The Student Room Group

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Reply 1

agreed

Reply 2

It was easy. That means the curve is gonna be high. :|

Reply 3

What did you get for P? What did you get for the angle for the impulse question and the angle for the centre of mass question?

Reply 4

impulse 63 and COM 14 (cant remember dp)

Reply 5

imasillynarb
What did you get for P? What did you get for the angle for the impulse question and the angle for the centre of mass question?


0.4mg < P < 13.6mg

The angle for the impulse question was somewhere around 60-80, and the angle for the CM question was somewhere between 15-25. I don't really remember. :tongue:

Reply 6

shift3
< P < 13.6

The angle for the impulse question was somewhere around 60-80, and the angle for the CM question was somewhere between 15-25. I don't really remember. :tongue:


I got 14.6 I think for the COM question, and 40.3 I think for the impulse angel...

How did you find P?

Reply 7

i didnt!

Reply 8

imasillynarb
What did you get for P? What did you get for the angle for the impulse question and the angle for the centre of mass question?


P is greater than or equal to 0.4mg

angle for impulse: 63.43 degrees

angle for C of M: 14.6 degrees


(i think)

Reply 9

yea that sounds like the angles i got

Reply 10

imasillynarb
I got 14.6 I think for the COM question, and 40.3 I think for the impulse angel...

How did you find P?


Assume two cases, one where the friction is in the same direction as P, and one where it's opposite.

friction < 0.6 * 11mg (6.6mg)

Reply 11

and t=5/2 in the vectors. t=3 in the last question. distance 48m in last question. 0.7 for the first...COM was at 5/6 from BC

Reply 12

Ermmm... P is between 1.95mg and 6.3mg....

I'm certainly wrong on that one then!!!

Anyway, ce la vis!
I did well enough on the rest of it!

Reply 13

sam99
angle for impulse: 63.43 degrees

angle for C of M: 14.6 degrees


Yup

Reply 14

Impulse question was quite easy to make a mistake on I reckon..

P = 0.4mg, I got that as well, except theyre not going to give you 7 marks for rearraging F + P = N

So that ones wrong.

As with the impulse question, it wanted the angle of projection. The balls comes in at 30i and is projected at root(5^2 + 25^2???), so you wanted the angle between the resultant and the 30i, not the angle between 5 and 25, I think :smile:

Reply 15

Sapphira
and t=5/2 in the vectors. t=3 in the last question. distance 48m in last question. 0.7 for the first...COM was at 5/6 from BC


Yeah, and the last part of question 7 (the last question) was ~23.6.

For the first question I got a=1.2, I don't know where I messed up. :frown:

Reply 16

shift3
Assume two cases, one where the friction is in the same direction as P, and one where it's opposite.

friction < 0.6 * 11mg (6.6mg)


Err, why do you assume 2 cases? You know which way the friction acts. Its not a 'maybe' ...

Reply 17

lol dont worry was prob me that messed up, im particularly crap at maths :redface:

Reply 18

Sapphira
lol dont worry was prob me that messed up, im particularly crap at maths :redface:


Your answer is right, all my friends got 0.7.

Reply 19

imasillynarb
Err, why do you assume 2 cases? You know which way the friction acts. Its not a 'maybe' ...


Friction acts opposite to motion. In the default case, it acts opposite to the motion of the ladder, i.e. to the right. However, when P becomes strong enough to break equilibrium, friction acts opposite to it, i.e. to the left.