| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Circular Motion 2 |
| Type | Projectile motion after leaving circle |
| Difficulty | Challenging +1.2 This is a two-part circular motion problem requiring energy conservation to find speed at B, then circular motion dynamics for normal force, followed by projectile motion. While it involves multiple steps and combines topics, each individual component uses standard A-level techniques without requiring novel insight. The geometry is clearly specified, making this a methodical but straightforward application problem, slightly above average difficulty due to the multi-stage nature and combination of mechanics topics. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} \times 0.4v_B^2 = \frac{1}{2} \times 0.4 \times 0.8^2 + 0.4g \times 1.4(\cos 30° - \cos 60°)\) | M1 | Conservation of energy from A to B |
| \(v_B^2 = 10.68...\) | A1 | soi |
| \(R - 0.4g \cos 30° = 0.4 \times \frac{v_B^2}{1.4}\) | M1 | Use of Newton II radially, 3 terms |
| Magnitude of contact force at B is 6.45 N | A1 [5] | Correct equation, with their \(v_B\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| Vertical velocity component at B is \(v_B \sin 30°\) | B1Ft | (1.634...) |
| Height of B above plane is \(2.5 + 1.4(1 - \cos 30°)\) | M1 | soi; (2.687...) |
| Time of flight given by \(-2.687 = 1.634t - \frac{1}{2}gt^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) vertically |
| \(t = 0.9259...\) | A1 | Correct equation |
| Horizontal distance covered in flight is \((v_B \cos 30°)t\) | M1 | Using their values of \(v_B\) and \(t\); (2.6209...) |
| \(FH = 1.4\sin 30° + 2.6209... = 3.32\) m | A1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Height of B above plane is \(2.5 + 1.4(1 - \cos 30°)\) | B1 | (2.687...) |
| Projection velocity is \(v_B\) at 30° above horizontal | B1Ft | soi; ft their value of \(v_B\) |
| Equation of trajectory is \(y = x\tan 30° - \frac{gx^2}{2v_B^2 \cos^2 30°}\) | M1 | oe, if origin other than B is used |
| \(0.61152x^2 - 0.57735x - 2.68756 = 0\) | M1 | Formation of explicit quadratic in x |
| \(x = 2.62...\) | A1 | |
| \(FH = 1.4\sin 30° + 2.6209... = 3.32\) m | A1 [7] |
## (i)
$\frac{1}{2} \times 0.4v_B^2 = \frac{1}{2} \times 0.4 \times 0.8^2 + 0.4g \times 1.4(\cos 30° - \cos 60°)$ | M1 | Conservation of energy from A to B
$v_B^2 = 10.68...$ | A1 | soi
$R - 0.4g \cos 30° = 0.4 \times \frac{v_B^2}{1.4}$ | M1 | Use of Newton II radially, 3 terms
Magnitude of contact force at B is 6.45 N | A1 [5] | Correct equation, with their $v_B$; cao
## (ii)
Vertical velocity component at B is $v_B \sin 30°$ | B1Ft | (1.634...)
Height of B above plane is $2.5 + 1.4(1 - \cos 30°)$ | M1 | soi; (2.687...)
Time of flight given by $-2.687 = 1.634t - \frac{1}{2}gt^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ vertically
$t = 0.9259...$ | A1 | Correct equation
Horizontal distance covered in flight is $(v_B \cos 30°)t$ | M1 | Using their values of $v_B$ and $t$; (2.6209...)
$FH = 1.4\sin 30° + 2.6209... = 3.32$ m | A1 [7] |
**Alternative solution:**
Height of B above plane is $2.5 + 1.4(1 - \cos 30°)$ | B1 | (2.687...)
Projection velocity is $v_B$ at 30° above horizontal | B1Ft | soi; ft their value of $v_B$
Equation of trajectory is $y = x\tan 30° - \frac{gx^2}{2v_B^2 \cos^2 30°}$ | M1 | oe, if origin other than B is used
$0.61152x^2 - 0.57735x - 2.68756 = 0$ | M1 | Formation of explicit quadratic in x
$x = 2.62...$ | A1 |
$FH = 1.4\sin 30° + 2.6209... = 3.32$ m | A1 [7] |
---7 A smooth track $A B$ is in the shape of an arc of a circle with centre $O$ and radius 1.4 m . The track is fixed in a vertical plane with $A$ above the level of $B$ and a point $C$ on the track vertically below $O$. Angle $A O C$ is $60 ^ { \circ }$ and angle $C O B$ is $30 ^ { \circ }$. Point $C$ is 2.5 m vertically above the point $F$, which lies in a horizontal plane. A particle of mass 0.4 kg is placed at $A$ and projected down the track with an initial velocity of $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The particle first hits the plane at point $H$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{a8c9d007-e67f-4637-9e74-630ba9a91442-5_767_1265_488_415}\\
(i) Find the magnitude of the contact force between the particle and the track when the particle is at $B$. [5]\\
(ii) Find the distance $F H$.
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q7 [12]}}