Question 11:
11 | (a) | 0.2u 0.6u 0.2 4 0.6  2 
A B
420.5 u u 
A B
u 9.5i2j
A
u 2.5i3j
B | M1*
A1
M1*
A1
M1dep*
A1
A1
[7] | 3.3
1.1
3.3
1.1
2.1
2.2a
1.1 | Attempt at conservation of linear
momentum – correct number of terms
Attmpt at NLR – correct number of
terms and speed of approach must be
multiplied by coefficient of restitution
oe 420.5u u . Must be
A B
consistent with CLM e.g.
0.2u 0.6u 0.240.62and
A B
420.5u u would score the
A B
first 4 marks
Solving simultaneous equations and
correct j components (u and u must
A B
be of the form ij)
ISW if speeds found after correct
velocities | Let u u iv j and
A A A
u u iv j
B B B
u
No components of
A
u
and for this mark
B
(unless done correctly)
11 | (b) | Velocity of B after impact is 2ei3j
1  0.6  2232    2e232   1.152
 
2
e0.2 | B1
M1
A1
[3] | 3.1b
3.4
1.1 | Condone 2ei3j (or for dividing their
answer by 2 if using v for 2e)
1
Or for  0.6  222e21.152or
 
2
1  
for  0.6  2e232 2.748or for
2
1
 0.6 2e2 0.048 - allow slip in
2
one power only – must be using 2e
(either seen or implied by dividing by
2) | May be seen in an
equation for e
Must be using speeds
after collision (so must
be using 2i + 3j or just
2i) and correct mass – if
using say v for 2e then
this mark can only be
earned when dividing
by 2
11 | (c) | 2 3
tan or tan
3 2
2e 3
tan or tan
3 2e
1  1 2e
Deflected angle = tan 2 tan
3 3
41.3 | B1
B1ft
M1
A1
[4] | 1.1
1.1
3.1b
1.1 | In degrees: 33.6900675...
With their e - if correct for reference
7.59464336...
 tan1  tan1 
or 180 3 3
2 2e
41.284710… (correct to 1 d.p.) | Radians:
0.5880026...
0.1325515...
0.72055413…