Question 12:
--- 12(a) ---
12(a) | Velocity perpendicular to XY does not change and is
zero after (so therefore A must have been moving
along XY before the collision) | B1
[1] | 2.4 | Accept ‘perpendicular impulse is zero and A is moving to
the right before collision’ oe – as a minimum must refer
to either the vertical component of velocity is zero or
unchanged or that the perpendicular impulse is zero
--- 12(b) ---
12(b) | 3a+5=5bor 3a+5(2cos60)=5b
b=−0.8+0.8aor b−0=−0.8(2cos60−a)
Speed of A before is 9 (m s−1)
b=6.4
Speed of B after is b2 +(2sin60)2 =6.63 (m s−1) | M1
A1
M1
A1
A1
A1
A1
[7] | 3.3
1.1
3.3
1.1
1.1
2.1
1.1 | CLM - correct number of terms but allow sign errors and
sin/cos mix with the 60-degree angle, where a is the
component of the velocity of A before collision, and b is
the component of the velocity of B after collision – for b
accept vcosθ but no angle associated with a
NEL; correct number of terms – allow sign errors and
sin/cos mix with the 60-degree angle but e must be with
the approach speed - for b accept vcosθ but no angle
associated with a
Must be consistent with CLM
Condone just leaving answer as a = 9
Or vcosθ=6.4- this mark can be implied by a correct
speed of B
awrt 6.63 or allow 6.6 www
--- 12(c) ---
12(c) | 2sin60
tanθ=
b
θ=15.1

Angle turned through is 44.9( ) | M1
A1
A1FT
[3] | 3.1b
1.1
3.2a | Their value of b from part (a) must be substituted or for
2sin60 b
sinθ= with their 6.63 or cosθ= with their
6.63 6.63
b
b and their 6.63 – allow tanθ= with their b
2sin60
awrt 15.1 or 15 www – this mark can be implied by a
correct final answer of 44.9
Follow through 60−θ provided 60−θ>0(so θ must be
less than 60 for this FT mark) and M1 awarded – awrt
44.9 or allow 45 www
--- 12(d) ---
12(d) | Component of velocity perp to XY after collision is
e(2sin60)
KE before collision is 1×5×(6.42 +(2sin60)2) (=109.9 )
2
or speed component squared before collision is
6.42 +(2sin60)2
or speed component squared after collision is
6.42 +(2esin60)2
1×5×(6.42 +(e×2sin60)2)=0.95×1×5×(6.42 +(2sin60)2)
2 2
e=0.517 | B1
B1FT
M1
A1
[4] | 3.1b
1.1
3.4
1.1 | soi
Follow through their value of b (the 6.4) from part (b)
only
Correct use of either KE after = 0.95(KE before) or KE
before = 0.95(KE after), condone lack of 1×5on both
2
sides (but M0 if wrong mass used), allow their value of b
(the 6.4) but everything else must be correct
awrt 0.517 or allow 0.52 www