coconut64
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Size:  439.4 KB hello, I am stuck on 6e and have no idea how to do this. It is to do with f=ma and I did draw a diagram for part d but I still don't know where to go from there as the answer for 6e should be 400N but I have no idea how to do it. Thank you for helping, I know this is a really long question. Attachment 488225488227
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coconut64
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Also , anyone mind explaining why the velocity is unchanged in 1b?Name:  1451246282642-1928341670.jpg
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rayquaza17
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(Original post by coconut64)
Also , anyone mind explaining why the velocity is unchanged in 1b?Name:  1451246282642-1928341670.jpg
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Hi, I've moved this to maths for you. Could you also post your working out for both questions?
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coconut64
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(Original post by rayquaza17)
Hi, I've moved this to maths for you. Could you also post your working out for both questions?
Wasn't it already in the Mathis section? I have uploaded a force diagram for 6d. But I will try to upload the other one then. Thanks
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coconut64
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(Original post by coconut64)
Wasn't it already in the Mathis section? I have uploaded a force diagram for 6d. But I will try to upload the other one then. Thanks
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the bear
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(Original post by coconut64)
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this is fine.
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Absent Agent
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(Original post by coconut64)
Also , anyone mind explaining why the velocity is unchanged in 1b?Name:  1451246282642-1928341670.jpg
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I believe it's because since the child is standing on the sledge, to throw the snowball the child has to exert a force in opposite direction, increasing the momentum of the sledge and hence its velocity in the opposite direction at which the snowball was thrown, but as soon as the snowball is dropped at the back of the sledge it will give an equal but opposite momentum to the sledge, decreasing the velocity to the initial velocity so overall the velocity is unchanged.
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Onlineslayer
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(Original post by Mehrdad jafari)
I believe it's because since the child is standing on the sledge, to throw the snowball the child has to exert a force in opposite direction, increasing the momentum of the sledge and hence its velocity in the opposite direction at which the snowball was thrown, but as soon as the snowball is dropped at the back of the sledge it will give an equal but opposite momentum to the sledge, decreasing the velocity to the initial velocity so overall the velocity is unchanged.
OMG really?
And errr.....standing? errrr........throwing?
LoL...I love that spirit.

My two pence...

Still in the spirit of conservation of momentum, the velocity is unchanged because the total mass remains unchanged.

Both the child and every bit of the snowball still remain on the sledge and there is no external HORIZONTAL force introduced.
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(Original post by Onlineslayer)
OMG really?
And errr.....standing? errrr........throwing?
LoL...I love that spirit.

My two pence...

Still in the spirit of conservation of momentum, the velocity is unchanged because the total mass remains unchanged.

Both the child and every bit of the snowball still remain on the sledge and there is no external HORIZONTAL force introduced.
Hmm, I tried to 'explain' why that was the case and not provide the original poster with rules (although that's what will be required in the actual exam).

Also, the user asked for help (http://www.thestudentroom.co.uk/show...1&postcount=35) that's why you see my post here because otherwise I have decided not to spend much more time on TSR to deal with users like you whose username should be onlinesaver. Where have you been so far?


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Onlineslayer
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(Original post by Mehrdad jafari)
Not knowledge? Perhaps your criticisms then.

Even if I have errors in my reasoning you were not the person to have corrected them. You only stated your own reasoning. To correct someone you should state the reasons as to why they are wrong.

Good luck!


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I am qualified enough to correct a few of your errors and vice versa...I am open to learning.

It is always nice to add a bit of humour... it helps, you know?

Otherwise I could have bluntly said you went totally off point with the words "standing" and "throwing" as against "sitting" and "dropping" used in the question, not to mention your assumptions about the velocity going up and returning down which totally defeats your conclusion that "the velocity remained UN-CHANGED". Everything about that response was just wrong.

Good-luck
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(Original post by Onlineslayer)
I am qualified enough to correct a few of your errors and vice versa...I am open to learning.

It is always nice to add a bit of humour... it helps, you know?

Otherwise I could have bluntly said you went totally off point with the words "standing" and "throwing" as against "sitting" and "dropping" used in the question, not to mention your assumptions about the velocity going up and returning down which totally defeats your conclusion that "the velocity remained UN-CHANGED". Everything about that response was just wrong.

Good-luck
Humour without actually stating your reasons to my 'errors'? A qualified person is not only qualified to know what is incorrect, but also why it is incorrect.

I used the word standing to point out the physical contact between the body of the child and the sledge because during the 'throw' of the snowball the force exerted on the sledge is transferred through this physical contact. Also, I believe it would be throwing rather than dropping because otherwise there is no need to question whether there is a change in velocity. In fact the question has stated the the child "drops it over the back". What do you think that means?

