A Level Mathematics 9709 Paper 42, Oct/Nov 2023

$\pm \frac{1}{2}\times 15\times 2^2[=\pm 30]\pm \frac{1}{2}\times 15\times 4^2[=\pm 120]$

$\pm 15g\times 1.6[=\pm 240]$

$240+30=120+W$

Work done = 150J

Attempt to resolve in at least one direction to form an equation.

$T\sin 30+T\sin 40-2=0$

$T\cos 30-T\cos 40-mg=0$

Attempt to solve for $T$ or $m$

Tension $T=1.75$, $m=0.0175$

Work done by $120\text{ N}$ force $=120\times 5\cos 20[=563.81557\dots ]$

$(\text{PE change}=)10g\times 5\sin 30[=250]$

Attempt at work energy equation

$120\times 5\cos 20-10g\times 5\sin 30-200=\frac{1}{2}\times 10\times v^2$ $[563.815\cdots -250-200=5v^2]$

Speed $=4.77{\text{ ms}}^{-1}$

$R=0.2g$

$1.2=\mu \times 0.2g$

$\mu =0.6$

$1.2-0.3\times 0.2g=0.2a$

$a=3$

$s_3=0+\frac{1}{2}\times 3\times 3^2[=13.5]s_2=0+\frac{1}{2}\times 3\times 2^2[=6]$

Distance $=13.5-6=7.5\text{ m}$

$v^2={25}^2+2(-g)\times 20$ OR $\frac{1}{2}\times 0.5\times v^2=\frac{1}{2}\times 0.5\times {25}^2-0.5\times g\times 20$

Speed $=15{\text{ ms}}^{-1}$

Taking up as positive direction: $0.5\times 15+0.3\times (-32.5)=0.5v+0$ or Taking down as positive direction: $0.5\times (-15)+0.3\times 32.5=0.5v+0$

[Taking up as positive direction: velocity of A $=-4.5{\text{ ms}}^{-1}$] [Taking down as positive direction: velocity of A $=4.5{\text{ ms}}^{-1}$] Speed $=4.5{\text{ m s}}^{-1}$ direction downwards

Downwards to be positive, for A $20=4.5t_A+\frac{1}{2}g{t}_A^2$ and solve for $t_A$ Upwards to be positive, for A $-20=-4.5t_A-\frac{1}{2}g{t}_A^2$ and solve for $t_A$

For B $20=0+\frac{1}{2}g{t}_B^2[t=2]$ and solve for $t_B$

$t_A=1.6$ or $t_B=2$

Difference $=0.4\text{ s}$ only

Attempt at Newton's second law at least once; correct number of relevant terms; allow sign errors; allow sin/cos mix; allow $g$ missing; a value for $\alpha$ or $\sin \alpha$ must be substituted.

Any equations correct.

Two equations correct.

Solve for $T$ or $a$

Acceleration $=0.3{\text{ ms}}^{-2}$ and Tension $=43000\text{ N}$

Driving force, $DF=\frac{4500000}{30}[=150000]$

Attempt to resolve parallel to the track once if using system equation, twice if using equations for engine and coach separately

System: $150000-120000g\sin \beta -60000g\sin \beta -22000-13000=0$ or for Engine: $150000-120000g\sin \beta -22000-T^{\mathrm{\prime }}=0$ and Coach: $T^{\mathrm{\prime }}-60000g\sin \beta -13000=0$

Solve to get $\beta ={3.7}^{\circ }$

Attempt to differentiate $v$

$(a=\frac{\mathrm{d}v}{\mathrm{d}t}=)3\times -0.1t^{3-1}+2\times 1.8t^{2-1}-6t^{1-1}=-0.3t^2+3.6t-6$

Setting $a=\frac{\mathrm{d}v}{\mathrm{d}t}=0$ and attempt to solve a 3 term quadratic for $t$. $[a=\frac{\mathrm{d}v}{\mathrm{d}t}=0\Rightarrow 3t^2-36t+60=0\Rightarrow t^2-12t+20=0]$

$p=2,q=10$

Velocities are $0{\text{ ms}}^{-1}$ and $25.6{\text{ ms}}^{-1}$

Curve with single minimum turning point followed by single maximum turning point

All correct in 1st and 4th quadrant.

Attempt to integrate $v$

$(s=)-\frac{0.1}{4}{t}^{3+1}+\frac{1.8}{3}{t}^{2+1}-\frac{6}{2}{t}^{1+1}+5.6t(+c)=-0.025t^4+0.6t^3-3t^2+5.6t[+c]$

Attempt distance from $t=0$ to $t=14$ $[=176.4]$

Attempt distance from $t=14$ to $t=15$ $[=(-)8.025]$

Total distance $=176.4+8.025=184.425\text{ m}=\frac{7377}{40}$