Mechanics 1 P42 2014; Discussion!

[wizteddy13]

you lst only two freakin marks i lost 6

Lol, I lost 2 more, I couldn't show for no 5, the last part. I realized that one 5 minutes after handing in the paper. Such sorrow :/

 

[A star]

Lol, I lost 2 more, I couldn't show for no 5, the last part. I realized that one 5 minutes after handing in the paper. Such sorrow :/

there is still a big difference betwen 40 and 46 one can be a B and the other minumum A*

 

[17194]

71.3

It was not 71.3, the answer was around 40.9 if I recall correctly. First you had to calculate the distance covered during the motion with linear acceleration (from t=0 to t=8). Then you had to integrate the expression for velocity and apply the limits 19 and 0 (27-8 = 19). Add the two distances which were 8m and 40.something for the final answer.

 

[Oyessorzo]

It was not 71.3, the answer was around 40.9 if I recall correctly. First you had to calculate the distance covered during the motion with linear acceleration (from t=0 to t=8). Then you had to integrate the expression for velocity and apply the limits 19 and 0 (27-8 = 19). Add the two distances which were 8m and 40.something for the final answer.

Are you sure about this? because i did the same, nd now i dont know if i was right :/

 

[wizteddy13]

It was not 71.3, the answer was around 40.9 if I recall correctly. First you had to calculate the distance covered during the motion with linear acceleration (from t=0 to t=8). Then you had to integrate the expression for velocity and apply the limits 19 and 0 (27-8 = 19). Add the two distances which were 8m and 40.something for the final answer.

YES EXACTLY! Whew, glad I'm not the only one who did that. As I thought the equation was applicable from t=8 and onwards, so till t=27, the time is 19s. I integrated and put in values for t=19s, which is 40. something, and added 8 to it. I think I ended up with 48.9 or something.

 

[wizteddy13]

I swear, the paper may have been do-able, but this is one of the trickiest papers I've ever given. *sigh*

 

[mshabir007]

Salam hi i have seen the 7th question in my M1 Textbook but still messed it up M1 BOOK Chapter 7 Exercise B

 

[17194]

YES EXACTLY! Whew, glad I'm not the only one who did that. As I thought the equation was applicable from t=8 and onwards, so till t=27, the time is 19s. I integrated and put in values for t=19s, which is 40. something, and added 8 to it. I think I ended up with 48.9 or something.

Sorry, 48.9 not 40.9. Yes the distance for linear acceleration was 8 and for variable acceleration it was 40.9.

 

[wizteddy13]

hi i have seen the 7th question in my M1 Textbook but still messed it up M1 BOOK Chapter 7 Exercise B

I freaked out when I saw that one...the wall of text is such a deterrent.

 

[wizteddy13]

Sorry, 48.9 not 40.9. Yes the distance for linear acceleration was 8 and for variable acceleration it was 40.9.

Whoopee, glad mine matches with someone else!
Now I just hope we're correct...

 

[mshabir007]

Salam
did any one solved the question about a stone is dropped and i think find resistance and also the second part about the tension if so please reply and also Q2 Thank You

 

[Snowysangel]

Quoting from the mark scheme:

Marks are of the following three types:

"M Method mark, awarded for a valid method applied to the problem. Method marks are
not lost for numerical errors, algebraic slips or errors in units. However, it is not
usually sufficient for a candidate just to indicate an intention of using some method or
just to quote a formula; the formula or idea must be applied to the specific problem in
hand, e.g. by substituting the relevant quantities into the formula. Correct application
of a formula without the formula being quoted obviously earns the M mark and in some
cases an M mark can be implied from a correct answer.

"A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated method mark is earned (or
implied).

"B Mark for a correct result or statement independent of method marks."

In other words, there's no ECF, so to speak. But you do get the marks for your process, so it's a... partial ECF, I suppose.
Understand?

What if its a one mark question? And what of its a 2 mark one?

 

[wizteddy13]

Salam
did any one solved the question about a stone is dropped and i think find resistance and also the second part about the tension if so please reply and also Q2 Thank You

Question about stone was discussed someplace in pages 5 or 6 I believe. R was 15.5 (I think, forgot exact), and T was 17.59 (maybe, again, forgot exact). If question 2 is the collision question, I think the T is 5s.

 

[17194]

Salam
did any one solved the question about a stone is dropped and i think find resistance and also the second part about the tension if so please reply and also Q2 Thank You

For question two you had to form two equations of motions for both particles P and Q in displacement 's'. Then you had to equate s1 = 10-s2. This would give you a quadratic equation in t, and the answer was t=5s.

 

[bigboy]

Okay.
The tension answer thing with two strings was T2 = 11.9, T2 = 0.5. Checked a thousand times. The particle had weight too which you had to consider.

yees i got the same as you but most people didnt consider the weight,i hope we're right in shaa allah

 

[raysonzaffar]

Sorry, 48.9 not 40.9. Yes the distance for linear acceleration was 8 and for variable acceleration it was 40.9.

So I got the answer for the first 8 seconds right but then I used limits of 27 and 8. how many marks will i lose for this?
This part was for how many marks?

 

[raysonzaffar]

P was 6150W
because at constant speed the forward force will be equal to the resistive force. So using P =FV, where F=410 and v=15

I got F as 410. But my power was different. How many marks were both the parts for? And how much will I lose for getting the Power wrong?

 

[wizteddy13]

I got F as 410. But my power was different. How many marks were both the parts for? And how much will I lose for getting the Power wrong?

I have no idea, but the question was just 1 mark...so :/

 

[raysonzaffar]

I have no idea, but the question was just 1 mark...so :/

The one asking for power was for 1 mark?
So i got 410. Is that correct? part i was for 4 marks right?

 

[bbbbgf]

It was not 71.3, the answer was around 40.9 if I recall correctly. First you had to calculate the distance covered during the motion with linear acceleration (from t=0 to t=8). Then you had to integrate the expression for velocity and apply the limits 19 and 0 (27-8 = 19). Add the two distances which were 8m and 40.something for the final answer.

it was 71.3 confirmed
the first distance was 8 the rest was found by integrating from 27 to 8 it came as 63.3 , cheers !