A Level Mathematics 9709 Paper 43, Oct/Nov 2023

Use of suvat to find expressions for $s$

$s=3u-5\times 9$ $s=4u-5\times 16$

$u=35,s=60$

Attempt at conservation of momentum [$1.2v=(1.2+0.004)\times 40$]

$v=\frac{602}{15}$

$0^2=(40{)}^2+2\times 0.04\times a[a=-20000]$ or $0.04=\frac{0+40}{2}t\text{ gets }t=0.002,\text{ so }0=40+0.002a[a=-20000]$

Attempt to use Newton’s Second Law vertically. $[-R+(1.2+0.004)g=(1.2+0.004)\times a]$ $[-R+12.04=1.204a]$

$R=24100\text{ N }[24092.04=\frac{602301}{25}]$

Correct force diagram with 3 forces in the correct directions.

$R=8g\cos 30[=40\sqrt{3}=69.282\dots ]$

Resolving parallel to the plane and attempt to apply Newton’s second law. $[8g\sin 30-F=8\times 2.4⟹F=20.8]$

Use of $F=\mu R$ to get an equation in $\mu$ only. $[8g\sin 30-8g\mu \cos 30=8\times 2.440-40\sqrt{3}\mu =19.2]$

$\mu =0.3[0\dots ]\left[\text{May first see }\frac{20.8}{40\sqrt{3}}\text{ or }\frac{20.8}{69.282\dots }\right]$

[$v^2=2\times 2.4\times 3⟹\text{greatest speed}=]3.79{\text{ ms}}^{-1}=\frac{6\sqrt{10}}{5}$

$P=480\times 24$ or, e.g. $\frac{P}{24}-480=0$

$P=11.52$ [kW]

[${\text{KE}}_{\text{before}}$] $=\frac{1}{2}\times 1600\times {24}^2[=460800]$ [${\text{KE}}_{\text{after}}$] $=\frac{1}{2}\times 1600\times {32}^2[=819200]$

[${\text{PE}}_{\text{loss}}$] $=1600g\times 280\times 0.09[=1600g\times 25.2=403200]$

Total WD $=12000\times 10[=120000]$

Work done against resistance = or $280F=$ or WD = or W = oe $12000\times 10+1600g\times 280\times 0.09-\frac{1}{2}\times 1600\times {32}^2+\frac{1}{2}\times 1200\times {24}^2$ $[=120000+403200-819200+460800]$

WD $=164800$ [J]

Attempt to resolve in one direction and form equation.

$T\sin \theta =32$ and $T\cos \theta =80$ or $0=80\sin \theta -32\cos \theta$ and $T=80\cos \theta +32\sin \theta$

Attempt to solve for $T$ or $\theta$

$T=86.2$ [N 86.1626….] or $16\sqrt{29}$ or $\sqrt{7424}$ and $\theta =21.8$ [21.801…]

Attempt to resolve in one direction and form equation

$120\sin \theta =P$ and $120\cos \theta =80$ or $0=80\sin \theta -P\cos \theta$ and $120=80\cos \theta +P\sin \theta$

Attempt to solve for $P$ or $\theta$

$P=89.4$ [89.4427] or $40\sqrt{5}$ or $\sqrt{8000}\theta =48.2$ [48.1896…]

Attempt to integrate $a$

$[v=]36t-3t^2[+c]$ or $[v=]36t-\frac{6t^2}{2}[+c]$

$0=36t-3t^2-33$ $[27=36\times 2-3\times 2^2+c⟹c=-33]$

Solve $0=36t-3t^2-33$ to get $t=1$ and $t=11$

Attempt to integrate an expression of the form $at+b{t}^2[+c]$ with non-zero $a$ and $b$. If correct $[s=]\frac{36t^2}{2}-\frac{3t^3}{3}-33t[+c^{\mathrm{\prime }}]$ or $[s=]18t^2-t^3-33t[+c^{\mathrm{\prime }}]$

Attempt to evaluate their $[18t^2-t^3-33t]$ for $t=0$ to $t=1$ or $t=1$ to $t=11$ or $t=11$ to $t=12$ $0\text{ to }1:-16-0=-161\text{ to }11:484-(-16)=50011\text{ to }12:468-484=-16$

For all three

Distance $=[16+500+16]=532\text{ m}$

Resolving for both particles or for the system to form equation(s)

Either $T-F-2.4g\sin 30=0$ AND $3.3g-T=0$ Or $3.3g-F-2.4g\sin 30=0$

$R=2.4g\cos 30[=12\sqrt{3}=20.7846\dots ]$

Use of $F=\mu R$ to get an equation in $\mu$ only $[3.3g-2.4g\mu \cos 30-2.4g\sin 30=0]$

$\mu =1.01$ [sight of 1.01036… or 1.0104]

Using Newton’s second law for both particles or the system

Either $3.3g-T=3.3a$ and $T-F-2.4g\sin 20=2.4a[T-22.778-8.208=2.4a]$ or $[T-30.986=2.4a]$ or $3.3g-F-2.4g\sin 20=(2.4+3.3)a[2.013367\cdots =5.7a]$

$F=1.01\times 2.4g\cos 20[=22.778]$

Attempt to solve for $aa=0.353$ [0.353222…]

$v^2=2\times 0.353\times 1[=0.706444\dots ]$ or $[v=0.841]$ Or $1=0+\frac{1}{2}\times 0.353t^2⟹t=2.3795⟹v=0.353\times 2.38$

Using Newton’s second law on A after B reaches the ground $-F-2.4g\sin 20=2.4a[-1.01\times 2.4g\cos 20-2.4g\sin 20=2.4a][-22.78814\cdots -8.20848\cdots =2.4a]$

Use of suvat to find $s$ $[0=\text{their }{0.841}^2+2\times \text{their }-12.911\cdots \times s⟹s=0.027358\dots ]$

Total distance = 1.03 m