A Level Mathematics 9709 Paper 43, May/June 2026

Setting up an equation in $u$ only e.g. $u-g\times 5=0$ or $u\times 4+\frac{1}{2}\times (-g)\times 4^2=u\times 6+\frac{1}{2}\times (-g)\times 6^2(\Rightarrow 4u-80=6u-180)$

$u=50$

$0^2=(\text{their }50{)}^2+2\times (-g)\times s$

Greatest height = 125 m

$v=2t^2-6t-3\Rightarrow a=4t-6$

1.5 s

For attempt at integration

$s=\frac{2}{3}{t}^3-3t^2-3t[+c]$

$\frac{2}{3}\times 3^3-3\times 3^2-3\times 3[-0][=-18]$ or $[0]-(\frac{2}{3}\times 3^3-3\times 3^2-3\times 3)[=18]$

Distance = 18 m

$R=mg\cos 30[=5m\sqrt{3}]$

$20-F-mg\sin 30=m\times 2$

$20-0.4\times mg\cos 30-mg\sin 30=m\times 2[\Rightarrow 20-2\sqrt{3}m-5m=2m]$

$m=1.91$

$170=32t+\frac{1}{2}a{t}^2[\Rightarrow 340=64t+a{t}^2]$

$280=32\times 2t-\frac{1}{2}a(2t{)}^2[\Rightarrow 280=64t-2a{t}^2]$ or $280=(32-2at)\times 2t+\frac{1}{2}a(2t{)}^2$ or $32=\frac{280}{t}-32+a(2t)$ or $280=(\frac{280}{t}-32)\times 2t+\frac{1}{2}a(2t{)}^2$ or $32=\sqrt{1024-560a}+2at[\Rightarrow 4a{t}^2-128t+560=0]$ or $32-2at=\frac{140}{t}-a\times t$

Forming an equation in $a$ or $t$ only

$t=5$

$a=0.8$

$80000=25F[\Rightarrow F=3200]$

Use of N2L on lorry or trailer or system

$F-2400-T=13500a$ $T-400=2500a$ $F-2400-400=16000a$

Attempt to solve for $T$

$[a=0.025\Rightarrow ]\text{Tension}=462.5\text{ N}$

Change in KE = $\pm (\frac{1}{2}\times (13500+2500)\times {30}^2-\frac{1}{2}\times (13500+2500)\times {27}^2)[\Rightarrow \pm (7200000-5832000)=\pm 1368000]$

Change in PE = $\pm (13500+2500)g\times x\sin 5$

Work done against resistance = $\pm (2500+500)x[\Rightarrow \pm 3000x]$

$55000+16000g\times x\sin 5-3000x=\frac{1}{2}\times 16000\times {30}^2-\frac{1}{2}\times 16000\times {27}^2[\Rightarrow 55000+(\mathrm{13944.91...})x-3000x=7200000-5832000]$

Distance $AB=120\text{ m}$

Attempt to resolve in any direction to form an expression for either the vertical or horizontal component

Vertical component $=\pm (40\sin 30+50\sin 40-65\sin 60)[=\mp \mathrm{4.15227...}]$

Horizontal component $=\pm (-40\cos 30+50\cos 40+65\cos 60)[=\pm \mathrm{36.16120...}]$

For attempt to find resultant or direction

Resultant = 36.4 N e.g. 6.6° below negative x-axis or bearing of 263° or 173° clockwise from the positive x-axis or 6.6° anticlockwise from the negative x-axis or 6.6° below the west

$F\sin 30+50\sin 40=65\sin 60$

$\pm (50\cos 40+65\cos 60-(\text{their }F)\times \cos 30)$

29 N in the negative x-direction

For P: $R=2g\times 0.96[\Rightarrow 19.2]$

$F=0.2\times (0.96\times 2g)[\Rightarrow 0.2\times 19.2=3.84]$

For P: $2g\times 0.28-F=2a[\Rightarrow 2g\times 0.28-3.84=2a]$

$a=0.88{\text{ ms}}^{-2}$

For Q: $-5g\times 0.28-0.2\times (0.96\times 5g)=5a$ or $5g\times 0.28+0.2\times (0.96\times 5g)=5a[\Rightarrow -5g\times 0.28-9.6=5a\Rightarrow a=-4.72]$

$s_P=8\times 0.25+\frac{1}{2}\times 0.88\times {0.25}^2[=2.0275]$ or $s_Q=u\times 0.25+\frac{1}{2}\times (-4.72)\times {0.25}^2[=0.25u-0.1475]$

$2.0275+0.25u-0.1475=3$

$u=4.48$

Velocity of P before impact: $=\pm (8+0.88\times 0.25)[\Rightarrow \pm 8.22]$

$5\times 3.3-2\times 8.22=-5\times 0.5+2\times v[\Rightarrow v=1.28]$

$-2g\times 0.28-0.2\times (0.96\times 2g)=2a[-2g\times 0.28-3.84=2a\Rightarrow a=-4.72]$ or $2g\times 0.28+0.2\times (0.96\times 2g)=2a[2g\times 0.28+3.84=2a\Rightarrow a=4.72]$

$0={1.28}^2+2\times (-4.72)\times s$ or $0=1.28+(-4.72)\times t\Rightarrow t=\frac{16}{59}(=\mathrm{0.27118...})$ followed by $s=1.28\times \mathrm{0.2711...}+\frac{1}{2}\times (-4.72)\times (\mathrm{0.2711...}{)}^2$

Distance = 0.174 m