Y1_UniMaths
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Please can someone help me on question 9. The “half a second later” aspect is confusing me.
https://imgur.com/a/g8317nr
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Snoozinghamster
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(Original post by Y12_FurtherMaths)
Please can someone help me on question 9. The “half a second later” aspect is confusing me.
https://imgur.com/a/g8317nr
You’ve made a good start. You know how long after dropping the first ball the collision occurs. Try to use that to work out how long the second ball is travelling for. (Remember the half second delay between balls)
That should give you enough to find the initial velocity
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Y1_UniMaths
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(Original post by Snoozinghamster)
You’ve made a good start. You know how long after dropping the first ball the collision occurs. Try to use that to work out how long the second ball is travelling for. (Remember the half second delay between balls)
That should give you enough to find the initial velocity
Are you saying that the 2nd ball with be travelling for 1.75-0.5=1.25 seconds?
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Muttley79
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(Original post by Y12_FurtherMaths)
Please can someone help me on question 9. The “half a second later” aspect is confusing me.
https://imgur.com/a/g8317nr
If you call the time 't' for the first then, in terms of t, what is the time fir the second?
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Snoozinghamster
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(Original post by Y12_FurtherMaths)
Are you saying that the 2nd ball with be travelling for 1.75-0.5=1.25 seconds?
Yeh, then it should be ok from there!
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RDKGames
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(Original post by Y12_FurtherMaths)
Please can someone help me on question 9. The “half a second later” aspect is confusing me.
https://imgur.com/a/g8317nr
Well the second ball is half a second behind, so its time variable would be T = t - 0.5.
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Y1_UniMaths
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(Original post by Muttley79)
If you call the time 't' for the first then, in terms of t, what is the time fir the second?
t-0.5? I think that’s right and I feel really stupid
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Y1_UniMaths
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(Original post by RDKGames)
Well the second ball is half a second behind, so its time variable would be T = t - 0.5.
Yeah facepalm thanks
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RDKGames
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(Original post by Y12_FurtherMaths)
Please can someone help me on question 9. The “half a second later” aspect is confusing me.
https://imgur.com/a/g8317nr
Also your acceleration needs to be -9.8 for both if you are taking upwards as the +ve direction.
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Y1_UniMaths
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(Original post by RDKGames)
Also your acceleration needs to be -9.8 for both if you are taking upwards as the +ve direction.
I’m taking downwards as positive? And it wouldn’t make a difference would it, as it’s always 9.8 down and -9.8 up?
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Muttley79
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(Original post by RDKGames)
Well the second ball is half a second behind, so its time variable would be T = t - 0.5.
Surely you did not need to say that much detail?
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RDKGames
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(Original post by Y12_FurtherMaths)
I’m taking downwards as positive? And it wouldn’t make a difference would it, as it’s always 9.8 down and -9.8 up?
No, gravity always acts in one direction.
Once you choose whether that direction is +ve or not, that gravitational acceleration has its sign fixed for ALL bodies it affects, regardless of the direction in which they are going.

If you are taking downwards as positive, then your second ball needs a = 9.8
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Y1_UniMaths
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(Original post by RDKGames)
No, gravity always acts in one direction.
Once you choose whether that direction is +ve or not, that gravitational acceleration has its sign fixed for ALL bodies it affects, regardless of the direction in which they are going.

If you are taking downwards as positive, then your second ball needs a = 9.8
If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now
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Muttley79
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(Original post by Y12_FurtherMaths)
If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now
Think about 's' - it is a vector .... that's the issue I see.
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RDKGames
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(Original post by Y12_FurtherMaths)
If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now
Not quite.

Suppose you're on the ground and you take downwards as positive.

You throw a ball upwards with some velocity, and since this velocity acts upwards at the moment of your throw, it is a -ve value. This acceleration of 9.8m/s^2 is the rate at which the velocity of the ball will change as it goes up. Since velocity is initially -ve, every second 9.8m/s are added on top of it. It will eventually reach 0 velocity (hence reach the max point of the throw) and then the velocity will increase into the +ve's which means the direction of motion is now towards the ground (ie. downward)

As you'd expect.

Not sure where you got the idea from that acceleration is different for objects under the same gravitational field depending on the direction they are going.

The only case where it would be acceptable to say a=9.8 for first object and a=-9.8 for second object is if you decide all of a sudden to take UPWARDS as +ve after calculating your t value from the first object... but just as it sounds, it's unnecessarily more complicated.
Last edited by RDKGames; 2 years ago
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Y1_UniMaths
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(Original post by RDKGames)
Not quite.

Suppose you're on the ground and you take downwards as positive.

You throw a ball upwards with some velocity, and since this velocity acts upwards at the moment of your throw, it is a -ve value. This acceleration of 9.8m/s^2 is the rate at which the velocity of the ball will change as it goes up. Since velocity is initially -ve, every second 9.8m/s are added on top of it. It will eventually reach 0 velocity (hence reach the max point of the throw) and then the velocity will increase into the +ve's which means the direction of motion is now towards the ground (ie. downward)

As you'd expect.

Not sure where you got the idea from that acceleration is different for objects under the same gravitational field depending on the direction they are going.

