Question 8:
--- 8(a) ---
8(a) | 4mgx
3mg =
0.4
Distance is 0.7 (m) | M1
A1
[2] | 3.3
1.1 | 4mg(x−0.4)
Correct application of Hooke’s law (or 3mg = )
0.4
soi
cao
--- 8(b) ---
8(b) | 3mg−T =3mx
4mg(0.3+x)
3mg− =3mx
0.4
x+10g x=0which represents SHM
3
Use of 2.5= Aω
A = 0.437 (m) | M1*
A1
A1
M1dep*
A1
[5] | 3.1b
1.1
2.1
3.4
2.2b | Use of N2L with correct number of terms, allow sign errors
and a slip in one mass (e.g. allow mass m on the RHS but
not m on both sides) and allow a etc. for the acceleration –
allow T for tension (or an incorrect expression for T)
oe – allow un-simplified and allow a for acceleration
Must be a comment that this is SHM and must be using x
d2x
or - allow x=−10g x
dt2 3
Must have substituted their value of ω from an equation of
the form x+ω2x=0 (could be from x= Asin(ωt)but must
be equivalent to applying the equation 2.5= Aω with their
ω)
5 6
Exact or awrt 0.437 or allow 0.44 www
28
--- 8(c) ---
8(c) | 4mg(0.3)2
1(3m)(2.5)2 + =...
2 2(0.4)
3mg(0.7−d)
Smallest distance is 0.231 (m) | B1
B1
B1
[3] | 3.1b
1.1
1.1 | Setting PE as zero at the equilibrium position (A)
Correct initial KE and EPE terms added together (ignore
the addition of any initial PE term as this is covered in the
next mark)
Correct change in PE – allow 3mghonly if d =0.7−h
implied by later working – to award the first two B marks
there must be an equation that would lead to the correct
answer – condone consistent omission of m throughout for
full marks
awrt 0.231 or allow 0.23 www
Alternative solution 1 | Setting PE as zero at lowest point of motion
4mg(0.3+ A)2
=...
2(0.4) | B1FT | B1FT | Correct EPE term using their A from part (b) | Correct EPE term using their A from part (b)
3mg(0.7+ A−d) | B1FT | Correct change in PE term using their A from part (b) –
allow3mghonly if d =0.7+ A−h implied by later working
– to award the first two B marks there must be an equation
that would lead to the correct answer following through
their value of A (provided their answer for d would be
positive) – condone consistent omission of m throughout
for full marks
Smallest distance is 0.231 (m) | B1 | awrt 0.231 or awrt 0.232 or allow 0.23 www
Correct change in PE term using their A from part (b) –
allow3mghonly if d =0.7+ A−h implied by later working
– to award the first two B marks there must be an equation
that would lead to the correct answer following through
their value of A (provided their answer for d would be
positive) – condone consistent omission of m throughout
Alternative solution 2
10g
v2 = ×(0.4372 −0.32)
3 | 10g
v2 = ×(0.4372 −0.32)
3 | B1FT* | B1FT* | Correct use of v2 =ω2(A2 −x2)with their ω and A from
part (b) and x=0.3 to find the speed or speed2 when the
string becomes slack – for reference if correct: v=1.81... | Correct use of v2 =ω2(A2 −x2)with their ω and A from
part (b) and x=0.3 to find the speed or speed2 when the
0=1.81...2 −2g(0.4−d) | B1dep* | Correct use of v2 =u2 +2as with v=0,a=−gand their
initial speed (which if correct is 1.81…) which leads to a
positive value of d only – allow 0=1.81...2 −2ghonly if
d =0.4−himplied by later working
Smallest distance is 0.231 (m) | B1 | awrt 0.231 or awrt 0.232 or allow 0.23 www
Correct use of v2 =u2 +2as with v=0,a=−gand their
initial speed (which if correct is 1.81…) which leads to a
positive value of d only – allow 0=1.81...2 −2ghonly if