Question 7:
--- 7(a) ---
7(a) | (m)a=(m)gsin=(m)g0.6 a=6  | M1 | For attempt at Newton’s second law; 2 terms; allow
sign errors; allow sin/cos mix, allow =36.9 or
better.
Allow =37. Allow for a=6 seen/used.
v2 =02+260.75 v=3
  | A1 | AG.
3= 0 +6t t=0.5 | A1 | AG.
Allow A2 for use of a=6 with any 2 of v=3,
t=0.5 and s=0.75 to get the third value.
Alternative Method 1 for Question 7(a)
(m)a=(m)gsin=(m)g0.6 a=6  | M1 | For attempt at Newton’s second law; 2 terms; allow
sign errors; allow sin/cos mix, allow =36.9 or
better.
Allow =37. Allow for a=6 seen/used.
0+60.5=3 | A1 | AG.
1
00.5+ 60.52 =0.75
2
1
or 30.5− 60.52 =0.75
2 | A1 | AG.
1
Allow A2 for 30.5− 60.52 =0.75.
2
Alternative Method 2 for Question 7(a)
1
(m)v2 =(m)g0.750.6
2 | M1 | Attempt at conservation of energy, 2 terms,
dimensionally correct, allow sin/cos mix, allow
=36.9 or better. Allow=37.
v=3 | A1 | AG.
Use constant acceleration to get t=0.5 | A1 | AG. Must show a=6.
Question | Answer | Marks | Guidance
7(a) | Alternative Method 3 for Question 7(a)
1
(s=) (0+3)0.5
2
1
or 0.75= (0+v)0.5
2
1
or 0.75= (0+3)t
2 | M1 | 1
Use s= (u+v)t with u=0 with any 2 of v=3,
2
t=0.5 and s=0.75.
Correctly identifies one value. | A1
(s=)0.75
or v=3
or t=0.5 | A1 | AG.
Question | Answer | Marks | Guidance
7(a) | Alternative Method 4 for Question 7(a)
32 = ( 02 + ) 2a0.75 →a=6
or 3=(0+)0.5a →a=6
1
or 0.75=(00.5+) a0.52 →a=6
2
1
or 0.75=30.5− a0.52 →a=6
2 | B1 | Use constant acceleration to get an equation in a
only, using u=0 and any 2 of v=3, t=0.5 and
s=0.75.
Or using v=3, t=0.5 and s=0.75.
1 1
3=(0+)6t , 0.75=(0t+) 6t2, 0.75=3t− 6t2
2 2
1 1
or s=(00.5+) 60.52, s=30.5− 60.52
2 2
or v=(0+)0.56; v2 = ( 02 + ) 260.75;0.75=0.5v− 1 60.52
2
1
or 3=u+60.5, 32 =u2 +260.75, 0.75=0.5u+ 60.52
2 | M1 | Use constant acceleration formula in an attempt to
find the third unused value using a=6.
Or use constant acceleration formula in an attempt to
find u using a=6.
Get correct unused value | A1 | Or u=0 if using all 3 values to find a=6.
AG.
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Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | For BC (m)a=(m)gsin−(m)g0.80.25= (m)g0.6−(m)g0.80.25 a=4  | *M1 | For attempt at Newton’s second law; 3 terms; allow
sign errors; allow g missing; allow sin/cos mix, allow
=36.9 or better.
1 1
For P, s =3t+ 4t2 For Q, s = 4(t+0.5)2
P Q
2 2
1 1
Or for P, s =3(T −0.5)+ 4(T −0.5)2 For Q, s = 4T2
P Q
2 2 | A1 | For either s or s .
P Q
t is the time after P arrives at B.
T is the time from when both particles released.
1 1
3t+ 4t2 = 4(t+0.5)2 3t+2t2 =2t2 +2t+0.5
 
2 2
1 1
Or 3(T −0.5)+ 4(T −0.5)2 = 4T2
2 2
3T −1.5+2 ( T2 −T +0.25 ) =2T2
  | DM1 | For use of s =s ; s and s of correct form.
P Q P Q
Using their atheir 6 from part (a), ag.
Using same a in both s and s .
P Q
t = 0.5, so total time = 1 s or time T=1s | A1
Question | Answer | Marks | Guidance
7(b) | Alternative Method for Question 7(b)
For BC (m)a=(m)gsin−(m)g0.80.25= (m)g0.6−(m)g0.80.25 a=4  | *M1 | For attempt at Newton’s second law; 3 terms; allow
sign errors; allow g missing; allow sin/cos mix, allow
=36.9 or better.
1
For P s =3t+ 4t2
P
2
1
Or Q has speed 2 m s–1 after 0.5 s, so s =2t+ 4t2
Q
2 | A1 | For either. t is the time after P arrives at B.
1 1 1
3t+ 4t2 =2t+ 4t2 + 40.52
2 2 2 | DM1 | 1
For use of s =s their 40.52; s or s of
P Q P Q
2
correct form.
Using their atheir 6 from part (a), ag.
Using same a in both s or s .
P Q
t = 0.5 so total time = 1 s | A1
Alternative Method 2 for Question 7(b): Using relative velocity
For BC (m)a=(m)gsin−(m)g0.80.25= (m)g0.6−(m)g0.80.25 a=4  | *M1 | For attempt at Newton’s second law; 3 terms; allow
g missing; allow sign errors; allow sin/cos mix.
[In 0.5 s] Q has speed 2ms–1 and has moved 0.5m | A1 | For both.
0.5
t=
3−2 | DM1 | Attempt at time from relative velocity using their 0.5
m and their 2ms–1.
t = 0.5 so total time = 1s | A1
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Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) |  1
Immediately before the collision speed of P =3+4 =5 ms−1
 
 2
Speed of Q =0+41 =4m s–1 | B1 | For either.
5(m)+4(m)= (m)+(m) v v=4.5 
  | *M1 | Use of conservation of momentum; 4 non-zero
terms; allow sign errors; allow their 4m s–1 (0 or 2)
and 5m s–1 (3).
Use of mg then withhold final A mark.
1
Distance from B at collision = 412 =2 m
2
1
OR =30.5+ 40.52 =2 m
2 | B1 | May be implied by 1.25m.
1
1 .25=4.5t+ 4t2
2 | DM1 | Use of constant acceleration with u=their 4.5,
a=their 4 (g) and s=3.25−their 2,
s3.25, s0.75, s4, s2.75, s0.5.
If using two formulae, must be a complete method to
get an equation in t only.
t=0.25s only | A1 | A0 if from use of mg in conservation of momentum.
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