A Level Mathematics 9709 Paper 42, May/June 2024

Initial $\text{KE}=\frac{1}{2}\times 72\times 8^2[=2304]$ OR Final $\text{KE}=\frac{1}{2}\times 72\times {16}^2[=9216]$ OR Work done against resistance $=28\times 100[=2800]$

Attempt at work-energy equation $[\frac{1}{2}\times 72\times 8^2+\text{WD}=\frac{1}{2}\times 72\times {16}^2+28\times 100]$

$\text{WD}=9712\text{ J}$

Attempt to differentiate given $v$

$[44-12t>0\Rightarrow ]t<\frac{11}{3}$

Attempt to integrate given $v$

$(s=)\frac{44}{1+1}{t}^{1+1}-\frac{6}{2+1}{t}^{2+1}-36t(+c)=22t^2-2t^3-36t(+c)$

$[22t^2-2t^3-36t=0\Rightarrow ]t=2,9(\text{and }0)\text{ ONLY}$

Resolving either direction to get an equation

$10\cos 25=2\cos 40+16\sin \theta$ $\left[\begin{array}{rl}{\displaystyle 9.06307787}& {\displaystyle =1.532088886+16\sin \theta }\\ {\displaystyle 7.530988984}& {\displaystyle =16\sin \theta }\\ {\displaystyle \sin \theta }& {\displaystyle =0.4706868115}\end{array}\right]$

$P=10\sin 25+16\cos \theta +2\sin 40$ $\left[\begin{array}{rl}{\displaystyle P}& {\displaystyle =4.226182617+16\cos \theta +1.285575219}\\ {\displaystyle P}& {\displaystyle =5.511757837+16\cos \theta }\end{array}\right]$

Attempt to solve for $\theta ={\sin }^{-1}\left(\frac{10\cos 25-2\cos 40}{16}\right)$

Attempt to solve for $P=10\sin 25+16\cos (\text{their }\theta )+2\sin 40$

$\theta =28.1\text{ AND }P=19.6$

$[k\times {24}^2=480\Rightarrow ]k=\frac{5}{6}$

Attempt at Newton's second law $[\text{DF}=480+1400g\times 0.12\Rightarrow \text{DF}=2160]$

$\text{Power}=(\text{their }2160)\times 24$

$51840\text{ W}$

$\text{DF}=\frac{54000}{v}\text{ and DF}=\left(\text{their }\frac{5}{6}\right)v^2$

Get an equation of the form $a{v}^3=b$ and attempt to solve for $v$ to get a positive value

$\text{Speed}=40.2{\text{ ms}}^{-1}$

Attempt at resolving perpendicular to plane to get an equation

$R+T\sin 35=0.8g\cos 28[R+(0.57357\dots )T=7.06358]$

Attempt to resolve parallel to plane for one of the possible cases to get an equation

$T\cos 35=0.8g\sin 28+F[(0.81915\dots )T=3.75577\cdots +F]$

$T\cos 35=0.8g\sin 28-F[(0.81915\dots )T=3.75577\cdots -F]$

Use of $F=0.2R$ to get an equation in $T$ only

Solve to get $T=5.53$

Solve to get $T=3.33$

Attempt at conservation of momentum for the 1st collision $[5\times 6=(5+1)v_D]$

Attempt at conservation of momentum for the 2nd collision $[(5+1)(\text{their }{v}_D)-2v=(5+1+2)v_E]$

$(v_E=)\frac{15-v}{4}$

${\text{KE}}_{\text{initial}}=\frac{1}{2}\times 5\times 6^2+\frac{1}{2}\times 2\times v^2[=90+v^2]$ ${\text{KE}}_{\text{final}}=\frac{1}{2}(5+1+2){\left(\frac{15-v}{4}\right)}^2$

Attempt difference in KE is 63 to get an equation $[\frac{1}{2}\times 5\times 6^2+\frac{1}{2}\times 2\times v^2-\frac{1}{2}(5+1+2){\left(\frac{15-v}{4}\right)}^2=63]$

Solve algebraically $3v^2+30v-117=0\text{ OE to get }v=3\text{ ONLY}$

Time $A$ to $B=6\text{ s}$

$\text{Distance }BC=98-3\times (\text{their }6)[=80]$

Use sum of distance moved by $D$ and distance moved by $C$ is $80\text{ m}$ $[(\text{their }5)t+3t=\text{their }80]$ OR use distance moved by C divided by relative velocity $\left[\frac{80}{(\text{their }5)+3}\right]$

$\text{Time}=10\text{ s}$

$[6+10+\frac{3\times 10+3\times 6}{3}=]32\text{ s}$

$R=2.5g\cos 30[=\frac{25\sqrt{3}}{2}=21.65063509]$

Attempt at Newton's second law

EITHER: $T-0.5g=0.5a\text{ AND }2.5g\sin 30-F-T=2.5a$ OR: $2.5g\sin 30-F-0.5g=(2.5+0.5)a$

Use of $F=0.2R$ to get an equation in $a$ only $[2.5g\sin 30-0.2\times 2.5g\cos 30-0.5g=(2.5+0.5)a]$

$a=1.06{\text{ ms}}^{-2}$

$[x\sin 30=2-x\Rightarrow ]x=\frac{4}{3}$ OR $[(2-y)\sin 30=y\Rightarrow ]y=\frac{2}{3}$ OR $[\frac{1}{2}\times 1.056\cdots \times t^2+\frac{1}{2}\times 1.056\cdots \times t^2\times \sin 30=2\Rightarrow t=1.5886\dots ]\Rightarrow x[=\frac{1}{2}\times 1.056\cdots \times (1.5886\dots {)}^2\sin 30]=\frac{2}{3}$ OR $\Rightarrow y[=\frac{1}{2}\times 1.056\cdots \times (1.5886\dots {)}^2]=\frac{4}{3}$

$\text{Change in PE}=\pm (2.5g(\text{their }x)\sin 30-0.5g(\text{their }x))[=\pm \frac{3}{4}g(\text{their }x)]$ OR $=\pm (2.5g(2-\text{their }y)\sin 30-0.5g(2-\text{their }y))[=\pm (3(\text{their }y)-1)g]$

$\text{WD against friction}=0.2\times 2.5g\cos 30\times (\text{their }x)[=4.33(\text{their }x)]$ OR $\text{WD against friction}=0.2\times 2.5g\cos 30\times (2-(\text{their }y))$

$\frac{1}{2}\times 2.5\times v^2+\frac{1}{2}\times 0.5\times v^2=2.5g\left(\frac{4}{3}\right)\sin 30-0.5g\left(\frac{4}{3}\right)-0.2\times 2.5g\cos 30\times \left(\frac{4}{3}\right)$ OR $\frac{1}{2}\times 2.5\times v^2+\frac{1}{2}\times 0.5\times v^2=2.5g(2-\frac{2}{3})\sin 30-0.5g(2-\frac{2}{3})-0.2\times 2.5g\cos 30\times (2-\frac{2}{3})$

$v=1.68$