Question 7:
--- 7(a) ---
7(a) | T −3g =3a
5g−T =5a
5g−3g =(3+5)a | *B1 | For one correct equation.
*B1 | For any two correct consistent equations. If tensions T and T
A B
both stated must at some point state or imply that they are equal to
score the second B1.
Attempt to solve for T | DM1 | May find a first a=2.5  Must get to 'T='. Dep on both B
marks.
75
T =37.5N or N
2 | A1 | Allow without working.
Correct use of suvat with their a and solve for t | DM1 | E.g. 2=0.52.5t2 Must get to 't='. Dep on both B marks or
the B1 for the equation 5g−3g =(3+5)a if this used to find a,
but not on first M1 mark.
0+their 10
Could find v= 10 then use 2= t .
2
40 2 10
t=1 .26 or or
5 5 | A1 | t=1.2649 If candidates do not try to find T but do attempt to
find the time, they can score B1B1M0A0M1A1. Do not allow if
also give negative answer and do not discard. Note:
t = 1.6only, scores A0 .
Question | Answer | Marks | Guidance
7(a) | Alternative for final 2 marks, even if nothing scored earlier
Use of energy to find velocity at plane and then suvat to find t | M1 | 1 1
PE loss = KE gain: 5g2−3g2 = 5v2 + 3v2,
2 2
Or PE loss – WD by tension = KE gain: 5g2−their 37.52 =
1
5v2 ,
2
Or WD by tension – PE gain = KE gain:
1
their 37.52−3g2= 3v2 ,
2
0+v
v= 10 or v=3.16 then 2= t,
2
Dependent on both B marks only if candidate uses tension, but
otherwise not dependent on either B mark.
40 2 10
t= 1.6 =1.26 or or
 
5 5 | A1 | t=1.2649
6
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Correct use of suvat before B hits the plane or when A has risen 2
m, to attempt to find velocity of A when string becomes slack
Using their a and/or their t. | *M1 | v2 =0+22.52. Must use s=2 and u=0,
OR v=0+2.5 1.6 Must use u=0 ,
1
OR 2= (0+v) 1.6 Must use s=2 and u=0.
2
Must be complete method to find v or v2 v= 10 .
 
Could have found v= 10 in part (a) and give M1 if used in part
(b).
Correct use of suvat for motion of A (between height of 3m and
3.25 m), to form an equation in t, using a=−g | *DM1 | 3.25−3= ( their 10 ) t+0.5(−10)t2.
Solving a 3 term quadratic for t to get at least one (unsimplified)
value using their 2.5 (or using any other correct method) | DM1 | 10− 5 10+ 5
If correct should get t=0.093,0.540 or , ,
10 10
0.09262… or 0.53978…
Could use formula and realise that the time for at least 3.25 m,
b2 −4ac 10−450.25
=2 =2 which gets DM1.
2a 25
5 1
Time = 0.447 s or s or
5 5 | A1 | Time =0.53978−0.09262 = 0.44721…
 10 10− 5
2 −  = 0.44721…
 
 1 0 10 
Question | Answer | Marks | Guidance
7(b) | Alternative for last 3 marks of Q7(b) finding max height
For attempt to find max height using correct suvat with a=−g | *DM1 | 02 = ( their 10 )2 +2(−10)s Where s is distance above 3
metres.
(which leads to a maximum height of 3.5 m).
For attempt to find time from .25 m below top to top | DM1 | 5
t= 0.224 or [0.22360….] probably from
10
3.5−3.25=0t+0.510t2. Dep on both previous M1s.
 
5
Time = 0.447 s or s
5 | A1 | For doubling.
Alternative for last 3 marks of Q7(b) by finding velocity at height of 3.25 m
Correct use of suvat for motion of A (between height of 3m and
3.25 m), to form an equation to find speed at height of 3.25 m | *DM1 | w2 =their 10 2 +2(−g)0.25. w2 =5 or w= 5.
   
For correct use of suvat to find time to max height | DM1 |  5
0= 5+(−g)t t =  .
 10 
5
Time = 0.447 s or s
5 | A1 | For doubling.
Question | Answer | Marks | Guidance
7(b) | Alternative using energy
Correct use of energy before B hits the plane or when A has risen 2
m, to attempt to find velocity of A when string becomes slack
using their T if necessary. | *M1 | 1 1
PE loss = KE gain: 5g2−3g2 = 5v2 + 3v2,
2 2
Or PE loss – WD by tension = KE gain: 5g2−their 37.52 =
1
5v2 ,
2
Or WD by tension – PE gain = KE gain:
1
their 37.52−3g2= 3v2 .
2
Must be complete method to find v or v2 v= 10 or 3.162.
 
Do not allow sign errors.
Correct use of energy to find velocity of A at height of 3.25 m | *DM1 | 1 2 1
PE gain = KE loss: 3g0.25= 3 10 − 3w2 .
2 2
Must be complete method to find w or w2 w= 5 or 2.236
Do not allow sign errors.
For attempt to find time from .25 m below top to top | DM1 | 5
t= 0.224 or [0.22360….] probably from 0= 5−10t .
10
Dep on both previous M1s.
Do not allow sign errors.
5
Time = 0.447 s or s
5 | A1 | For doubling.
4