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EDEXCEL MATHS M1 14 JUNE 2017 - Unofficial Mark Scheme

This poll is closed

How many marks do you think you got?

0-10 2%

11-20 1%

21-30 1%

31-40 3%

41-50 6%

51-60 16%

61-70 41%

71-7529%

Total votes: 1961

Redcoats

1)p = 5/2
q=-3/2

2a) speed of p after collision: 2u
b) Direction HAS been reversed
c) impulse = 12mu

3) x = 2.3 (2sf)
Uniform plank = (i) Center of mass is at the center of plank (i.e. weight acts
directly at center of plank)
(ii) Mass is evenly distributed across the plank or plank is rigid and does not bend or buckle

4) mew = 0.47

5) thrust in rod = 6

6) acceleration of cyclist = 0.8
time when reached midpoint of AB = 2.8 seconds

7) bearing: 103
p = (position vector)+t(velocity)
q = same as above but with respective position and velocity vectors
show QP was the given vector
t = 0 of t= 0.32 when ships are 10km apart

8) motion of A: T - 2m(mew)g = 2ma
motion of B: mg-T=ma
b) show acceleration is g/3 (1-2(mew))
c) speed when b hits floor: sqrt(2g/3(1-2(mew))h)
d) d = 4/3h
e) if mew = 1/2, A would not slide along the table and the system would remain in rest

No profanity. Reply with edits.

(edited 6 years ago)

Scroll to see replies

Reply 1

eliteoppresser

what did u get for u , was it 0.49

Reply 2

Redcoats

Original post by eliteoppresser

what did u get for u , was it 0.49

do you mean mew for the question with the inclined plane?

I got 0.43

Reply 3

Lemuelc14

1) p= 5/2 and q=-3/2
2) 2u
b) direction reversed
c12um Ns
3) x=34/15 and 2.27
b) weight acts through centre
rod comment
4) 0.473
5) 6N
6) 0.8 ms-2
b) 2.81
7) 103
b) position vectors you know dat ****
c) proof duh
d) t=0 and t=0.32 hours
8) equations of motion
b) proof
c) root :frown: