Edexcel M3 2015 June — Question 7 15 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2015
SessionJune
Marks15
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, finding the leaving condition (normal reaction = 0), then projectile motion. It involves multiple connected parts with algebraic manipulation and requires understanding of when contact is lost. More demanding than typical M3 questions but follows standard patterns for this topic.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall
  1. A solid smooth sphere, with centre \(O\) and radius \(r\), is fixed to a point \(A\) on a horizontal floor. A particle \(P\) is placed on the surface of the sphere at the point \(B\), where \(B\) is vertically above \(A\). The particle is projected horizontally from \(B\) with speed \(\frac { \sqrt { g r } } { 2 }\) and starts to move on the surface of the sphere. When \(O P\) makes an angle \(\theta\) with the upward vertical and \(P\) remains in contact with the sphere, the speed of \(P\) is \(v\).
    1. Show that \(v ^ { 2 } = \frac { g r } { 4 } ( 9 - 8 \cos \theta )\).
    The particle leaves the surface of the sphere when \(\theta = \alpha\).
  2. Find the value of \(\cos \alpha\). After leaving the surface of the sphere, \(P\) moves freely under gravity and hits the floor at the point \(C\). Given that \(r = 0.5 \mathrm {~m}\),
  3. find, to 2 significant figures, the distance \(A C\).
(a)
\(\frac{1}{2}mv^2 = \frac{1}{2}m \cdot mgr\left(1-\cos\theta\right)\)
M1A1A1
\(v^2 = \frac{rg}{4}(9 - 8\cos\theta)\)
A1 (4)
(b)
\(R - mg\cos\theta = \frac{mv^2}{r}\)
M1A1
\(R = mg\cos\theta + \frac{mg}{r} \cdot 9 - 8\cos\theta\)
DM1
\(R = 0\) when \(mg\cos\theta = \frac{9}{4}\cos\theta\)
\(\cos\theta = \frac{3}{4}\) or \(0.75\)
A1 (4)
(c)
Initial vertical component of speed \(= \frac{3g}{8}\sin\theta = \frac{3g}{8}\sqrt{1 - \left(\frac{3}{4}\right)^2} = 1.2679...\) m/s
M1A1
\(\frac{7}{8} = 1.2679...t - \frac{1}{2}gt^2\)
M1
\(7 - 10.143...t - 39.2t^2 = 0\)
DM1
\(39.2t^2 - 10.143...t - 7 = 0\)
\(t = \frac{10.143... + \sqrt{10.143...^2 + 4 \times 39.2 \times 7}}{2 \times 39.2}\)
A1
\(t = 0.3125...\) s
Horizontal speed \(= \frac{3g}{8}\cos\theta = \frac{3g}{8} \cdot \frac{3}{4} = \frac{27g}{32}\)
M1A1cso (7) [15]
\(AC = 0.3125 \times r\sin\theta + 0.4493 \times 0.3307 = 0.78\) m
Guidance Notes:
AnswerMarks
M1Using energy to get the speed at the floor. Can be from the top or the point of leaving the surface
A1Correct speed at floor
M1Using the horizontal component of the speed and Pythagoras to obtain the vertical component at the floor
M1Using \(v = u + at\) vertically to get \(t\)
A1Correct \(t\)
M1A1Complete as main method
\(\frac{dv}{dt} = 900 - \frac{63000}{kt^2}\)
M1
\(\int_0^{10.5} dv = \int_0^4 \frac{70}{1+kt^2} dt\)
AnswerMarks
DM1A1Integration, limits not needed
\(10.5 - 0 = \frac{70}{4k} - \frac{70}{k}\)
AnswerMarks
M1Substitute limits
\(k = 5\)
AnswerMarks
A1Correct value
\(v = t - \frac{14}{t}\)
AnswerMarks
A1Integrate again with limits as shown
\(v = 14 - \frac{1}{t}\)
AnswerMarks
A1Obtain GIVEN answer
Alternative Solution:
\(\frac{dv}{dt} = 900 - \frac{63000}{kt^2}\)
M1
\(\int_v \int_t dv = \int_0^t \frac{70}{1+kt^2} dt\)
AnswerMarks
DM1A1Integration, limits not needed
\(v - v_0 = \frac{70}{k}\left(\frac{1}{1+kt} - \frac{1}{kt}\right)\)
AnswerMarks
M1Substitute limits and \(v = 10.5\), \(t = 4\)
\(k = 5\)
AnswerMarks
A1Correct value
\(v = \frac{70}{5}\left(1 - \frac{1}{t}\right)\)
AnswerMarks
A1Substitute
$v = 14 - \frac
**(a)**

