| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 11 |
| Topic | Circular Motion 2 |
| Type | Projectile motion after leaving circle |
| Difficulty | Challenging +1.2 This is a multi-step Further Mechanics question combining circular motion with projectile motion. Part (a) requires applying the standard circular motion equation with energy conservation to find tension. Part (b) involves projectile motion from the break point. While it requires careful coordinate work and multiple techniques, the individual steps follow standard procedures without requiring novel insight. The Further Mechanics context and multi-stage nature place it above average difficulty. |
| Spec | 3.02i Projectile motion: constant acceleration model6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}m \times 3.5^2 = \frac{1}{2}mv^2 + mg \times 0.8\left(1 - \cos\frac{1}{3}\pi\right)\) | M1 [3.1b] | Conservation of energy (could be general angle for RHS) |
| \(v^2 = 3.5^2 - 0.8g = 4.41\) | A1 [1.1] | Or \(v = 2.1\) |
| \(T - 1.2g\cos\frac{1}{3}\pi = \frac{1.2v^2}{0.8}\) | M1 [3.1b] | Resolving weight and use of NII with correct centripetal acceleration |
| So tension when string breaks is awrt 12.5 N | A1 [3.2a] | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_y = 2.1\sin\frac{1}{3}\pi\) upwards | M1 [3.1b] | Direction can be implied by eg use of \(-9.8\) |
| \(P\) starts freefall \(1.5 + 0.8(1 - \cos\frac{1}{3}\pi)\) above plane | B1 [3.1b] | \((= 1.9)\) |
| \(-(1.5 + 0.8(1 - \cos\frac{1}{3}\pi)) = (2.1\sin\frac{1}{3}\pi)t - \frac{1}{2}gt^2\) | M1 [1.1] | Setting up 3 term quadratic equation in \(t\) with adjustment to \(\pm 1.5\) as displacement |
| \(-1.9 = (1.05\sqrt{3})t - 4.9t^2\) or \(4.9t^2 - 1.8186K \cdot t - 1.9 = 0\) | A1 [1.1] | Correct equation soi |
| \(t = \text{awrt } 0.835\) | A1 [1.1] | BC |
| Horizontal distance in freefall is \(2.1\cos\frac{1}{3}\pi \times 0.835...\) | M1 [1.1] | 0.877... |
| \(0.8\sin\frac{1}{3}\pi + 2.1\cos\frac{1}{3}\pi \times 0.835... =\) awrt 1.57 m | A1 [2.2a] | horizontal distance while attached plus horizontal distance in freefall. |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Projection velocity is 2.1 m s\(^{-1}\) at \(\frac{1}{3}\pi\) above horizontal | B1ft [3.1b] | soi; ft their 2.1 |
| \(P\) starts freefall \(1.5 + 0.8(1 - \cos\frac{1}{3}\pi)\) above plane | B1 [3.1b] | \((= 1.9)\) |
| Equation of trajectory: \(y = x\tan\frac{1}{3}\pi - \frac{gx^2}{2 \times 2.1^2\cos^2(\frac{1}{3}\pi)}\) | M1 [3.4] | Or suitably adjusted if not using \(P\)'s position when string breaks as the origin |
| \(-1.9 = x\sqrt{3} - \frac{40}{9}x^2\) or \(4.444K x^2 - 1.732K x - 1.9 = 0\) | M1 [3.4] | oe: formulation of explicit quadratic equation in \(x\) |
| \(x = \text{awrt } 0.877\) | A1 [1.1] | BC |
| \(0.8\sin\frac{1}{3}\pi + 0.877... =\) awrt 1.57 m | A1 [2.2a] | |
| [7] |
### (a)
$\frac{1}{2}m \times 3.5^2 = \frac{1}{2}mv^2 + mg \times 0.8\left(1 - \cos\frac{1}{3}\pi\right)$ | M1 [3.1b] | Conservation of energy (could be general angle for RHS)
$v^2 = 3.5^2 - 0.8g = 4.41$ | A1 [1.1] | Or $v = 2.1$
$T - 1.2g\cos\frac{1}{3}\pi = \frac{1.2v^2}{0.8}$ | M1 [3.1b] | Resolving weight and use of NII with correct centripetal acceleration
So tension when string breaks is awrt 12.5 N | A1 [3.2a]
| [4]
### (b)
$v_y = 2.1\sin\frac{1}{3}\pi$ upwards | M1 [3.1b] | Direction can be implied by eg use of $-9.8$
$P$ starts freefall $1.5 + 0.8(1 - \cos\frac{1}{3}\pi)$ above plane | B1 [3.1b] | $(= 1.9)$
$-(1.5 + 0.8(1 - \cos\frac{1}{3}\pi)) = (2.1\sin\frac{1}{3}\pi)t - \frac{1}{2}gt^2$ | M1 [1.1] | Setting up 3 term quadratic equation in $t$ with adjustment to $\pm 1.5$ as displacement
$-1.9 = (1.05\sqrt{3})t - 4.9t^2$ or $4.9t^2 - 1.8186K \cdot t - 1.9 = 0$ | A1 [1.1] | Correct equation soi
$t = \text{awrt } 0.835$ | A1 [1.1] | BC | $\frac{\sqrt{331} + 3\sqrt{3}}{28}$ or $\frac{3\sqrt{331} + 9\sqrt{3}}{80}$
Horizontal distance in freefall is $2.1\cos\frac{1}{3}\pi \times 0.835...$ | M1 [1.1] | 0.877...
$0.8\sin\frac{1}{3}\pi + 2.1\cos\frac{1}{3}\pi \times 0.835... =$ awrt 1.57 m | A1 [2.2a] | horizontal distance while attached plus horizontal distance in freefall.
| [7]
**Alternative solution:**
Projection velocity is 2.1 m s$^{-1}$ at $\frac{1}{3}\pi$ above horizontal | B1ft [3.1b] | soi; ft their 2.1
$P$ starts freefall $1.5 + 0.8(1 - \cos\frac{1}{3}\pi)$ above plane | B1 [3.1b] | $(= 1.9)$
Equation of trajectory: $y = x\tan\frac{1}{3}\pi - \frac{gx^2}{2 \times 2.1^2\cos^2(\frac{1}{3}\pi)}$ | M1 [3.4] | Or suitably adjusted if not using $P$'s position when string breaks as the origin
$-1.9 = x\sqrt{3} - \frac{40}{9}x^2$ or $4.444K x^2 - 1.732K x - 1.9 = 0$ | M1 [3.4] | oe: formulation of explicit quadratic equation in $x$
$x = \text{awrt } 0.877$ | A1 [1.1] | BC | $\frac{3\sqrt{331} + 9\sqrt{3}}{80}$
$0.8\sin\frac{1}{3}\pi + 0.877... =$ awrt 1.57 m | A1 [2.2a]
| [7]
---5 One end of a light inextensible string of length 0.8 m is attached to a fixed point, $O$. The other end is attached to a particle $P$ of mass $1.2 \mathrm {~kg} . P$ hangs in equilibrium at a distance of 1.5 m above a horizontal plane. The point on the plane directly below $O$ is $F$.\\
$P$ is projected horizontally with speed $3.5 \mathrm {~ms} ^ { - 1 }$. The string breaks when $O P$ makes an angle of $\frac { 1 } { 3 } \pi$ radians with the downwards vertical through $O$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-3_776_910_1242_244}
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the tension in the string at the instant before the string breaks.
\item Find the distance between $F$ and the point where $P$ first hits the plane.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q5 [11]}}