| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Vector form projectile motion |
| Difficulty | Standard +0.3 This is a standard M2 projectile motion question using vector notation. Part (a) is routine derivation of trajectory equation from given initial velocity, part (b) requires finding velocity components at a specific x-value, and part (c) involves solving simultaneous equations with parametric forms. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02d Constant acceleration: SUVAT formulae3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| Horizontal distance: \(x = 7t\) | B1 |
| Vertical distance: \(y = 7\sqrt{3}t - \frac{1}{2}gt^2\) | M1A1 |
| Sub for \(t\): \(y = 7\sqrt{3} \times \frac{x}{7} - \frac{g}{2} \times \frac{x^2}{49} = \sqrt{3}x - \frac{g}{98}x^2\) | DM1A1 (5) *Given Answer* |
| Notes: B1 for \(x = 7t\) seen or implied; M1 for vertical motion equation \(y = 7\sqrt{3}t - \frac{1}{2}gt^2\) need correct no. of terms, but condone sign errors; First A1 for a correct equation; Second DM1, dependent on first M1, for substituting for \(t\); Second A1 for given answer. |
| Answer | Marks |
|---|---|
| Differentiate to find gradient: \(\frac{dy}{dx} = \sqrt{3} - \frac{2gx}{98}\) | M1A1 |
| Sub \(x = 20\) & use tan: \(\tan^{-1}\left(\sqrt{3} - \frac{40g}{98}\right)\) | DM1 |
| \(= 66.2°\) or 66⁰ below the horizontal oe | A1 (4) |
| Or in the direction of ( parallel to is A0) \((7i - 16j)\) or \((7i - 15.9j)\) | |
| Notes: First M1 for attempt to differentiate path equation (both powers going down); First A1 for a correct expression; Second DM1, dependent on first M1, for putting \(x = 20\) and using \(\tan^{-1}\) to obtain an angle; Second A1 for 66⁰ or 66.2⁰ below the horizontal or 24⁰ or 23.8⁰ to the downward vertical; OR the angle marked on a clear diagram with an arrow. |
| Answer | Marks |
|---|---|
| First M1 for using \(x = 20\) in horizontal motion equn to obtain \(t = \frac{20}{7}\) and using \(v = u + at\) to obtain vertical speed or any other complete method e.g. put \(x = 20\) in equation of path to obtain \(y\), which is then used in \(v^2 = u^2 + 2as\) vertically to obtain vertical speed. | |
| First A1 for a correct expression for vertical speed | |
| Second DM1, dependent on first M1, for using ratio of components and \(\tan^{-1}\) to obtain an angle | |
| OR finding the velocity vector AND referring to its direction ('parallel to' can score M1 but not the A1). | |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or in the direction of ( parallel to is M1A0) \((7i - 16j)\) or \((7i - 15.9j)\) or a multiple. |
| Answer | Marks |
|---|---|
| Use the x/y ratio to form an equation in \(T\): | M1A1 |
| \(7T = 14\sqrt{3}T - gT^2\) | |
| Solve for \(T\): \(T = \frac{14\sqrt{3} - 7}{g}(= 1.8)\) (1.76) | DM1A1 (4) |
| Notes: First M1 for using \(\frac{2\lambda}{\lambda} = \frac{7T}{7\sqrt{3}T - gT^2}\) to obtain an equation in \(T\) only; Allow M1 if they have the reciprocal of RHS; First A1 for a correct equation; Second DM1 for solving the equation for \(T\) (N.B. if incorrect answer, need to see at least one line of working for this DM mark); Second A1 for \(T = 1.76\) or 1.8 |
| Answer | Marks |
|---|---|
| First M1 for using the point \((2\lambda, \lambda)\) and the path equation to obtain an equation in \(\lambda\) only (or \(x\)) | |
| First A1 for a correct equation | |
| Second DM1 for solving for \(\lambda\) or \(x\) and using it to obtain a value for \(T\) | |
| Second A1 for \(T = 1.76\) or 1.8 | |
| N.B. Allow interchange of \(\lambda\) and \(2\lambda\) for the method marks. |
## 6a
| Horizontal distance: $x = 7t$ | B1 |
| Vertical distance: $y = 7\sqrt{3}t - \frac{1}{2}gt^2$ | M1A1 |