Well, the velocity does increase before the snowball has dropped, and I mentioned that "overall" the velocity is unchanged. I don't even think you are being pedantic.


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Onlineslayer
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(Original post by Mehrdad jafari)
Humour without actually stating your reasons to my 'errors'? A qualified person is not only qualified to know what is incorrect, but also why it is incorrect.

I used the word standing to point out the physical contact between the body of the child and the sledge because during the 'throw' of the snowball the force exerted on the sledge is transferred through this physical contact. Also, I believe it would be throwing rather than dropping because otherwise there is no need to question whether there is a change in velocity. In fact the question has stated the the child "drops it over the back". What do you think that means?

Well,the velocity does increase before the snowball has dropped, and I mentioned that "overall" the velocity is unchanged. I don't even think you are being pedantic.


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Again you are mixing things up...in physics, the word "dropping" and "throwing" are totally different especially when considered from the perspective of a projectile where one has an initial velocity and the other does not.

But also and most importantly, in the context of the question, the word horizontally and horizontal were used emphatically, Now let us for once assume your assumption about "throwing" is correct...the only force introduced by the child is in the vertical and that does not in any way affect the horizontal balance or the horizontal velocity.

But if your argument is strictly about the split seconds for which the "thrown" snow ball was "flying in the air", then it is important to also note that if the time of flight were any significant as to cause a fall in the sledge's velocity, additional complexities regarding the landing of such "projectile" will be involved.

Suffice to say that the question was only interested in a simple explanation that says:
"Velocity will remain unchanged because no mass was lost or gained and the momentum must be conserved"
(Original post by Mehrdad jafari)
I believe it's because since the child is standing on the sledge, to throw the snowball the child has to exert a force in opposite direction, increasing the momentum of the sledge and hence its velocity in the opposite direction at which the snowball was thrown, but as soon as the snowball is dropped at the back of the sledge it will give an equal but opposite momentum to the sledge, decreasing the velocity to the initial velocity so overall the velocity is unchanged.
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(Original post by Onlineslayer)
Again you are mixing things up...in physics, the word "dropping" and "throwing" are totally different especially when considered from the perspective of a projectile where one has an initial velocity and the other does not.
That's true, and I used the word throwing to mention the throw of the snowball by the scoop over the back of the sledge and the word drop for when it has fallen completely, unless I'm fallen into error with the term drop over.

But also and most importantly, in the context of the question, the word horizontally and horizontal were used emphatically, Now let us for once assume your assumption about "throwing" is correct...the only force introduced by the child is in the vertical and that does not in any way affect the horizontal balance or the horizontal velocity.
Not if the snow ball has a horizontal velocity when dropped.

But if your argument is strictly about the split seconds for which the "thrown" snow ball was "flying in the air", then it is important to also note that if the time of flight were any significant as to cause a fall in the sledge's velocity, additional complexities regarding the landing of such "projectile" will be involved.
Not exactly, but I'm talking about the time during the thrown of the ball and when it left the scoop.

Suffice to say that the question was only interested in a simple explanation that says:
"Velocity will remain unchanged because no mass was lost or gained and the momentum must be conserved"
Yes, and I didn't deny that but, as I said before, I wanted to explain why that was the case. Also, your reasoning that the mass was unchanged is not always true. There can be a case when the mass is added/removed from a uniformly moving body without any change of velocity of the body, that is, when the mass is added/removed from an external body moving uniformly with the same velocity next to the targeted body. You would be true only if the mass added/removed was landed/thrown with a component velocity to the velocity of the sledge. For example, the velocity of the sledge wouldn't change if the snowball is let go off the sledge without giving it any velocity in any directions (i.e. if it was left to fall under gravity outside of the sledge).
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rayquaza17
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Guys can we please try and help the OP in a friendly manner?

(I've removed some posts that don't really contribute to this.)

Also OP: You posted in the general maths/science section. I moved it to the specific maths one where the maths people lurk more often!
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(Original post by rayquaza17)
Guys can we please try and help the OP in a friendly manner?
Sure! I apologise for any inconvenience I may have caused.

(Original post by coconut64)
Also , anyone mind explaining why the velocity is unchanged in 1b?Name:  1451246282642-1928341670.jpg
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I have just found the marking scheme and I think in order to get the marks it would be better to follow what is in the mark scheme rather than what I tried to explain.

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coconut64
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(Original post by Mehrdad jafari)
Sure! I apologise for any inconvenience I may have caused.



I have just found the marking scheme and I think in order to get the marks it would be better to follow what is in the mark scheme rather than what I tried to explain.

