Edexcel M3 2018 June — Question 7 14 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2018
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeParticle on outer surface of sphere
DifficultyStandard +0.8 This is a substantial M3 circular motion problem requiring energy conservation, Newton's second law in polar form, and projectile motion after leaving the sphere. Part (a) is standard 'show that' mechanics requiring careful application of radial equation and energy. Part (b) requires insight that the particle must leave before the equator (α < 90°). Part (c) combines the leaving condition with projectile motion to find a specific value. The multi-stage reasoning and integration of several mechanics topics makes this harder than average, though the techniques are all standard M3 material.
Spec3.02h Motion under gravity: vector form6.05e Radial/tangential acceleration

7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2273ca38-1e16-44ab-ae84-f4c576cbb8f9-24_575_821_214_566} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} A smooth solid sphere, with centre \(O\) and radius \(r\), is fixed with its lowest point on a horizontal plane. A particle is placed on the surface of the sphere at the highest point of the sphere. The particle is then projected horizontally with speed \(u\) and starts to move on the surface of the sphere. The particle leaves the surface of the sphere at the point \(A\) where \(O A\) makes an angle \(\alpha , \alpha > 0\), with the upward vertical, as shown in Figure 4.
  1. Show that \(\cos \alpha = \frac { 1 } { 3 g r } \left( u ^ { 2 } + 2 g r \right)\)
  2. Show that \(u < \sqrt { g r }\) After leaving the surface of the sphere, the particle strikes the plane with speed \(3 \sqrt { \frac { g r } { 2 } }\)
  3. Find the value of \(\cos \alpha\).

7(a)
AnswerMarks
M1A1A1Energy equation from top to reaching the plane. Difference of KE terms and a PE term needed. \(m\) in all terms or none. Deduct 1 each error.
[7]
7(b)
AnswerMarks
M1A1csoUse their expression for \(u^2\) in the result given in (a) to obtain a numerical value for \(\cos\theta\). Complete to \(\frac{5}{6}\). Accept \(0.83\) or better.
[2]
7(c)
AnswerMarks
M1A1For circular motion, \(R \geq 0\)
A1\(\cos = \frac{5}{6}\)
dM1A1caoUse energy from A to the plane
[5]
7(c) ALT 1
AnswerMarks
M1Energy equation with correct number of terms and dimensionally correct
A1A1Deduct 1 each error
M1A1caoUse their equation of motion from (a) to find an expression for \(v^2\) and complete to a numerical value for \(\cos\theta\). Complete to \(\frac{5}{6}\). Accept \(0.83\) or better.
[5]
7(c) ALT 2: Using SUVAT equations
AnswerMarks
M1A1A1Find the speed at A, obtain horizontal and vertical components of this speed. Obtain the vertical component of the speed at the plane using SUVAT
M1A1caoObtain the resultant speed at the plane, equate to \(\sqrt{\frac{9gr}{2}}\) and eliminate \(v_A\). For the equation obtained following the steps described above. Deduct 1 each error in this equation. Equation need not be simplified. Deduct one mark if the horizontal and vertical components at A have been interchanged.
[5]
[14]
**7(a)**

M1A1A1 | Energy equation from top to reaching the plane. Difference of KE terms and a PE term needed. $m$ in all terms or none. Deduct 1 each error.

[7]

**7(b)**

M1A1cso | Use their expression for $u^2$ in the result given in (a) to obtain a numerical value for $\cos\theta$. Complete to $\frac{5}{6}$. Accept $0.83$ or better.

[2]

**7(c)**

M1A1 | For circular motion, $R \geq 0$

A1 | $\cos = \frac{5}{6}$

dM1A1cao | Use energy from A to the plane

[5]

**7(c) ALT 1**

M1 | Energy equation with correct number of terms and dimensionally correct

A1A1 | Deduct 1 each error

M1A1cao | Use their equation of motion from (a) to find an expression for $v^2$ and complete to a numerical value for $\cos\theta$. Complete to $\frac{5}{6}$. Accept $0.83$ or better.

[5]

**7(c) ALT 2: Using SUVAT equations**

M1A1A1 | Find the speed at A, obtain horizontal and vertical components of this speed. Obtain the vertical component of the speed at the plane using SUVAT

M1A1cao | Obtain the resultant speed at the plane, equate to $\sqrt{\frac{9gr}{2}}$ and eliminate $v_A$. For the equation obtained following the steps described above. Deduct 1 each error in this equation. Equation need not be simplified. Deduct one mark if the horizontal and vertical components at A have been interchanged.

[5]

[14]
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{2273ca38-1e16-44ab-ae84-f4c576cbb8f9-24_575_821_214_566}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

A smooth solid sphere, with centre $O$ and radius $r$, is fixed with its lowest point on a horizontal plane. A particle is placed on the surface of the sphere at the highest point of the sphere. The particle is then projected horizontally with speed $u$ and starts to move on the surface of the sphere. The particle leaves the surface of the sphere at the point $A$ where $O A$ makes an angle $\alpha , \alpha > 0$, with the upward vertical, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \alpha = \frac { 1 } { 3 g r } \left( u ^ { 2 } + 2 g r \right)$
\item Show that $u < \sqrt { g r }$

After leaving the surface of the sphere, the particle strikes the plane with speed $3 \sqrt { \frac { g r } { 2 } }$
\item Find the value of $\cos \alpha$.

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2018 Q7 [14]}}