Edexcel Further Mechanics 1 - UNOFFICIAL MARK SCHEME [20th June 2019]

Given that the other "unofficial markscheme" is a troll with all the incorrect answers, here's what I got:

1a) Show that
1b) T = 15/2u when d=3

1a) v = 3sqrt(13)/5 = 2.16ms^-1

3) Possible impulses were (-2, +-1.5), (+-1.5, -2) - 4 in total

4a) Show that?
4b) a = 1.38ms^-1
4c) 22.7m (although I believe this to be wrong)

5a) 1/9<e<=1
5b) e=2/3

6a) v = (-11/3, 2)
6b) |I| = 4/3
6c) 117.7?

7b) a = 5g/3
7c) vmax = 5/2 sqrt(ga)

I've also made this thread so that we can actually have one with a poll of how people did

4c was 25.2, rest are right i think

For the last one I believe is 5/6root3ga

not max speed for 7c and angle needs to be nearest degree

yeah, everyones getting different answers for last one. i got 118 and 25.2, just forgot answer said to nearest degree. for maximum speed me and my friend got ((25/12)ga)^1/2, which I'm assuming is the same as what you got expanded???

Thanks guys, fixed. Looks like I did worse than I wanted to though

Here is what i got, can't guarantee all correct
1.T=15/2u
2.v=3sqrt(13)/5=2.16
3. -2 1.5, interchange position and change signs, 4 answers in total
4. a=1.987 d=25.17
5.e>1/9 e=2/3
6. v=(-11/3 2) I=2/15 angle=118
7.a=5g/3 v=5sqrt(3ag)/6

Me and some other people in my class got e= 0.49 for q5

Me and some other people in my class got e= 0.49 for q5

Given that the other "unofficial markscheme" is a troll with all the incorrect answers, here's what I got:

1a) Show that
1b) T = 15/2u when d=3

1a) v = 3sqrt(13)/5 = 2.16ms^-1

3) Possible impulses were (-2, +-1.5), (+-1.5, -2) - 4 in total

4a) Show that?
4b) a = 1.38ms^-1
4c) 25.2 m

5a) 1/9<e<=1
5b) e=2/3

6a) v = (-11/3, 2)
6b) |I| = 4/3
6c) 118 degrees

7b) a = 5g/3
7c) vmax = 5/6 sqrt(3ga)

I've also made this thread so that we can actually have one with a poll of how people did

And for 6b my impulse was 10/3 cos m= 0.5 and v-u was -20/3

quite sure it was i = mv - m(-u) which was mv+mu, which in the end got 4/3