$\frac{1}{2}mv^2 = \frac{1}{2}m \cdot mgr\left(1-\cos\theta\right)$

M1A1A1

$v^2 = \frac{rg}{4}(9 - 8\cos\theta)$

A1 (4)

**(b)**

$R - mg\cos\theta = \frac{mv^2}{r}$

M1A1

$R = mg\cos\theta + \frac{mg}{r} \cdot 9 - 8\cos\theta$

DM1

$R = 0$ when $mg\cos\theta = \frac{9}{4}\cos\theta$

$\cos\theta = \frac{3}{4}$ or $0.75$

A1 (4)

**(c)**

Initial vertical component of speed $= \frac{3g}{8}\sin\theta = \frac{3g}{8}\sqrt{1 - \left(\frac{3}{4}\right)^2} = 1.2679...$ m/s

M1A1

$\frac{7}{8} = 1.2679...t - \frac{1}{2}gt^2$

M1

$7 - 10.143...t - 39.2t^2 = 0$

DM1

$39.2t^2 - 10.143...t - 7 = 0$

$t = \frac{10.143... + \sqrt{10.143...^2 + 4 \times 39.2 \times 7}}{2 \times 39.2}$

A1

$t = 0.3125...$ s

Horizontal speed $= \frac{3g}{8}\cos\theta = \frac{3g}{8} \cdot \frac{3}{4} = \frac{27g}{32}$

M1A1cso (7) [15]

$AC = 0.3125 \times r\sin\theta + 0.4493 \times 0.3307 = 0.78$ m

---

**Guidance Notes:**

M1 | Using energy to get the speed at the floor. Can be from the top or the point of leaving the surface

A1 | Correct speed at floor

M1 | Using the horizontal component of the speed and Pythagoras to obtain the vertical component at the floor

M1 | Using $v = u + at$ vertically to get $t$

A1 | Correct $t$

M1A1 | Complete as main method

---

$\frac{dv}{dt} = 900 - \frac{63000}{kt^2}$

M1

$\int_0^{10.5} dv = \int_0^4 \frac{70}{1+kt^2} dt$

DM1A1 | Integration, limits not needed

$10.5 - 0 = \frac{70}{4k} - \frac{70}{k}$

M1 | Substitute limits

$k = 5$

A1 | Correct value

$v = t - \frac{14}{t}$

A1 | Integrate again with limits as shown

$v = 14 - \frac{1}{t}$

A1 | Obtain GIVEN answer

---

**Alternative Solution:**

$\frac{dv}{dt} = 900 - \frac{63000}{kt^2}$

M1

$\int_v \int_t dv = \int_0^t \frac{70}{1+kt^2} dt$

DM1A1 | Integration, limits not needed

$v - v_0 = \frac{70}{k}\left(\frac{1}{1+kt} - \frac{1}{kt}\right)$

M1 | Substitute limits and $v = 10.5$, $t = 4$

$k = 5$

A1 | Correct value

$v = \frac{70}{5}\left(1 - \frac{1}{t}\right)$

A1 | Substitute

$v = 14 - \frac
\begin{enumerate}
  \item A solid smooth sphere, with centre $O$ and radius $r$, is fixed to a point $A$ on a horizontal floor. A particle $P$ is placed on the surface of the sphere at the point $B$, where $B$ is vertically above $A$. The particle is projected horizontally from $B$ with speed $\frac { \sqrt { g r } } { 2 }$ and starts to move on the surface of the sphere. When $O P$ makes an angle $\theta$ with the upward vertical and $P$ remains in contact with the sphere, the speed of $P$ is $v$.\\
(a) Show that $v ^ { 2 } = \frac { g r } { 4 } ( 9 - 8 \cos \theta )$.
\end{enumerate}

The particle leaves the surface of the sphere when $\theta = \alpha$.\\
(b) Find the value of $\cos \alpha$.

After leaving the surface of the sphere, $P$ moves freely under gravity and hits the floor at the point $C$.

Given that $r = 0.5 \mathrm {~m}$,\\
(c) find, to 2 significant figures, the distance $A C$.

\hfill \mbox{\textit{Edexcel M3 2015 Q7 [15]}}
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