| Sub for $t$: $y = 7\sqrt{3} \times \frac{x}{7} - \frac{g}{2} \times \frac{x^2}{49} = \sqrt{3}x - \frac{g}{98}x^2$ | DM1A1 (5) *Given Answer* |
| **Notes:** B1 for $x = 7t$ seen or implied; M1 for vertical motion equation $y = 7\sqrt{3}t - \frac{1}{2}gt^2$ need correct no. of terms, but condone sign errors; First A1 for a correct equation; Second DM1, dependent on first M1, for substituting for $t$; Second A1 for given answer. |
## 6b
| Differentiate to find gradient: $\frac{dy}{dx} = \sqrt{3} - \frac{2gx}{98}$ | M1A1 |
| Sub $x = 20$ & use tan: $\tan^{-1}\left(\sqrt{3} - \frac{40g}{98}\right)$ | DM1 |
| $= 66.2°$ or 66⁰ below the horizontal oe | A1 (4) |
| Or in the direction of ( parallel to is A0) $(7i - 16j)$ or $(7i - 15.9j)$ | |
| **Notes:** First M1 for attempt to differentiate path equation (both powers going down); First A1 for a correct expression; Second DM1, dependent on first M1, for putting $x = 20$ and using $\tan^{-1}$ to obtain an angle; Second A1 for 66⁰ or 66.2⁰ below the horizontal or 24⁰ or 23.8⁰ to the downward vertical; OR the angle marked on a clear diagram with an arrow. |
## 6b alt
| First M1 for using $x = 20$ in horizontal motion equn to obtain $t = \frac{20}{7}$ and using $v = u + at$ to obtain vertical speed or any other complete method e.g. put $x = 20$ in equation of path to obtain $y$, which is then used in $v^2 = u^2 + 2as$ vertically to obtain vertical speed. | |
| First A1 for a correct expression for vertical speed | |
| Second DM1, dependent on first M1, for using ratio of components and $\tan^{-1}$ to obtain an angle | |
| OR finding the velocity vector AND referring to its direction ('parallel to' can score M1 but not the A1). | |
| Second A1 for 66⁰ or 66.2⁰ below the horizontal or in the direction of ( parallel to is M1A0) $(7i - 16j)$ or $(7i - 15.9j)$ or a multiple. | |
## 6c
| Use the x/y ratio to form an equation in $T$: | M1A1 |
| $7T = 14\sqrt{3}T - gT^2$ | |
| Solve for $T$: $T = \frac{14\sqrt{3} - 7}{g}(= 1.8)$ (1.76) | DM1A1 (4) |
| **Notes:** First M1 for using $\frac{2\lambda}{\lambda} = \frac{7T}{7\sqrt{3}T - gT^2}$ to obtain an equation in $T$ only; Allow M1 if they have the reciprocal of RHS; First A1 for a correct equation; Second DM1 for solving the equation for $T$ (N.B. if incorrect answer, need to see at least one line of working for this DM mark); Second A1 for $T = 1.76$ or 1.8 |
## 6c alt
| First M1 for using the point $(2\lambda, \lambda)$ and the path equation to obtain an equation in $\lambda$ only (or $x$) | |
| First A1 for a correct equation | |
| Second DM1 for solving for $\lambda$ or $x$ and using it to obtain a value for $T$ | |
| Second A1 for $T = 1.76$ or 1.8 | |
| N.B. Allow interchange of $\lambda$ and $2\lambda$ for the method marks. | |
---6. [In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in a vertical plane, $\mathbf { i }$ being horizontal and $\mathbf { j }$ being vertically upwards.]
At $t = 0$ a particle $P$ is projected from a fixed point $O$ with velocity ( $7 \mathbf { i } + 7 \sqrt { 3 } \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$. The particle moves freely under gravity. The position vector of a point on the path of $P$ is $( x \mathbf { i } + y \mathbf { j } ) \mathrm { m }$ relative to $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$y = \sqrt { 3 } x - \frac { g } { 98 } x ^ { 2 }$$
\item Find the direction of motion of $P$ when it passes through the point on the path where $x = 20$
At time $T$ seconds $P$ passes through the point with position vector $( 2 \lambda \mathbf { i } + \lambda \mathbf { j } ) \mathrm { m }$ where $\lambda$ is a positive constant.
\item Find the value of $T$.\\
DO NOT WIRITE IN THIS AREA
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2016 Q6 [13]}}