# Question 5

## 5(a)

Using horizontal motion, $s = ut$, with 35 resolved

$$x = 35\cos\theta \cdot t$$ | M1 | 3.3

| A1 | 1.1b

Using vertical motion, $s = ut + \frac{1}{2}at^2$, with 35 resolved

$$y = 35\sin\theta \cdot t - \frac{1}{2}gt^2$$ | M1 | 3.4

| A1 | 1.1b

Eliminate $t$:

$$y = 35\sin\theta \cdot \frac{x}{35\cos\theta} - \frac{1}{2}g\left(\frac{x}{35\cos\theta}\right)^2$$ | DM1 | 3.1b

$$y = \frac{3}{4}x - \frac{1}{160}x^2$$ | A1* | 1.1b

**N.B.** No marks available if they just quote the equation of the path.

**ALTERNATIVE:** they do (b) and/or (c) first using a suvat method

Assume $y = ax^2 + bx + c$

Use any three of $(0,0)$, $(120,0)$ from part (b), $(60,22.5)$ from part (c)

or $\frac{dy}{dx} = \frac{3}{4}$ at $x = 0$ to find $a$, $b$, and $c$.

$\text{M1A1, M1A1, DM1A1}^*$ for finding each of $a$, $b$ and $c$ and stating final answer in correct form.

**N.B.** If they realise that $c = 0$, and just use $y = ax^2 + bx$, that could score M1A1.

Enter marks on ePEN in the order in which $a$, $b$ and $c$ are found:

$x = 0, y = 0 \Rightarrow c = 0$ | M1A1

$x = 0, \frac{dy}{dx} = 2ax + b = \frac{3}{4} \Rightarrow b = \frac{3}{4}$ | M1A1

$x = 120, y = 0 \Rightarrow 0 = 120^2a + 120 \cdot \frac{3}{4} \Rightarrow a = -\frac{1}{160}$ | DM1

$$y = \frac{3}{4}x - \frac{1}{160}x^2$$ | A1*

**(6 marks)**

## 5(b)

**ALT 1**

$$0 = \frac{3}{4}x - \frac{1}{160}x^2$$ and solve for $x$

**ALT 2**

$$\frac{dy}{dx} = \frac{3}{4} - \frac{x}{80} = 0 \Rightarrow x = 60$$ and $OA = 2 \times 60$

**ALT 3**

A complete suvat method to find $OA$:

e.g. $0 = 35\sin\theta \cdot t - \frac{1}{2}gt^2$

or $0 = 35\sin\theta - \frac{1}{2}gt$

or $35\sin\theta = 35\sin\theta - gt$

to find $t = \frac{70\sin\theta}{g}$

AND $OA = 35\cos\theta \cdot t = \frac{35\cos\theta \cdot 70\sin\theta}{g}$ | M1 | 3.1b

**N.B.** OR use the calculator to input the equation of the path which then gives $y_{\max} = \frac{45}{2}$ when $x = 60$ with no working, so $OA = 2 \times 60$

$OA = 120$ (m) | A1 | 1.1b

**(2 marks)**

## 5(c)

**ALT 1**

$$H = \frac{3}{4} \cdot 60 - \frac{1}{160} \cdot 60^2$$

**ALT 2**

$$y = \frac{1}{160}(x^2 - 120x) = 22.5 - \frac{1}{160}(x - 60)^2$$ so $\max y = \frac{45}{2}$ or $22.5$

**ALT 3**

$$\frac{dy}{dx} = \frac{3}{4} - \frac{2x}{160} = 0 \Rightarrow x = 60$$ then find $y$ when $x = 60$

**ALT 4**

A complete suvat method:

e.g. $H = \frac{(35\sin\theta)^2}{2g}$

or $0 = 35\sin\theta - gt$ to find the time to top, $t = \frac{35\sin\theta}{g}$, or use half their time they found in (b)

AND $H = 35\sin\theta \cdot t - \frac{1}{2}g\left(\frac{35\sin\theta}{g}\right)^2$ or $\frac{35\sin\theta + 0}{2} \cdot \frac{35\sin\theta}{g}$

**N.B.** OR use the calculator to input the equation of the path which then gives $y_{\max} = \frac{45}{2}$ (when $x = 60$) with no working. | M1 | 3.1b

$H = 22.5$ Accept 23 | A1 | 1.1b

**(2 marks)**

## 5(d)

$H$ is greater (or $K$ is smaller), as air resistance would slow the particle down or equivalent. | B1 | 3.5a

**(1 mark)**

## 5(e)

e.g. the inaccuracy of using $9.8 \text{ m s}^{-2}$ for $g$ | B1 | 3.5b

**(1 mark)**

**(12 marks total)**

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## Notes

**5a**

| M1 | Correct terms but condone $\sin/\cos$ confusion and sign errors. Available if they use $s$ instead of $x$

| A1 | Correct equation in $x$ and $t$. N.B. they may have the wrong value for $\cos\theta$

| M1 | Correct terms but condone $\sin/\cos$ confusion and sign errors. Available if they use a different letter for $y$ provided it's not the same as they've used for $x$. N.B. M0 if they subsequently use a value for $y$ e.g. 0

| A1 | Correct equation in $y$ and $t$. N.B. they may have the wrong value for $\sin\theta$. They may have $t$ in terms of $x$, from their first equation.

| DM1 | Dependent on the two previous M marks for eliminating $t$ to give an equation in $y$ and $x$ only.

| A1* | Given answer correctly obtained, with at least one further line of working with trig ratios and $g = 9.8$ explicitly seen or $4.9$ or equivalent used for $0.5g$. Allow $y = \frac{3x}{4} - \frac{x^2}{160}$ and with $y$ on the RHS

**5b**

| M1 | ALT 1: Use of $y = 0$ in equation of path and solve for $x$. ALT 2: Use calculus to find the $x$-coordinate of the max point and double it. ALT 3: Any other complete suvat method to