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M1 Kinematics

A boy projects a ball vertically upwards with speed 10ms^-1 from a point X, which is 50m above the ground. T seconds after the first ball is projected upwards, the boy drops a second ball from X. Initially the second ball is at rest. The balls collide 25m above the ground. Find the value of T.

-I have tried creating an expression for t in terms of t and T and then I equated expressions of the displacements of both balls in terms of t.
I tried to solve for T by solving for t (because my expression for the 2nd ball was t+T), but this didnt work...
please help :colondollar:
I assume the underlined bit was to suggest furthermore than u=0 for the 2nd ball.

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Reply 1
Original post by MathMeister
A boy projects a ball vertically upwards with speed 10ms^-1 from a point X, which is 50m above the ground. T seconds after the first ball is projected upwards, the boy drops a second ball from X. Initially the second ball is at rest. The balls collide 25m above the ground. Find the value of T.

-I have tried creating an expression for t in terms of t and T and then I equated expressions of the displacements of both balls in terms of t.
I tried to solve for T by solving for t (because my expression for the 2nd ball was t+T), but this didnt work...
please help :colondollar:
I assume the underlined bit was to suggest furthermore than u=0 for the 2nd ball.


you must have a common origin and well defined direction for positive and negative in this type of problem
Original post by TeeEm
you must have a common origin and well defined direction for positive and negative in this type of problem

common origin? Anyway I did set downwards as positive. I had the accelerations as opposite though.
Reply 3
Original post by MathMeister
common origin? Anyway I did set downwards as positive. I had the accelerations as opposite though.


My last remark can be ignored as there are no simultaneous equations here.

The plan is

find how long it takes the first ball to get to the collision point.

Then take away T seconds from this time.

Then start a new equation for the second ball


(as a matter of interest is g=9.8 or 10 in this problem?)
Original post by TeeEm

(as a matter of interest is g=9.8 or 10 in this problem?)

In M1/2- Edexcel- g is always taken to be equal to 9.8. Nothing else. :smile:
Reply 5
Original post by MathMeister
In M1/2- Edexcel- g is always taken to be equal to 9.8. Nothing else. :smile:


all good
Original post by TeeEm
My last remark can be ignored as there are no simultaneous equations here.
find how long it takes the first ball to get to the collision point.
Then take away T seconds from this time.
Then start a new equation for the second ball

I don't get why. Let's say the 2nd ball (the one being dropped) is dropped 1 second after the 1st ball is projected- then the 1st ball will have t= (t+1) or (t) and the second ball with have t= (t) or (t-1) respectively? Surely?
So why is the first ball having T taken from it in this example?

Plus- when do simultaneous equations ever come into Kinematics M1?
Reply 7
Original post by MathMeister
I don't get why. Let's say the 2nd ball (the one being dropped) is dropped 1 second after the 1st ball is projected- then the 1st ball will have t= (t+1) or (t) and the second ball with have t= (t) or (t-1) respectively? Surely?
So why is the first ball having T taken from it in this example?

Plus- when do simultaneous equations ever come into Kinematics M1?


FORGET about the second ball !

Find how long it takes the first ball to get 25 metres above ground.


(simultaneous equations in kinematics? Look at the first question of M1. June 2009)
the second ball has not been in flight for t+T seconds

it only starts falling after T seconds, which is why you need to take T away
Thank you both of you.
Ive been doing Kinematics all day- that's why my brain is so fried lol
:smile:
Reply 10
Original post by MathMeister
Thank you both of you.
Ive been doing Kinematics all day- that's why my brain is so fried lol
:smile:


did you get an answer at the end?
Original post by TeeEm
did you get an answer at the end?

Yeah- the answer is 1.2 seconds(2s.f)
but I had to use the negative value from the quadratic to put into the second quadratic.
So in total it was a bit tedious and I'm just thinking about why I have to chose the negative root to put into the 2nd quadratic to solve for T because otherwise in an exam i'm stuffed lol.
Reply 12
Original post by MathMeister
Yeah- the answer is 1.2 seconds(2s.f)
but I had to use the negative value from the quadratic to put into the second quadratic.
So in total it was a bit tedious and I'm just thinking about why I have to chose the negative root to put into the 2nd quadratic to solve for T because otherwise in an exam i'm stuffed lol.


I will "borrow" this question so when I have it ready I will post you the solution.
(Exactly the same, just different numbers)
Original post by TeeEm
I will "borrow" this question so when I have it ready I will post you the solution.
(Exactly the same, just different numbers)

What did you mean by common origin for the sim equation question?

It also doesn't make sense why I have to use the negative value of t to solve the quadratic in T and t- because I'm defining t to be the time (in my equation) that it takes ball 1 to go -25 m (upwards is defined as positive) with u=10 ms^-1. so in an exam im stuffed?
thanks for the upcoming solution :biggrin:
(edited 9 years ago)
Reply 14
Original post by MathMeister
What did you mean by common origin for the sim equation question?

It also doesn't make sense why I have to use the negative value of t to solve the quadratic in T and t- because I'm defining t to be the time (in my equation) that it takes ball 1 to go -25 m (upwards is defined as positive) with u=10 ms^-1. so in an exam im stuffed?
Thank you so much


Your problem was simple (but very interesting) so when I first read it I assumed it is the usual "simultaneous type scenario" which is a lot harder.

So ignore common origin.

Wait for my question and see if this makes sense about the negative.
Original post by TeeEm
Your problem was simple (but very interesting) so when I first read it I assumed it is the usual "simultaneous type scenario" which is a lot harder.
So ignore common origin
Wait for my question and see if this makes sense about the negative.

thanks
Original post by TeeEm
Your problem was simple (but very interesting) so when I first read it I assumed it is the usual "simultaneous type scenario" which is a lot harder.

So ignore common origin.

Wait for my question and see if this makes sense about the negative.

is your question/answer imaginary?
Reply 17
Original post by MathMeister
is your question/answer imaginary?


imaginary?
Original post by TeeEm
imaginary?

i would say complex but there's nothing real about it :tongue:
math joke- sorry lol
Reply 19
Original post by MathMeister
i would say complex but there's nothing real about it :tongue:
math joke- sorry lol


give me half hour

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