Question 11:
--- 11(a) ---
11(a) | 1×gcos20=1×a
a=9.21 (m s−2) | M1
A1
[2] | 3.3
1.1 | Use of F = ma tangentially soi – allow without explicitly
seeing m = 1 and award this mark for m×gcos20=m×a-
condone sign errors and sin/cos mix only
awrt 9.21or allow 9.2 www
--- 11(b) ---
11(b) | 1×v2
32−1×gsinθ=
0.5
1(1)(3.2)2 = 1(1)v2 −1(g)(0.5sinθ)
2 2
Substitute for v2 in energy equation
1(1)(3.2)2 = 1(1)(16−0.5×1×gsinθ)−1(g)(0.5sinθ)
2 2
θ=23.1 | M1*
M1*
M1dep*
A1
A1
[5] | 3.3
1.1
3.4
1.1
1.1 | Use of N2L radially – correct number of terms – allow
sign errors and sin/cos mix - allow without explicitly
seeing m = 1 stated/used– must be using a= mv2 with
r
correct values of m and r but allow T for 32 for this mark
Use of conservation of energy (correct number of terms
with h=0.5sinθ or h=0.5cosθ only (if taking the initial
position as PE = 0)) – allow sign errors
Or solve simultaneous equations in v2 and sinθ (oe) – all
values must have been substituted for this mark
Correct equation in θ only – note that
32=29.4sinθ+20.48(oe) scores the first four marks
AG allow awrt 23.1 (for reference: 23.068…) should be
from sinθ= 96 or 0.3918… - condone 23.1 being stated
245
after a correct equation (as AG must check this
carefully) – any wrong working seen scores A0 – note
that θmight be defined as the angle to the vertical – if
90−θis then considered then allow for full marks
--- 11(c) ---
11(c) | Vertical distance to fall is 1.55 or 1.75−0.5sin23.1
Velocity when string breaks is 3.75 (m s-1)
1V2 = 1×3.752 +g(1.75−0.5sin23.1)or
2 2
V2 =(3.75×sin23.1)2 +(3.75×cos23.1)2 +2×g×1.55
V =6.67(m s-1) | B1
B1
M1
A1
[4] | 3.1b
1.1
3.4
1.1 | Allow 1523or an answer in the interval [1.544, 1.554] –
980
3 sf (e.g. 1.54 or 1.55) or better
Allow for v2 = 352 or v= 4 22 or awrt 3.75 for v or a
25 5
value in the interval [14.077, 14.085] for v2– 3 sf (e.g.
3.75) or better (allow un-simplified e.g.
v= 10.24+gsin23.1 or v= 16−0.5×gsin23.1)
Conservation of Energy with correct number of terms but
allow sign errors – with their 14.08 or 3.75 (but M0 if
using 3.2) – for h allow from 1.75−0.5sin23.1 or
1.75−0.5cos23.1 only. Their 3.75 must come from an
equation with the correct number of relevant terms. If
using v2 =u2 +2as to find V, then must be using the
equivalent of their 3.75 and not just a single component
of their 3.75 (must also be using a=±g )
If using exact values may see V2 = 2227 - awrt 6.67 – 3 sf
50
or better
Alternative solution
M1 | Conservation of Energy (loss of PE = gain in KE) –
setting up an equation with the correct number of terms
using the 1.75, 3.2 and the speed of P at A only
1×g×1.75= 1×1×(v2 −3.22)
2 | A1 A1 | A1 for correct LHS, A1 for correct RHS
v=6.67(m s-1) | A1 | awrt 6.67 – 3 sf or better
--- 11(d) ---
11(d) | 1.75−0.5sin23.1=(3.75cos23.1)t+0.5gt2
Time to A is 0.312 (s)
Horizontal distance travelled = 0.31...×(3.75sin23.1)
Distance = 0.459
(Which is very close to) 0.5cos23.1=0.460(so A is
almost directly below O) | M1*
A1
M1dep*
A1
A1
[5] | 3.3
1.1
3.4
1.1
3.5a | For vertical motion – if correct expect to see
1.55...=3.45t+0.5gt2 - using their 3.75 ×cos23.1 or
their 3.75 ×sin23.1(from part (c)). Allow sign errors on
RHS, but LHS must be 1.75−0.5sin23.1only. Must be
using a=±g.
Watch out for other complete methods that lead to the
time of flight e.g. finding the two vertical components of
velocity and applying v=u+atwith a= g e.g. if correct
expect to see 6.51−3.45= gtor 6.50−3.45= gt (oe)
awrt 0.312 or awrt 0.311 or allow 0.31 www (if using exact
then answer is 0.3121…) – also allow correct un-
6.51(or 6.50)−3.45
simplified expression for t e.g. (but
g
must be a correct expression for t and not just an equation
containing t)
Use of s = ut horizontally with their t – allow sin/cos mix
but must be using the correct value of 3.75 now
awrt 0.459 or 0.458 www (but not just 0.46) – so must
have been using a correct t
Must either explicitly state that 0.5cos23.1 = 0.459… or
0.46 (or better) or show the difference between the two
values is very close to zero (if either value is not
explicitly stated) – as a minimum must show explicitly
both values (but no further comment required) www
(dependent on all previous marks)