# Mark Scheme

## Question 1
**M1** Use $F = ma$ to get $3t + 1 = 2a$, so $a = \frac{3t + 1}{2}$

**M1** Use $v = u + \int a \, dt$ or equivalent method

**M1** $v = 1 + \int_0^t \frac{3t + 1}{2} dt = 1 + \left[\frac{3t^2}{4} + \frac{t}{2}\right]_0^t$

**A1** $v = 1 + \frac{3t^2}{4} + \frac{t}{2}$

**M1** Substitute $v = 5$ when $t = T$: $5 = 1 + \frac{3T^2}{4} + \frac{T}{2}$

**A1** $T = 2$ (or $T = -\frac{8}{3}$ rejected as $t \geq 0$)

## Question 2
**B1** Power has dimensions $[ML^2T^{-3}]$

**M1** Check $mgv\sin y$: $[MLT^{-2}][LT^{-1}] = [ML^2T^{-3}]$ ✓

**M1** Check $Rv$: $[MLT^{-2}][LT^{-1}] = [ML^2T^{-3}]$ ✓

**M1** Check $\frac{1}{2}mv^3\frac{\sin y}{h}$: $[M][L^3T^{-3}]\frac{1}{[L]} = [ML^2T^{-3}]$ ✓

**A1** All terms have correct dimensions

**A1** The formula is dimensionally consistent

## Question 3(a)
**M1** Horizontal component: $x = ut\cos\theta$, so $t = \frac{x}{u\cos\theta}$

**M1** Vertical component: $y = ut\sin\theta - \frac{1}{2}gt^2$

**M1** Substitute expression for $t$

**M1** $y = u \cdot \frac{x}{u\cos\theta} \cdot \sin\theta - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$

**A1** $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

**A1** $y = x\tan\theta - \frac{gx^2(1 + \tan^2\theta)}{2u^2}$ or equivalent

## Question 3(b)(i)
**M1** Basketball 0.5 m below and 5 m horizontally from basket, so substitute $x = 5$, $y = 0.5$

**M1** $0.5 = 5\tan\theta - \frac{g \cdot 25(1 + \tan^2\theta)}{2 \cdot 64}$

**M1** With $g = 9.8$: $0.5 = 5\tan\theta - 1.9141(1 + \tan^2\theta)$

**M1** Rearrange to: $1.9141\tan^2\theta - 5\tan\theta + 2.4141 = 0$

**A1** Solutions $\theta \approx 63.1°$ and $\theta \approx 32.6°$ (to 3 s.f.)

## Question 3(b)(ii)
**M1** At $\theta = 63.1°$, find $t$ when $x = 5$: $t = \frac{5}{8\cos(63.1°)} \approx 1.309$ s

**M1** Vertical velocity: $v_y = u\sin\theta - gt = 8\sin(63.1°) - 9.8(1.309)$

**M1** $v_y \approx -4.37$ m/s (downward)

**M1** Horizontal velocity: $v_x = 8\cos(63.1°) \approx 3.63$ m/s

**A1** Direction: $\tan^{-1}\left(\frac{4.37}{3.63}\right) \approx 50°$ below horizontal (or bearing) to nearest degree

## Question 3(c)
**B1** Air resistance is negligible / The basketball is a point mass / Rotation of basketball is negligible

## Question 4(a)
**M1** Conservation of momentum: $4um + 3m(2u) = mv_A + 3mv_B$

**A1** $10u = v_A + 3v_B$ ... (1)

**M1** Coefficient of restitution: $e = \frac{v_B - v_A}{4u - 2u}$

**A1** $2eu = v_B - v_A$ ... (2)

**M1** Solve (1) and (2) simultaneously

**A1** $v_A = u(5 - 3e)$ and $v_B = u(5 + e)/3$

## Question 4(b)
**M1** $v_B = \frac{u(5 + e)}{3} \leq 3u$ requires $5 + e \leq 9$

**A1** Since $e \leq 1$, we have $v_B \leq 2u < 3u$ or $v_B$ cannot exceed $3u$

## Question 4(c)
**M1** With $e = \frac{2}{3}$: $v_B = \frac{u(5 + 2/3)}{3} = \frac{17u}{9}$

**M1** Impulse = change in momentum of B: $J = 3m \cdot v_B - 3m(2u)$

**A1** $J = 3m\left(\frac{17u}{9} - 2u\right) = \frac{mu}{3}$ or equivalent

## Question 5(a)
**M1** Resolve perpendicular to plane: $g\cos\theta = $ component of $g$ perpendicular to plane

**M1** Perpendicular displacement = 0: $0 = u\sin\alpha \cdot T - \frac{1}{2}g\cos\theta \cdot T^2$

**M1** $u\sin\alpha = \frac{1}{2}g\cos\theta \cdot T$

**A1** $u = \frac{g T \cos\theta}{2\sin\alpha}$

## Question 5(b)
**M1** Distance along plane: $s = u\cos\alpha \cdot T + \frac{1}{2}g\sin\theta \cdot T^2$ where $T$ is time to return to plane

**M1** From part (a): $T = \frac{2u\sin\alpha}{g\cos\theta}$

**M1** Substitute and simplify: $s = u\cos\alpha \cdot \frac{2u\sin\alpha}{g\cos\theta} + \frac{1}{2}g\sin\theta \cdot \left(\frac{2u\sin\alpha}{g\cos\theta}\right)^2$

**M1** $s = \frac{2u^2\sin\alpha\cos\alpha}{g\cos\theta} + \frac{2u^2\sin\alpha^2\sin\theta}{g\cos^2\theta}$

**M1** $s = \frac{2u^2\sin\alpha}{g\cos^2\theta}(\cos\alpha\cos\theta + \sin\alpha\sin\theta)$

**A1** $OP = \frac{2u^2\sin\