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

Poll: How many marks do you think you got?
73-75 (68)
15.28%
70-72 (45)
10.11%
67-69 (74)
16.63%
64-66 (79)
17.75%
61-63 (52)
11.69%
58-60 (30)
6.74%
55-57 (28)
6.29%
52-54 (15)
3.37%
49-51 (16)
3.6%
46-48 (4)
0.9%
42-45 (12)
2.7%
39-41 (6)
1.35%
36-38 (3)
0.67%
33-35 (2)
0.45%
30-32 (5)
1.12%
Less than 30 (6)
1.35%
#1
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

2a) 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
Last edited by freebirdk; 4 weeks ago
4
4 weeks ago
#2
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
0
4 weeks ago
#3
some of your answers are wrong. 4c is 25.2 m. 6c is 118 degrees as it has to be to nearest degree. v max id 5/6 sqrt(3ga)
(Original post by freebirdk)
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
0
4 weeks ago
#4
For the last one I believe is 5/6root3ga
0
4 weeks ago
#5
not max speed for 7c and angle needs to be nearest degree
(Original post by UrBusted)
4c was 25.2, rest are right i think
0
4 weeks ago
#6
you are correct
(Original post by Cailiheng)
For the last one I believe is 5/6root3ga
0
4 weeks ago
#7
how many marks was 4b?
0
4 weeks ago
#8
(Original post by tbt27)
not max speed for 7c and angle needs to be nearest degree
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???
Last edited by UrBusted; 4 weeks ago
1
#9
Thanks guys, fixed. Looks like I did worse than I wanted to though
1
4 weeks ago
#10
yeah you are right. its just you said you agreed with his answer for that one which he got wrong.
(Original post by UrBusted)
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???
0
4 weeks ago
#11
seems like you have still done incredibly well even with those mistakes. you've probably still got an A* on this paper
(Original post by freebirdk)
Thanks guys, fixed. Looks like I did worse than I wanted to though
0
4 weeks ago
#12
Me and some other people in my class got e= 0.49 for q5
1
4 weeks ago
#13
I got 1.38 for 4.a)
And doesn’t e have to bigger than 1/6?
And I=0.2(11/3)+0.2(3)=4/3 surely?
(Original post by chimer)
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
0
4 weeks ago
#14
sorry but you are wrong
Me and some other people in my class got e= 0.49 for q5
2
4 weeks ago
#15
anyone remember the answer to the initial acceleration one? i got 1.987 but not sure
0
4 weeks ago
#16
What are we thinking for grade boundaries? Could be very high (70 for an A*?).
0
4 weeks ago
#17
Me and some other people in my class got e= 0.49 for q5
(1/2(m1)(u1)^2+1/2(m2)(u2)^2)/(1/2(m1(v1)^2 +1/2(m2)(v2)^2) = 0.25 and then solve is how i did it
0
4 weeks ago
#18
And for 6b my impulse was 10/3 cos m= 0.5 and v-u was -20/3
(Original post by freebirdk)
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
0
4 weeks ago
#19
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
3
4 weeks ago
#20
But because it's vectors you can just use vector notation so v =-11/3 and u = 3 ( along x component) so m(v-u) was -10/3?
(Original post by UrBusted)
quite sure it was i = mv - m(-u) which was mv+mu, which in the end got 4/3
0
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