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Hi, i didnt mean to get you involved in this , it wasnt inteneded as no one was helping so i just asked a few other people to help. Totally didnt expect to see this on tsr. Anway the markscheme is helpful and also thanks for trying to explain as well. I will ask my teacher for further explanation just to clarify. Could you quickly explain how to do 6e as i have drawn the digram but dont know where to go from there.Thanks again for helping.
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coconut64
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(Original post by Onlineslayer)
I am qualified enough to correct a few of your errors and vice versa...I am open to learning.

It is always nice to add a bit of humour... it helps, you know?

Otherwise I could have bluntly said you went totally off point with the words "standing" and "throwing" as against "sitting" and "dropping" used in the question, not to mention your assumptions about the velocity going up and returning down which totally defeats your conclusion that "the velocity remained UN-CHANGED". Everything about that response was just wrong.

Good-luck
Thanks for helping anyway.
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coconut64
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(Original post by the bear)
this is fine.
This is just an answer for part d but i have no idea how to do part d which is to work out the combined force thanks.
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(Original post by coconut64)
Hi, i didnt mean to get you involved in this , it wasnt inteneded as no one was helping so i just asked a few other people to help. Totally didnt expect to see this on tsr. Anway the markscheme is helpful and also thanks for trying to explain as well. I will ask my teacher for further explanation just to clarify. Could you quickly explain how to do 6e as i have drawn the digram but dont know where to go from there.Thanks again for helping.
No, no problem! I would be happy if I could help. I should also admit that our other friend Onlineslayer was also helpful as my reply was something of a quick response.

Part e was a bit more complicated than initially I thought it was. The assumption is that even if the braking force is not applied, both reindeer and the sleight will decelerate at the same rate of 0.25m/s^2 (This was the assumption you made when calculating the distance travelled by the reindeer and sleight as you essentially could consider the reindeer and sleigh as one body even though they were connected by a rigid harness). This means you have already taken account of the resistance to motion of the reindeer when calculating the deceleration (or the distance travelled) so you will have to bear this in mind for the next part.

Although the deceleration of the reindeer results from the tension in the harness as well as the resistance of the reindeer itself, you should not take into account the resistance of the reindeer when considering the tension in the harness calculated from the equation F=ma, where a will be the deceleration of the reindeer, as you have already considered this force when calculating the deceleration. You could also calculate the tension in the harness by considering the work done by the tension and the kinetic energy of the reindeer, that is, \dfrac{1}{2}mv^2 = f \times d. Again, note that the force in this case is only the tension in the harness without the resistance force because of the consideration of forces when calculating the distance trevelled.

I hope this is correct.

Edit: I've had a couple of glasses of wine (regardless of the fact that it's actually too late) so the thought that I might be wrong is crossing my mind.

Second edit (coconut64): indeed I was wrong in saying that the force in each case is the tension in the harness only, it is the tension as well as the resistance to the motion of the reindeer. Please do ask questions if you still have doubt about it.
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coconut64
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(Original post by Mehrdad jafari)
No, no problem! I would be happy if I could help. I should also admit that our other friend Onlineslayer was also helpful as my reply was something of a quick response.

Part e was a bit more complicated than initially I thought it was. The assumption is that even if the braking force is not applied, both reindeer and the sleight will decelerate at the same rate of 0.25m/s^2 (This was the assumption you made when calculating the distance travelled by the reindeer and sleight as you essentially could consider the reindeer and sleigh as one body even though they were connected by a rigid harness). This means you have already taken account of the resistance to motion of the reindeer when calculating the deceleration (or the distance travelled) so you will have to bear this in mind for the next part.

Although the deceleration of the reindeer results from the tension in the harness as well as the resistance of the reindeer itself, you should not take into account the resistance of the reindeer when considering the tension in the harness calculated from the equation F=ma, where a will be the deceleration of the reindeer, as you have already considered this force when calculating the deceleration. You could also calculate the tension in the harness by considering the work done by the tension and the kinetic energy of the reindeer, that is, \dfrac{1}{2}mv^2 = f \times d. Again, note that the force in this case is only the tension in the harness without the resistance force because of the consideration of forces when calculating the distance trevelled.

I hope this is correct.

Edit: I've had a couple of glasses of wine (regardless of the fact that it's actually too late) so the thought that I might be wrong is crossing my mind.

Second edit (coconut64): indeed I was wrong in saying that the force in each case is the tension in the harness only, it is the tension as well as the resistance to the motion of the reindeer. Please do ask questions if you still have doubt about it.
So how do I do this question then, so I just leave out the resistance? Thanks
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