Question 3:
3 | (a) | (i) | 12  6  u 2 − 4  3 .5 = 6 g  3 .5 s i n 2 5 
 u = 5 .8 ( 0 1 5 7 5 ) | M1
B1
A1
[3] | 3.3
1.1
1.1 | Attempt at WEP; must have 3
terms.
Correct GPE term.
(ii) | Energy from C to A:
6 g  ( 3 .5 + x ) s i n 2 5  − 4 ( 3 .5 + x ) = 12  6  5 .8 0 1 5 7 5 2
 x = 1 .3 ( 4 2 9 2 8 ) | M1
A1FT
A1FT | 3.3
1.1
1.1 | Attempt at WEP; must have 3
terms.
Correct equation in x FT their value
of u from part (a)
Follow through their value u
Or energy from B to A:
12  6  u 2 + 6 g x s i n 2 5  − 4 ( 7 + x ) = 12  6  u 2 | 12  6  u 2 + 6 g x s i n 2 5  − 4 ( 7 + x ) = 12  6  u 2 | M1
A1FT | M1 | 3.3 | Attempt at WEP; must have 4
terms.
A1FT | 1.1 | Correct equation in x FT their value
of u from part (a).
 x = 1 .3 ( 4 2 9 2 8 ) | A1FT | 1.1 | Follow through their value u
[3]
If energy methods not used for both (a) (i) and (a) (ii)
award for correct answers.
𝑢 = 5.8(01575…)
𝑥 = 1.3(4298…) | SC1
SC1
[2] | cao
cao
(b) | =2N
Frictional force
R = 6 g c o s 2 5 
So = 2 =0.0375298
6gcos25 | B1
M1
A1
[3] | 3.4
1.1
2.2b | Component of the normal contact
force – allow sin/cos mix
Awrt 0.038
(a) | Let the velocities of A and B after impact be v and v , in
A B
the direction A B .
3  1 = 3 v + 1 v
A B
v − v = e  1
B A
 3 = 3 v + v + e  v = 14 ( 3 − e )
A A A
v = 14 ( 3 − e ) + e = 34 ( 1 + e )
B | M1
M1
A1
A1
[4] | 3.3
3.3
3.4
1.1 | Attempt at COLM.
Attempt at NEL
AG (note speed requested)
(b) | 12  3   14 ( 3 − e )  2 + 12  1   34 ( 1 + e )  2 = 0 .7 9  12  3  1 2 
 e = 0 .4 | M1
A1FT
A1
[3] | 3.3
1.1
1.1 | Using the information to set up an equation in e with
their expressions.
Correct equation FT their expressions from part (a).
(c) | If they had different radii, the impact could not be direct. | B1
[1] | 3.5b | Accept explanations that there cannot be a vertical
impulse/force for motion to continue horizontally
only.
Let the velocities of C and D after impact be v and v , in
C D
the direction C D .
23 ( u + 1 ) = 2 ( v − ( − 1 ) )
D
 v = 13 u − 23
D
( )
1  u − 2  1 = 1  v + 2  13 u − 23
C
v = u − 2 − 23 u + 43 = 13 u − 23 which is the same as v so they
C D
have the same velocity afer impact.
Or | M1
A1
M1
A1 | 3.3
1.1
3.3
2.2a | Attempt at impulse-momentum equation: condone
incorrect signs.
Attempt at COLM: condone incorrect signs.
2
− (𝑢+1) = 𝑣 −𝑢
3 𝐶 | M1 | 3.3 | Attempt at impulse-momentum equation: condone
incorrect signs.
1 2
⇒ 𝑣 = 𝑢− which is the same as v so they have the
𝐶 3 3 D | A1 | 2.2a
same velocity afer impact.
Or
Let the velocities of C and D after impact be v and v , in
C D
the direction C D .
23 ( u + 1 ) = 2 ( v − ( − 1 ) )
D
𝑢−2 = 𝑣 +2𝑣
𝐶 𝐷 | Let the velocities of C and D after impact be v and v , in
C D | 3.3
3.3 | Attempt at impulse-momentum equation: condone
incorrect signs.
Attempt at COLM: condone incorrect signs.
the direction C D .
23 ( u + 1 ) = 2 ( v − ( − 1 ) )
D
M1
𝑢−2 = 𝑣 +2𝑣
𝐶 𝐷 | M1
𝑢 = 𝑣 +2𝑣 +2 Substitute into the Impulse-momentum
𝐶 𝐷
equation
3𝑣 = 𝑣 +2𝑣
𝐷 𝑐 𝐷 | A1 | 1.1 | For correct equation eliminating u
𝑣 = 𝑣
𝐶 𝐷 | A1 | 2.2a | AG For correct argument leading to equality
(e) | ( ) 2
12  3  13 u − 23
12  1  u 2 + 12  2  1 2
4 4
𝑢2−4𝑢+4 1− +
𝑢 𝑢2
= (= )
3𝑢2+6 6
3+
𝑢2
→ 13 u →  23
as so the fraction of energy lost tends to . | M1
M1
A1
[3] | 3.1b
1.1
2.2a | Attempt at formulating an expression in u to
represent the fraction of energy remaining or lost.
Simplifying to a form where the limit can be taken
or is apparent.
cao must follow correct expressions for
𝑉 𝑎𝑛𝑑 𝑉
𝐶 𝐷
3.3
3.3
Attempt at impulse-momentum equation: condone
incorrect signs.
Attempt at COLM: condone incorrect signs.
(a) | Total area = 12  ( 3 0 + 1 2 0 )  9 0 = 6 7 5 0
 x   2 0   4 5   8 0 
6 7 5 0 = 1 3 5 0 + 2 7 0 0 + 2 7 0 0
y 3 0 4 5 3 0
 x = 5 4
and y=36 | B1
B1
M1
A1
A1
[5] | 1.1
1.2
3.3
1.1
1.1 | Correct centre of mass for
triangle (implied by correct
position vector in either of first
or third term on RHS).
Fully correct equation for
either x or y.
AG: some manipulation
required in order to obtain
answer.
(b) | If M is the centre of mass, it must lie within BCD, so in the
extreme cases it lies on either BD or CD.
If it lies on BD, then 35 64 −− 0d = 96 00 −− 35 64 (  d = 5 0 )
or in form 𝑦 = 9𝑥−450
If it lies on CD, then 35 64 −− 0d = 93 00 −− 35 64 (  d = 7 0 )
9 315
or in form 𝑦 = − 𝑥+
4 2
So 5 0  d  7 0 . | M1FT
A1FT
A1FT
[3] | 3.1b
1.1
2.2a | Follow through incorrect 𝑦̅
Considering gradients or
equations of lines.
One correct equation (implied
by one correct limit).
Range must be clearly stated,
in words if not using
inequalities or set notation.
Take moments about a line parallel to y-axis eg. x = 3 0 :
3 0 R + 3 3 R = 2 4  1 0 0
B D
𝑅 = 60 , 𝐷 R B = 1 4 and R C = 2 6 | M1
A1
[4] | 3.3
2.2a