Question 10:
10 | (a) | Initially, PE = m g ( r + r s i n 3 0 ) ( = 1 .5 m g r ) | B1 | 1.1 | Or if reference level is through O then for m g r s i n 3 0
At angle ,  PE = m g ( r r c o s )  − | B1 | 1.1 | Or if reference level is through O then for m g r c o s  −
Or B2 for mgr(sin30+cos)
12 m u 2 m g ( r r s i n 3 0 ) m g ( r r c o s ) 12 m v 2  + + = − + | M1* | 3.3 | Use of conservation of energy between initial position
and when P is at B (at least 2 KE and 2 PE terms)
v 2 u 2 g r 2 g r c o s  = + + | A1 | 1.1 | Correct expression for the speed or speed-squared
when P is at B (so must have been re-arranged or
implied by later working) – correct expression for v
or v2implies the first four marks
m v 2
R m g c o s  − =
r | M1* | 3.3 | N2L radially with correct number of terms and weight
resolved (but allow sin/cos confusion and sign errors)
2 v
– acceleration must be either or r2
r
m  
R m g c o s u 2 g r 2 g r c o s   − = + +
r | M1dep* | 3.4 | Substitute expression for v 2 (with correct number of
terms)
m u 2
R 3 m g c o s m g  = + +
r | A1 | 2.2a | Must be simplified to three terms
[7]
(b) | m u 2
At A, R + m g c o s 6 0 =
r | B1 | 3.3
m u 2 m u 2    
3 m g c o s m g m g c o s 6 0 4 m g  + + − − =
r r | M1* | 1.1 | The difference of two expressions for R (their
expression from (a) with three terms, and their two
term expression for R at A with the equivalent of a
single term for the resolved weight) equated to 4mg
3 c o s 32 4 c o s 56   + =  = | M1dep* | 1.1 | Obtain cos=kwhere 0  k  1 (or corresponding
value of )
Vertical distance = r − 56 r = 16 r | A1 | 2.2a
[4]
(c) | Diameter of the rim of the bowl is 2rcos30 (=r 3) | B1 | 1.1
 2 r 3 
r 3 = ( u c o s 6 0 ) t  t =
u | M1* | 3.1b | Realise that P leaves the inner surface with speed u
and applies s = u t horizontally with s = their diameter
(  r , 2 r ) and component of u
2
 2 r 3  1  2 r 3   6 g r 2 
y = ( u s i n 6 0 ) − g = 3 r −
u 2 u u 2 | M1dep* | 3.4 | Applies s = u t + 0 .5 a t 2 vertically with their value of t
6gr2
y03r− 0
u2 | M1 | 3.1b | Sets their expression for y (in terms of r, g and u) > 0
– dependent on both previous M marks (condone
‘equals’ or ‘greater than or equal to’ zero for this
mark)
( )
3 r u 2 − 2 g r  0  u 2  2 g r | A1 | 2.2a | AG (if using non-strict inequality or equals then
argument for strict inequality must be convincing)
[5]
Diameter of the rim of the bowl is 2rcos30 (=r 3) | B1
u c o s 3 0 t − 12 g t 2 = 0  t = ... | u c o s 3 0 t − 12 g t 2 = 0  t = ... | M1* | Finding the time (from s = u t + 0 .5 a t 2 with s = 0, a
component of u and a=g) when P is at the same
horizontal level as the rim of the bowl – if correct
u 3
then t=
g
u 3
x=(usin30)t= 1u 
2  g 
  | M1dep* | Using their t and s = ut horizontally (with a
component of u) to find the horizontal distance
travelled by P when it is at the same horizontal level
as the rim
u 3
1u r 3
2  g 
  | M1 | Sets their x > their r 3 (not just r or 2r) – condone
‘equals’ or ‘greater than or equal to’ – dependent on
both previous M marks
u2 2gr | A1 | AG | AG
R = F + 1 5 , P = F + R
A C A C | A1 | 1.1 | Both correct
(Least value of P implies) F =0.5R , F =0.5R
A A C C | B1* | 3.1b | For both (no justification required)
Taking moments to form an equation containing all relevant
forces | M1* | 3.3 | Taking moments about e.g. A, B, C etc.– correct
number of terms (at least 4 terms if A or C and 5
terms if B) – one for each force), dimensionally
consistent – allow sin/cos confusion. Must have
numerical values for all distances and each term must
be resolved with a term in sin/cos
P(3cos)+4(15cos)=1(15sin)+12(R sin)−12(F cos)
C C
P 8 (1 5 c o s ) 1 (1 5 s i n ) (1 2 s i n 3 c o s )     + + −
R (1 2 c o s ) F (1 2 s i n )   = +
A A | A2 | 1.1
1.1 | Award A1 for any two terms correct
A2 for a fully correct equation
P ( 3 c o s ) 1 5 1 7 s in ( a r c ta n 4 ) 1 2 ( R s in ) 1 2 ( F c o s )     = − + −
C C
3 P c o s 6 0 c o s 1 5 s i n    + −
4 P 3 0 4 P 3 0  −   − 
1 2 s i n 6 c o s   = −
5 5 | M1dep* | 1.1 | Deriving an equation containing P and only
(Reference:R =0.5R +152(P−R )=0.5R +15
A C C C
4 P − 3 0
 R = )
C
5
120cos+285sin 40cos+95sin
P= =
48sin−39cos 16sin−13cos | A1 | 2.2a | AG
Possible intermediate step is
15Pcos+300cos−75sin
=12(4P−30)sin−6(4P−30)cos
[8]