Question 11:
--- 11(a)(i) ---
11(a)(i) | T = 3mg sin α [= 2mg] | B1 | Find T by resolving forces along plane on P
T = kmg (5a/4 – a) / a [= ¼ kmg] | B1 | Find T using Hooke’s Law
k = 8 | B1 | Combine using sin α = ⅔ to find k
Total: | 3
--- 11(a)(ii) ---
11(a)(ii) | EITHER:
± 2m d2OQ/dt2 = 2mg sin α – kmg (OQ – a) / a | (M1 A1 | Apply Newton’s law at general point (e.g. below E)
d2OQ/dt2 = (4g/a) (7a/6 – OQ) | A1 | Substitute values of k and sin α
d2x/dt2 = – (4g/a) x where x = OQ – 7a/6 | A1) | Derive standard SHM form (requires minus sign)
OR:
2mg sin α = kmg (e – a) / a, e = 7a/6 | (M1 | Find new equilibrium distance e from O
± 2m d2x/dt2 = 2mg sin α – kmg (e + x – a) / a | M1 A1 | Apply Newton’s law at general point (e.g. below E)
d2x/dt2 = – (4g/a) x | A1) | Derive standard SHM form (requires minus sign)
Centre is 7a/6 (or 1⋅17 a) from O | B1 | State centre of motion
Period is π√(a/g) or 0⋅993√a | B1 | State period in simplified form, allowing g = 10
Total: | 6
--- 11(a)(iii) ---
11(a)(iii) | x = 5a/4 – e = a/12
0 | B1 | Find amplitude x of motion
0
T = kmg (5a/4 – 2x – a) / a = 2 mg/3
min 0 | M1 A1 | Find least tension
(d2x/dt2) = [±] (4g/a) x = [±] ⅓ g
max 0 | M1 A1 | Find maximum acceleration (accepting either sign)
Total: | 5
Question | Answer | Marks | Guidance
--- 11(b)(i) ---
11(b)(i) | x= (1/250) Σ x f(x) = 414/250 = 1⋅656AG | B1 | Verify given mean
Total: | 1
--- 11(b)(ii) ---
11(b)(ii) | p = x/6 = 0⋅276, q = 0⋅724 | M1 A1 | Use 250 6C q6-ipiand find pandq
i
a=250 6C q4 p2 = 78⋅49 ±0⋅01 (to 2 d.p.)
2 | A2 | Find either exp. value
b= 250 6C q2 p4 = 11⋅41 ±0⋅01 (to 2 d.p.)
4 | A1 | Find other exp. value(deduct single A1if either value
given to only 1 d.p.)
Total: | 5
--- 11(b)(iii) ---
11(b)(iii) | H : Distribution fits data ordistribution is binomial (AEF)
0 | B1 | State (at least) null hypothesis in full
Combine values consistent with all exp. values ⩾ 5
O: 48 69 78 32 23
i
E: 36⋅01 82⋅36 78⋅49 39⋅89 13⋅26 (±0⋅01)
i | M1FTA1 | (FTfor M1but not A1on values of a, b)
χ2 = 3⋅992 + 2⋅167 + 0⋅003 + 1⋅561 + 7⋅154 | M1 | Find χ2
=14⋅9 | A1
No. nof cells: 7 6 5 4 3
χ 2: 15⋅09 13⋅28 11⋅34 9⋅210 6⋅635
n-2, 0.99 | B1FT | State or use consistent tabular value χ 2(to 3 s.f.)
n-2, 0.99
[FT on number, n, of cells used to find χ2]
Accept H if χ2 > tabular value (AEF)
1
14⋅9 [±0⋅1] > 11⋅34 so distn. doesn’t fit [data] | M1 | State or imply valid method for conclusion
Conclusion (requires both values correct)
ormanager’s belief not justified (AEF) | A1
Total: | 8