichelvanBiezen Then, why didnt we take -20m sir?i am a bit confused.
It depends on where your reference point is. Here the ground is considered 0 height.
Biezen , my legend. 1:28 "eachother", what a great way to pronounce that word ! :D
Thank's
please can you do this if one of balls does a 2 dimensional fall?
Any horizontal motion is independent of any vertical motion. The only thing that matters in this type of problem is the vertical motion as explained in the video.
@@MichelvanBiezen okay thank you :)
Sir how did u taken y1=y2
Ritik Raj it is height from the ground when they meet
sir I didn't understand why the g is not negative...
g = - 9.8 m/sec^2 as used in this video
If one object is falling down and the other is thrown up, shouldn't one of the g= -9.8 and the other g=+9.8
No, the acceleration due to gravity is always negative.
why the acceleration of the ball 1 is equal to g? The sum of exterior forces on the ball 1 are the Force F used to push the ball and the Weight W of the ball F+W= m.a so isn't a=(F+W)/m ?
Im confused
+But why ?
Once the object is no longer being pushed or handled, then there is only on force acting on it, the force of gravity. Thus irregardless of what happened to the object before, once it is in the air, it will have an acceleration of g in the y-direction.
I can understand how their time is the same when they meet (t1=t2), but y1=y2 at 0:43? Isn't it B2 goes yb2 and B1 goes 20-yb2 a more logical assumption for their displacement at their meeting point from start points? Or would this y1=y2 assumption is based on the height from the ground when they meet? Sorry if I analyze poorly.
Thats what i thought