Question 11:
11 | (a) | mucosα=mv +mv
1 2 | M1* | 3.3 | Use of conservation of linear momentum
– correct number of terms | m is the mass of A and
B, v is the
1
component of the
velocity of A parallel
to the line of centres
after impact and v is
2
the equivalent
component for B
1
v −v =− ucosα
1 2
3 | M1* | 3.3 | Use of Newton’s experimental law –
correct number of terms and consistent
with conservation of linear momentum
1
v = ucosα
1
3 | A1 | 1.1
usinα
tanβ=
v
1 | M1dep* | 3.4 | Use of tan ratio for β with their v
1
usinα
tanβ= ⇒tanβ=3tanα
1
ucosα
3 | A1 | 2.2a | AG – sufficient working must be shown
as answer given
[5]
11 | (b) | The component of the velocity of A perpendicular to the
line of centres does not change | B1 | 3.5b
[1]
11 | (c) | 3tanα−tanα
tanγ=tan(β−α)=
1+(3tanα)(tanα) | M1 | 3.1b | Use of a correct compound-angle formula
tan(β±α)
for and substitute given
result from (a)
2tanα
tanγ=
1+3tan2α | A1 | 1.1
[2]
11 | (d) | M1* | 3.1b | Attempt to differentiate using quotient
rule
 dγ 
sec2γ =
 dα 
( )( ) −(2tanα)( )
1+3tan2α 2sec2α 6tanαsec2α
=0
( )
1+3tan2α | A1 | 1.1 | Correct derivative equated to zero
1
1+3tan2α−6tan2α=0⇒tanα=
3 | M1dep* | 1.1 | Find value of tanα or tan2α | π
α=
6
( )
2 3
3
tanγ= ⇒γ=...
2
 3
1+3 
 3  | M1 | 1.1 | Substitute their value for αor tanα into
their expression for tanγ- dependent on
both previous M marks | 3
tanγ=
3
π
γ=
6 | A1 | 1.1
[5]