| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse on inclined plane |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 1 oblique impact question requiring resolution of velocities parallel and perpendicular to the plane, application of Newton's experimental law (coefficient of restitution), and energy calculations. While it involves multiple parts and algebraic manipulation, the techniques are routine for FM1 students and the 'show that' format provides clear targets, making it moderately above average difficulty. |
| Spec | 6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CLM along the plane | M1 | 3.1a – Correct no. of terms, dimensionally correct, mass × velocity, condone sin/cos confusion |
| \((m)u\sin\alpha = (m)v\cos\alpha\) | A1 | 1.1b – Correct equation |
| Impulse-momentum perp to the plane | M1 | 3.1a – Dimensionally correct. Must be subtracting, condone wrong order and sin/cos confusion |
| \(I = m(v\sin\alpha - (-u\cos\alpha))\) | A1 | 1.1b – Correct unsimplified equation |
| \(I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha\) * | A1* | 2.2a – Given answer correctly obtained. Must be EXACT factorisation |
| Total | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Impulse-momentum vertically | M1 | 3.1a |
| [Diagram components resolved] | M1 | 3.1a |
| \(I\cos\alpha = m(0--u)\) | A1 | 1.1b |
| [Second equation] | A1 | 1.1b |
| \(I = mu\sec\alpha\) * | A1* | 2.2a |
| Total | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CLM along the plane | M1 | 3.1a |
| \(u\sin\alpha\) unchanged | A1 | 1.1b |
| Finds expression for \(e\): \(\tan\alpha = \dfrac{eu\cos\alpha}{u\sin\alpha} \Rightarrow e = \tan^2\alpha\) and \(I = m(eu\cos\alpha - (-u\cos\alpha))\) | M1 | 3.1a |
| \(I = m(u\cos\alpha\tan^2\alpha - (-u\cos\alpha))\) | A1 | 1.1b |
| \(I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha\) * | A1* | 2.2a |
| Total | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| CLM along the plane | M1 | 3.1a |
| \((m)u\sin\alpha = (m)v\cos\alpha\) leading to \(v = u\tan\alpha\) | A1 | 1.1b |
| Impulse-momentum as vector equation then Pythagoras: \(I = m\begin{pmatrix}-v\\u\end{pmatrix}\) and \( | I | = m\sqrt{v^2+u^2}\) |
| \( | I | = m\sqrt{u^2\tan^2\alpha + u^2}\) |
| \(I = m\sqrt{u^2(1+\tan^2\alpha)} = m\sqrt{u^2\sec^2\alpha} = mu\sec\alpha\) * | A1* | 2.2a |
| Total | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| NEL: \(eu\cos\alpha = v\sin\alpha\) | M1 | 3.4 – Attempt at NEL |
| Squaring and adding expressions for \(v\sin\alpha\) and \(v\cos\alpha\) | M1 | 1.1b – Squaring and adding to obtain \(v^2\) |
| \(v^2 = u^2(\sin^2\alpha + e^2\cos^2\alpha)\) * | A1* | 1.1b – Given answer correctly obtained. Must be EXACT |
| Total | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| KE loss \(= \dfrac{1}{2}mu^2 - \dfrac{1}{2}mu^2(\sin^2\alpha + e^2\cos^2\alpha)\) | M1 | 2.1 – Expression for difference of KE in terms of \(m\), \(u\), \(\alpha\) and \(e\) |
| Use \(\sin^2\alpha + \cos^2\alpha = 1\): KE loss \(= \dfrac{1}{2}mu^2(1-e^2)\cos^2\alpha\) * | A1* | 1.1b – Given answer correctly obtained. Factorisation must be EXACT |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\tan^2\alpha = e\) to eliminate \(\alpha\) from expression in (c) | M1 | 3.1a – Complete method to eliminate \(\alpha\). Any trig identity used must be correct e.g. \(\sec^2\alpha = 1+e\) or \(\cos^2\alpha = \dfrac{1}{1+e}\) |
| KE Loss \(= \dfrac{1}{2}mu^2(1-e)\) or \(\dfrac{1}{2}mu^2\dfrac{1}{1+e}(1-e^2)\) | A1 | 1.1b – Correct answer |
| Total | (2) |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| CLM along the plane | M1 | 3.1a – Correct no. of terms, dimensionally correct, mass × velocity, condone sin/cos confusion |
| $(m)u\sin\alpha = (m)v\cos\alpha$ | A1 | 1.1b – Correct equation |
