Question 13:
--- 13(a) ---
13(a) | T =4g
Tcosα+Rsinα=2g
4gcosα+Rsinα=2g
Tsinα−Rcosα=2(htanα)(2.8)2
4gsinα−Rcosα=2(htanα)(2.8)2
2g−4gcosα
4gsinα−cosα =15.68htanα
 sinα 
4cosα−2cos2α=1.6hsin2α
⇒4hsin2α+5cos2α=10cosα | B1*
M1*
A1
M1*
A1
M1dep*
A1
[7] | 1.1
3.3
1.1
3.3
1.1
3.4
2.1 | Resolving vertically for P (soi e.g. in an equation for Q)
Attempt to resolve vertically for Q – correct number of
relevant terms so must contain a component of T and R
but allow sign errors and sin/cos mix, must be
dimensionally consistent and using correct mass of 2 kg
but allow T instead of 4g
Where R is the normal contact force between Q and the
surface of the shell – must be using 4g for T now
Applying N2L radially for Q – correct number of
relevant terms so must contain a component of T and R
but allow sign errors and sin/cos mix on LHS, must be
dimensionally consistent and using correct mass of 2 kg
but allow T instead of 4g – must be using a=r(2.8)2with
h
eitherr =htanαor r = only (may have sub. for R)
tanα
oe e.g. 4gsinα−Rcosα=15.68htanα – must be using
4g for T now – note that substituting an incorrect
expression for R can still imply this mark
Eliminate R from the two equations to obtain an equation
in h and αonly
e.g. k =4,k =5 and k =10 (must be integers but accept
1 2 3
multiples of these but must be in this form) – values of k
do not need to be stated explicitly
Alternative Solution – applying N2L parallel to
the surface of the shell
T =4g | B1 | Resolving vertically for P (soi)
Acceleration component parallel to the plane is
2r(2.8)2sinα | B1 | soi – allow any trigonometric expression containing h and
αfor r but not just r
T −2gcosα=2(htanα)(2.8)2sinα | M1 | Applying N2L parallel to the slope for Q – correct
number of relevant terms (allow sign errors and sin/cos
mix on LHS), must be dimensionally consistent (so
component of Q’s mass but no component with T) and
using correct mass of 2 kg but allow T instead of 4g. Must
be using a=r(2.8)2sinα or a=r(2.8)2cosαwith either
h
r =htanαor r = only – if no component of
tanα
acceleration used then M0
4g−2gcosα=2(htanα)(2.8)2sinα | A3 | Award A2 for all terms correct but with sign errors only,
and A1 for sign errors only with a =(htanα)(2.8)2cosα
4gcosα−2gcos2α=15.68hsin2α
⇒4hsin2α+5cos2α=10cosα | A1
e.g. k =4,k =5 and k =10 (must be integers but accept
1 2 3
multiples of these but must be in this form) – values of k
do not need to be stated explicitly
[7]
Acceleration component parallel to the plane is
2r(2.8)2sinα
B1
αfor r but not just r
Applying N2L parallel to the slope for Q – correct
number of relevant terms (allow sign errors and sin/cos
mix on LHS), must be dimensionally consistent (so
component of Q’s mass but no component with T) and
using correct mass of 2 kg but allow T instead of 4g. Must
be using a=r(2.8)2sinα or a=r(2.8)2cosαwith either
h
r =htanαor r = only – if no component of
tanα
acceleration used then M0
T −2gcosα=2(htanα)(2.8)2sinα
M1
4gcosα−2gcos2α=15.68hsin2α
⇒4hsin2α+5cos2α=10cosα
A1
--- 13(b) ---
13(b) | 4gcosα=2g or 4gsinα=2(htanα)(2.8)2
e.g.
cosα= 1 ⇒4g× 3 =2h× 3×2.82or
2 2
cosα= 1 ⇒4h×3+5×1 =10×1
2 4 4 2
h= 5
4 | M1*
M1dep*
A1
[3] | 3.1b
2.1
2.2a | Setting R=0 or R>0or R0 in either their
4gcosα+Rsinα=2g or their
4gsinα−Rcosα=2(htanα)(2.8)2equation to obtain an
equation/inequality in α(and possibly h) only –
dependent on having scored the corresponding M
mark in part (a) or full working in this part (e.g. if
applying N2L parallel to the surface only in part (a))
Substituting their 4gcosα=2ginto either their
4gsinα=2(htanα)(2.8)2or their answer to part (a) oe
correct method that leads to an equation/inequality in h
only - dependent on having scored the corresponding
M mark(s) in part (a) or full working in this part
If h 5 or h< 5then must explicitly state that the
4 4
greatest value is 5 www
4
Response | Mark
The first three marks are for forming two equations that can be used to find both F and R - most candidates will resolve
A A
horizontally and vertically to the rod (see main scheme). However, they may resolve parallel and perpendicular to the rod or take
moments again. There are two marks for the first equation (M1 A1) and one mark (A1) for the second. Please mark to the benefit of
the candidate so one correct equation and one incorrect equation would score 2 out of 3. The same rule applies for the second M
mark in that their equations must contain the required number of relevant terms (so if they take moments again about a different
point and e.g. a force/distance is missing then M0 for the second M mark).
 1.5  1.5 1.5
M(peg): −2 ×3gcos25+ ×F sin25= ×R cos25
sin25  sin25 A sin25 A
 1.5 
M(com): −2 ×R +2×F sin25=2×R cos25
sin25  P A A
 1.5 
M(B):4−  ×R +4×R cos25=2×3gcos25+4×F sin25
 sin25 P A A
R(parallel): F cos25+R sin25=3gsin25
A A
R(perpendicular): R+R cos25= F sin25+3gcos25
A A
PMT
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