I think a much more elegant solution is as follows: Both particles have the same (downwards) acceleration, and thus both particles velocity changes by the same value in the same time interval. Thus the relative velocity (approach speed) between the particles is constant, and to begin with this is 23m/s so it remains 23m/s. The gap between them, 72m decreases at this speed. Thus using time = distance/speed, we get t = 72/23 = 3.13s as the collision time. We can then just work out the distance the top particle fell during this time: s = ut + 0.5at^2 = 6(3.13) + 0.5*(9.8)(3.13)^2 = 67m as required.
Question 1 As with any problem to do with the motion of point particles, it is first essential to ignore the problem’s tasks and first determine the reference frame for our problem. *Declaration of the Reference Frame* I say, let’s make the top of Tower A (0,0) in our cartesian system, and let the upwards vector be positive and the downwards vector be negative. As a consequence, gravitational effects will be treated as negatives and so will positions below the Cliff. The second Of our necessary preliminary tasks is to determine the motion of each particle. The meaning of this is to determine the y(t) - Displacement as a function of time v(t) - Velocity as a function of time a(t) - Acceleration as a function of time *Determination of the Motions of the Relevant Particles A and B* _For Particle A:_ y(t) = -g/2 t^2 -6t v(t) = -gt - 6 a(t) = -g _For Particle B_ y(t) = -g/2 t^2 +17t -72 v(t) = -gt + 17 a(t) -g Now with our preliminary work done, and remember this preliminary must be done for every problem on the motion of particles. Its a good habit in my opinion to do this without even reading the tasks of the problem. The tasks of the problem always require application of the mathematical apparatus we developed, so let’s do that now: *Issue - The distance from the cliff’s peak to the collision point* *Task 1 - Determine the Time at which the Rocks collide* This would mean that the displacements are the same, so lets set them equal to each other: -g/2 t^2 +17t -72 = -g/2 t^2 -6t 17t-72 = -6t 23t=72 *t=72/23 ~ 3.13(s)* Now we are prepared to determine the distance from the top of the cliff. That was our declared reference frame to begin with, by sheer coincidence. or was it? :) so we can just plug this time of collision into any of the two displacement time functions. Put it into both. The reason for this is, if the answer is different for both then we have made an error somewhere. It gives us more confidence if the answer is the same: B: -g/2 (3.13)^2 +17(3.13) -72 = -66.84m *distance is scalar hence: 66.84m* A: -g/2 (3.13)^2 -6(3.13) = -66.834 *distance is scalar hence: 66.83* Probably a little rounding error somewhere. You should always use exact answers when possible, or use 5 decimal points when recording answers
Time is universal. If they both get thrown at the same time, then they have both been moving for the same amount of time when they hit each other. Think of all the SUVAT variables. Each one is individual to each particle (velocity, displacement, acceleration...) apart from time! If they have both been in the air for 5 seconds, then t is 5 for both of them. It doesn't care about anything else!
I think a much more elegant solution is as follows:
Both particles have the same (downwards) acceleration, and thus both particles velocity changes by the same value in the same time interval. Thus the relative velocity (approach speed) between the particles is constant, and to begin with this is 23m/s so it remains 23m/s. The gap between them, 72m decreases at this speed. Thus using time = distance/speed, we get t = 72/23 = 3.13s as the collision time. We can then just work out the distance the top particle fell during this time: s = ut + 0.5at^2 = 6(3.13) + 0.5*(9.8)(3.13)^2 = 67m as required.
uh uh uhhhh
YESSS THANK YOU. KEEEP THIS DISGUSTING SERIES UP ITS SOOOOO GOOOOODDD
With that enthusiasm I'm gonna have to! 🔥
@@AITutor lies
Question 1
As with any problem to do with the motion of point particles, it is first essential to ignore the problem’s tasks and first determine the reference frame for our problem.
*Declaration of the Reference Frame*
I say, let’s make the top of Tower A (0,0) in our cartesian system, and let the upwards vector be positive and the downwards vector be negative.
As a consequence, gravitational effects will be treated as negatives and so will positions below the Cliff.
The second Of our necessary preliminary tasks is to determine the motion of each particle. The meaning of this is to determine the
y(t) - Displacement as a function of time
v(t) - Velocity as a function of time
a(t) - Acceleration as a function of time
*Determination of the Motions of the Relevant Particles A and B*
_For Particle A:_
y(t) = -g/2 t^2 -6t
v(t) = -gt - 6
a(t) = -g
_For Particle B_
y(t) = -g/2 t^2 +17t -72
v(t) = -gt + 17
a(t) -g
Now with our preliminary work done, and remember this preliminary must be done for every problem on the motion of particles. Its a good habit in my opinion to do this without even reading the tasks of the problem. The tasks of the problem always require application of the mathematical apparatus we developed, so let’s do that now:
*Issue - The distance from the cliff’s peak to the collision point*
*Task 1 - Determine the Time at which the Rocks collide*
This would mean that the displacements are the same, so lets set them equal to each other:
-g/2 t^2 +17t -72 = -g/2 t^2 -6t
17t-72 = -6t
23t=72
*t=72/23 ~ 3.13(s)*
Now we are prepared to determine the distance from the top of the cliff. That was our declared reference frame to begin with, by sheer coincidence. or was it? :)
so we can just plug this time of collision into any of the two displacement time functions.
Put it into both. The reason for this is, if the answer is different for both then we have made an error somewhere.
It gives us more confidence if the answer is the same:
B:
-g/2 (3.13)^2 +17(3.13) -72 = -66.84m
*distance is scalar hence: 66.84m*
A:
-g/2 (3.13)^2 -6(3.13) = -66.834
*distance is scalar hence: 66.83*
Probably a little rounding error somewhere. You should always use exact answers when possible, or use 5 decimal points when recording answers
wow bro you’re crazy good. what academic level are you studying at now?
Great vid, look forward to more of the disgusting series !!!
Please continue
but it's decelerating in a different direction, so shouldnt the the sign infront of acceleration be the same!
wait im a braindead idiot nevermind me !
i thought it was projectile motion and over-complicated it, otherwise from that; definitely some way harder questions out there for a level physics.
This is aimed for A Level maths I think, not physics.
@@TamilEelam1 the mechanics sections in both subjects are pretty identical
Thank you for the video this is just what i needed!
which exam board is this?
What a juxtaposing title!
Just keeping you on your toes 😉
Aren't you supposed to take the value of g as 10 instead of 9.8?
Na u don’t it don’t really matter
@@johnathanbradley6627 In mechanics, P4 or M1, you are supposed to. It says on the paper.
@@Mrmightymight I think this is more focused a level maths yk
@@johnathanbradley6627 It's written in the title, 'Mechanics Questions'.
@@Mrmightymight yes mechanics section of maths
is this AS level or A?
@@hhh11. who this
Legend
yh when u do physics none of these r hard lol physics mechanics killing us all
Why is t the same? I don't understand.
Time is universal. If they both get thrown at the same time, then they have both been moving for the same amount of time when they hit each other.
Think of all the SUVAT variables. Each one is individual to each particle (velocity, displacement, acceleration...) apart from time! If they have both been in the air for 5 seconds, then t is 5 for both of them. It doesn't care about anything else!
this better not come up