A Level Mathematics 9709 Paper 41, May/June 2024

Trapezium on the $t$-axis starting at $(0, 0)

Fully correct with correct labels

$\text{Area}=\frac{1}{2}(30+60)\times 20$ or $\text{Area}=\frac{1}{2}\times 10\times 20+30\times 20+\frac{1}{2}\times 20\times 20$

$=900\text{ m}$

$F=20\sin 60$

$=17.3\text{ N}$

For resolving in any direction

$(\text{Horizontal component}=X=R\cos \theta )=\pm (20\cos 60)[\pm 10]$ $(\text{Vertical component}=Y=R\sin \theta )=\pm (20\sin 60-10)[=\pm 7.3205\dots ]$

$\text{Magnitude}=\sqrt{(20\sin 60-10{)}^2+(20\cos 60{)}^2}[=12.393136\dots ]$

$\text{Angle}={\tan }^{-1}\left(\frac{20\sin 60-10}{20\cos 60}\right)[=36.206023\dots ]$

$\text{Magnitude}=12.4\text{ N}$ and $\text{Direction}={36.2}^{\circ }\text{ above (positive) }x\text{-axis}$

$\text{PE gained}=180000g\times 1500\sin 1.5[=70677760.4\dots ]$

${\text{KE}}_{\text{Initial}}=\frac{1}{2}\times 180000\times {45}^2[=182250000]$ ${\text{KE}}_{\text{Final}}=\frac{1}{2}\times 180000\times {40}^2[=144000000]$

For work energy equation: $\text{WD}=180000g\times 1500\sin 1.5+12000000-(\frac{1}{2}\times 180000\times {45}^2-\frac{1}{2}\times 180000\times {40}^2)$

$=44400\text{ kJ }[44427.7604\dots ]$

$\text{Driving force}=\frac{14000}{20}[=700]$

Attempt at Newton’s second law on car or trailer or system

$700-400-T=1700a$ $T-150=300a$ $700-400-150=(1700+300)a$

Solving for $a$ or $T$

$\text{Acceleration}=\frac{3}{40}=0.075{\text{ ms}}^{-2}$ $\text{Tension}=172.5\text{ N}$

$a=2.08[2.07911\dots ]$

$v^2=5^2+2a\times 60$

$\text{Speed}=16.6{\text{ ms}}^{-1}[16.567861\dots ]$

$R=mg\cos 12$

$mg\sin 12-F=ma$

For use of $F=0.03R$ to get equation in $a$ (and $m$) only

$60=5t+\frac{1}{2}a{t}^2$ and solve for $t$

$\text{Time}=5.86\text{ s }[5.86260\dots ]$

For an attempt at differentiation

$a=\frac{1}{2}k{t}^{\frac{1}{2}-1}-2t^{1-1}+0=\frac{1}{2}k{t}^{-\frac{1}{2}}-2$

$a=0\Rightarrow \frac{1}{2}k{t}^{-\frac{1}{2}}-2=0[t=\frac{k^2}{16}]$ or $k=4t^{\frac{1}{2}}$

$(v=)k{\left(\frac{k^2}{16}\right)}^{\frac{1}{2}}-2\frac{k^2}{16}-8=4.5$

$\Rightarrow \frac{k^2}{8}=12.5\Rightarrow k=10$

$10\times 1^{\frac{1}{2}}-2\times 1-8=0$ and $10\times {16}^{\frac{1}{2}}-2\times 16-8=0$ Or equate $v=0$ and solve for $t$ to get $t=1$ and $t=16$.

For integration (do not penalise missing $c$)

$(s=)\frac{10}{\frac{3}{2}}{t}^{\frac{3}{2}}-\frac{2t^2}{2}-8t[+c]=\frac{20}{3}{t}^{\frac{3}{2}}-t^2-8t[+c]$ or $(s=)\frac{k}{\frac{3}{2}}{t}^{\frac{3}{2}}-\frac{2t^2}{2}-8t[+c]=\frac{2k}{3}{t}^{\frac{3}{2}}-t^2-8t[+c]$

$(\frac{20}{3}{1}^{\frac{3}{2}}-1^2-8\times 1)-0[=-\frac{7}{3}]$ and $(\frac{20}{3}{16}^{\frac{3}{2}}-{16}^2-8\times 16)-(\frac{20}{3}{1}^{\frac{3}{2}}-1^2-8\times 1)=[\frac{128}{3}--\frac{7}{3}=45]$

For both set of limits applied correctly

$\frac{142}{3}\text{ m}$

$v^2={25}^2-2g\times 20$

$\Rightarrow \text{speed}=15{\text{ ms}}^{-1}$

$0.2\times 15-0.1\times 20=0+0.1v$ or $0.2\times 15-0.1\times 20=0-0.1V$

$v=10{\text{ ms}}^{-1}\text{ upwards}$

Speed of $P$ at impact is $20{\text{ ms}}^{-1}$

Time to when $P$ reaches ground = $2\text{ s}$

$s_P=10t+\frac{1}{2}\times -g\times t^2$

$s_Q=10t+\frac{1}{2}\times g\times t^2$

$s_P+s_Q=20\Rightarrow t=1$

$\text{Height}=5\text{ m}$