| Impulse-momentum perp to the plane | M1 | 3.1a – Dimensionally correct. Must be subtracting, condone wrong order and sin/cos confusion |
| $I = m(v\sin\alpha - (-u\cos\alpha))$ | A1 | 1.1b – Correct unsimplified equation |
| $I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha$ * | A1* | 2.2a – Given answer correctly obtained. Must be EXACT factorisation |
| **Total** | **(5)** | |
# Question 6(a) alt1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse-momentum vertically | M1 | 3.1a |
| [Diagram components resolved] | M1 | 3.1a |
| $I\cos\alpha = m(0--u)$ | A1 | 1.1b |
| [Second equation] | A1 | 1.1b |
| $I = mu\sec\alpha$ * | A1* | 2.2a |
| **Total** | **(5)** | |
# Question 6(a) alt2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| CLM along the plane | M1 | 3.1a |
| $u\sin\alpha$ unchanged | A1 | 1.1b |
| Finds expression for $e$: $\tan\alpha = \dfrac{eu\cos\alpha}{u\sin\alpha} \Rightarrow e = \tan^2\alpha$ and $I = m(eu\cos\alpha - (-u\cos\alpha))$ | M1 | 3.1a |
| $I = m(u\cos\alpha\tan^2\alpha - (-u\cos\alpha))$ | A1 | 1.1b |
| $I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha$ * | A1* | 2.2a |
| **Total** | **(5)** | |
# Question 6(a) alt3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| CLM along the plane | M1 | 3.1a |
| $(m)u\sin\alpha = (m)v\cos\alpha$ leading to $v = u\tan\alpha$ | A1 | 1.1b |
| Impulse-momentum as vector equation then Pythagoras: $I = m\begin{pmatrix}-v\\u\end{pmatrix}$ and $|I| = m\sqrt{v^2+u^2}$ | M1 | 3.1a |
| $|I| = m\sqrt{u^2\tan^2\alpha + u^2}$ | A1 | 1.1b |
| $I = m\sqrt{u^2(1+\tan^2\alpha)} = m\sqrt{u^2\sec^2\alpha} = mu\sec\alpha$ * | A1* | 2.2a |
| **Total** | **(5)** | |
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| NEL: $eu\cos\alpha = v\sin\alpha$ | M1 | 3.4 – Attempt at NEL |
| Squaring and adding expressions for $v\sin\alpha$ and $v\cos\alpha$ | M1 | 1.1b – Squaring and adding to obtain $v^2$ |
| $v^2 = u^2(\sin^2\alpha + e^2\cos^2\alpha)$ * | A1* | 1.1b – Given answer correctly obtained. Must be EXACT |
| **Total** | **(3)** | |
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| KE loss $= \dfrac{1}{2}mu^2 - \dfrac{1}{2}mu^2(\sin^2\alpha + e^2\cos^2\alpha)$ | M1 | 2.1 – Expression for difference of KE in terms of $m$, $u$, $\alpha$ and $e$ |
| Use $\sin^2\alpha + \cos^2\alpha = 1$: KE loss $= \dfrac{1}{2}mu^2(1-e^2)\cos^2\alpha$ * | A1* | 1.1b – Given answer correctly obtained. Factorisation must be EXACT |
| **Total** | **(2)** | |
# Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\tan^2\alpha = e$ to eliminate $\alpha$ from expression in (c) | M1 | 3.1a – Complete method to eliminate $\alpha$. Any trig identity used must be correct e.g. $\sec^2\alpha = 1+e$ or $\cos^2\alpha = \dfrac{1}{1+e}$ |
| KE Loss $= \dfrac{1}{2}mu^2(1-e)$ or $\dfrac{1}{2}mu^2\dfrac{1}{1+e}(1-e^2)$ | A1 | 1.1b – Correct answer |
| **Total** | **(2)** | |
---\begin{enumerate}
\item A particle $P$ of mass $m$ is falling vertically when it strikes a fixed smooth inclined plane. The plane is inclined to the horizontal at an angle $\alpha$, where $0 < \alpha \leqslant 45 ^ { \circ }$
\end{enumerate}
At the instant immediately before the impact, the speed of $P$ is $u$.\\
At the instant immediately after the impact, $P$ is moving horizontally with speed $v$.\\
(a) Show that the magnitude of the impulse exerted on the plane by $P$ is $m u \sec \alpha$
The coefficient of restitution between $P$ and the plane is $e$, where $e > 0$\\
(b) Show that $v ^ { 2 } = u ^ { 2 } \left( \sin ^ { 2 } \alpha + e ^ { 2 } \cos ^ { 2 } \alpha \right)$\\
(c) Show that the kinetic energy lost by $P$ in the impact is
$$\frac { 1 } { 2 } m u ^ { 2 } \left( 1 - e ^ { 2 } \right) \cos ^ { 2 } \alpha$$
(d) Hence find, in terms of $m$, $u$ and $e$ only, the kinetic energy lost by $P$ in the impact.
\hfill \mbox{\textit{Edexcel FM1 2023 Q6 [12]}}