The only case where it would be acceptable to say a=9.8 for first object and a=-9.8 for second object is if you decide all of a sudden to take UPWARDS as +ve after calculating your t value from the first object... but just as it sounds, it's unnecessarily more complicated.
Is the answer 18ms^-1 (2sf) because the book says 17 but my teacher said the book is wrong
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RDKGames
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(Original post by Y12_FurtherMaths)
If I’m taking downwards as positive then if a ball is dropped from a window it will accelerate down at 9.8ms^-2. And then if the ball is being projected up it will rise up with acceleration -9.8ms^-2. In really confused now
After another glance at what you wrote, I noticed you actually took the 'unnecessarily more complicated' route which is where your misunderstanding must be stemming from.

For the first object, you took the top of the building as your reference point and downwards as positive. This does imply that the object's displacement at time of collision is s = 15, and a = 9.8.

Then you switch focus to your second object. You scrap the point of reference you followed above and instead take the bottom of the building as your reference point, and now upwards as +ve as well. This, again, implies that at the point of collision, we have s = 15 and a = -9.8.

This is of course not a problematic approach for this question and you can follow it all you want as long as you understand what's going on...


But it does prove to be problematic in the following reshuffling of the question:

"A ball is dropped from a window 30m above ground. Half a second later, another ball is projected vertically upwards with speed  8\mathrm{m/s} from the ground, vertically below the window. Find the time at which the two objects collide."

The only difference is that I don't tell you they collide 15m above the ground, and I give you an initial speed of the second object. This prevents you from using your approach because you cannot use 'consider first object, then consider the second object inidividually' tactic. You must instead consider them both here.

The approach is to set their displacements equal to eachother since that's when they collide, and this leaves you with an equation in t.


Correct approach (taking downwards as +ve, for instance, with the bottom of the tower as the point of reference):

Spoiler:
Show

First object:  s_1 = s_0 + ut + \frac{1}{2}at^2 = -30+4.9t^2

Second object: s_2 = -8(t-0.5) + 4.9(t-0.5)^2.

At the moment of collision, s_1 = s_2... hence

-30 + 4.9t^2 = -8(t-0.5) + 4.9(t-0.5)^2 \implies t \approx 2.73



Incorrect approach (as you would do following the same style):

Spoiler:
Show

First object (taking top of tower as the point of ref, and downwards as +ve): s_1 = 4.9t^2.

Second object (taking bottom of tower as the point of ref, and upwards as +ve): s_2 = 8(t-0.5) - 4.9(t-0.5)^2.

At the moment of collision, s_1 = s_2 ... and such t doesn't even exist!
Last edited by RDKGames; 2 years ago
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RDKGames
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(Original post by Y12_FurtherMaths)
Is the answer 18ms^-1 (2sf) because the book says 17 but my teacher said the book is wrong
Pretty much. Approx. 18.1267...
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Y1_UniMaths
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(Original post by RDKGames)
After another glance at what you wrote, I noticed you actually took the 'unnecessarily more complicated' route which is where your misunderstanding must be stemming from.

For the first object, you took the top of the building as your reference point and downwards as positive. This does imply that the object's displacement at time of collision is s = 15, and a = 9.8.

Then you switch focus to your second object. You scrap the point of reference you followed above and instead take the bottom of the building as your reference point, and now upwards as ve as well. This, again, implies that at the point of collision, we have s = 15 and a = -9.8.

This is of course not a problematic approach for this question and you can follow it all you want as long as you understand what's going on...


But it does prove to be problematic in the following reshuffling of the question:

"A ball is dropped from a window 30m above ground. Half a second later, another ball is projected vertically upwards with speed  8\mathrm{m/s} from the ground, vertically below the window. Find the time at which the two objects collide."

The only difference is that I don't tell you they collide 15m above the ground, and I give you an initial speed of the second object. This prevents you from using your approach because you cannot use 'consider first object, then consider the second object inidividually' tactic. You must instead consider them both here.

The approach is to set their displacements equal to eachother since that's when they collide, and this leaves you with an equation in t.


Correct approach (taking downwards as ve, for instance, with the bottom of the tower as the point of reference):

Spoiler:
Show

First object:  s_1 = s_0   ut   \frac{1}{2}at^2 = -30 4.9t^2

Second object: s_2 = -8(t-0.5)   4.9(t-0.5)^2.

At the moment of collision, s_1 = s_2... hence

-30   4.9t^2 = -8(t-0.5)   4.9(t-0.5)^2 \implies t \approx 2.73



Incorrect approach (as you would do following the same style):

Spoiler:
Show

First object (taking top of tower as the point of ref, and downwards as ve): s_1 = 4.9t^2.

Second object (taking bottom of tower as the point of ref, and upwards as ve): s_2 = 8(t-0.5) - 4.9(t-0.5)^2.

At the moment of collision, s_1 = s_2 ... and such t doesn't even exist!
How do you know that they will collide when their displacements are equal and is that always the true? And what is s0?
Last edited by Y1_UniMaths; 2 years ago
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RDKGames
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(Original post by Y12_FurtherMaths)
How do you know that they will collide when their displacements are equal and is that always the true?
Well if you're measuring their displacements from the same point, then I'd think it is obvious that they collide when these displacements are the same?

s_1 = s_2 is always the condition in this consistent approach. If we measure displacement from different points, then TBH it just complicates things further and perhaps (?) makes the problem unsolvable.

s_0 means initial displacement.

(?) Unsure here because I can't be bothered to think about it much, and certainly this is going too beyond for you since you're new to this topic.
Last edited by RDKGames; 2 years ago
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