| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2020 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle with peg/obstacle |
| Difficulty | Challenging +1.2 This is a multi-part vertical circle problem with a collision/energy loss component. Part (a) is a standard 'show that' for tension using energy conservation and circular motion equations. Part (b) requires finding the rate of change of tension, which is more sophisticated. Part (c) involves analyzing motion after an inelastic collision with energy loss and finding when the string goes slack. While it requires multiple techniques (energy methods, circular motion, projectile motion after impact), the individual steps are fairly standard for M3 level, making it moderately above average difficulty but not exceptionally challenging. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05d Variable speed circles: energy methods |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO | |
| \includegraphics[max width=\textwidth, alt={}, center]{ace84823-db30-463e-b24b-f0cd7df73746-23_2255_50_314_34} | |||
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| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}m(8ag) + mg(8a) = \frac{1}{2}mv^2 + mg(8a\cos\theta)\) | M1A1A1 | Attempt at energy equation at general point. Must be dimensionally correct and contain two KE terms and a change in GPE. Correct unsimplified equation, -1 each error. |
| \(v^2 = 24ga - 16ga\cos\theta\) | ||
| \(T + mg\cos\theta = \frac{mv^2}{8a}\) | M1A1 | Attempt to resolve radially. Acceleration can be in either circular form. Correct equation, must be \(\frac{mv^2}{r}\) |
| \(T + mg\cos\theta = \frac{m(24ga - 16ga\cos\theta)}{8a}\) | DM1 | Eliminate \(v\) to produce equation in \(T\), \(m\), \(g\), \(\theta\). Dependent on previous 2 M marks. |
| \(T + mg\cos\theta = 3mg - 2mg\cos\theta\) | ||
| \(T = 3mg - 3mg\cos\theta = 3mg(1-\cos\theta)\) * | A1* | Reach given result with no errors seen. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At \(B\): \(v_B^2 = 24ga\) | B1 | Correct expression for speed (or speed squared) at \(B\). Will not be implied by a correct tension if they simply use the final result in (a). |
| \(T_1 = \frac{m(24ag)}{8a} = 3mg\) or \(T_2 = \frac{m(24ag)}{3a} = 8mg\) | B1 | Correct expression for tension at \(B\) for either radius. Can be found using result from (a). |
| \(\Delta T = 5mg\) | B1 | Correct expression for change in tension. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}mv_1^2 = \frac{1}{2}m(8ag) + mg(11a)\) | M1 | Attempt at energy equation at wall. Must include 2 KE terms and a change in GPE. |
| \(v_1^2 = 30ag\) | ||
| After impact: \(v_2^2 = 20ag\) | A1 | Correct speed (or speed squared, or KE) after impact. |
| \(\frac{1}{2}m(20ag) - mg(3a) = \frac{1}{2}mv_2^2 + mg(8a\cos\alpha)\) | M1A1 | Attempt at energy equation to \(\alpha\). Must include 2 KE terms and a change in GPE. Correct energy equation. |
| \(v_2^2 = 14ga - 16ga\cos\alpha\) | ||
| \(mg\cos\alpha = \frac{m(14ga - 16ga\cos\alpha)}{8a}\) | M1A1 | Attempt at radial equation. If \(T\) included, must be set to zero before this mark. Correct equation in \(\cos\alpha\) only. |
| \(mg\cos\alpha = \frac{7mg}{4} - 2mg\cos\alpha\) | ||
| \(\cos\alpha = \frac{7}{12}\) * | A1* | Solve to reach given result. |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}m(8ag) + mg(8a) = \frac{1}{2}mv^2 + mg(8a\cos\theta)$ | M1A1A1 | Attempt at energy equation at general point. Must be dimensionally correct and contain two KE terms and a change in GPE. Correct unsimplified equation, -1 each error. |
| $v^2 = 24ga - 16ga\cos\theta$ | | |
| $T + mg\cos\theta = \frac{mv^2}{8a}$ | M1A1 | Attempt to resolve radially. Acceleration can be in either circular form. Correct equation, must be $\frac{mv^2}{r}$ |
| $T + mg\cos\theta = \frac{m(24ga - 16ga\cos\theta)}{8a}$ | DM1 | Eliminate $v$ to produce equation in $T$, $m$, $g$, $\theta$. Dependent on previous 2 M marks. |
| $T + mg\cos\theta = 3mg - 2mg\cos\theta$ | | |
| $T = 3mg - 3mg\cos\theta = 3mg(1-\cos\theta)$ * | A1* | Reach given result with no errors seen. |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At $B$: $v_B^2 = 24ga$ | B1 | Correct expression for speed (or speed squared) at $B$. Will **not** be implied by a correct tension if they simply use the final result in (a). |
| $T_1 = \frac{m(24ag)}{8a} = 3mg$ or $T_2 = \frac{m(24ag)}{3a} = 8mg$ | B1 | Correct expression for tension at $B$ for either radius. Can be found using result from (a). |
| $\Delta T = 5mg$ | B1 | Correct expression for change in tension. |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv_1^2 = \frac{1}{2}m(8ag) + mg(11a)$ | M1 | Attempt at energy equation at wall. Must include 2 KE terms and a change in GPE. |
| $v_1^2 = 30ag$ | | |
| After impact: $v_2^2 = 20ag$ | A1 | Correct speed (or speed squared, or KE) after impact. |
| $\frac{1}{2}m(20ag) - mg(3a) = \frac{1}{2}mv_2^2 + mg(8a\cos\alpha)$ | M1A1 | Attempt at energy equation to $\alpha$. Must include 2 KE terms and a change in GPE. Correct energy equation. |
| $v_2^2 = 14ga - 16ga\cos\alpha$ | | |
| $mg\cos\alpha = \frac{m(14ga - 16ga\cos\alpha)}{8a}$ | M1A1 | Attempt at radial equation. If $T$ included, must be set to zero before this mark. Correct equation in $\cos\alpha$ only. |
| $mg\cos\alpha = \frac{7mg}{4} - 2mg\cos\alpha$ | | |
| $\cos\alpha = \frac{7}{12}$ * | A1* | Solve to reach given result. |7.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{ace84823-db30-463e-b24b-f0cd7df73746-20_808_542_264_703}
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\caption{Figure 5}
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A particle of mass $m$ is attached to one end of a light inextensible string of length 8a. The other end of the string is fixed to the point $O$ on the smooth horizontal surface of a desk. The point $E$ is on the edge of the desk, where $O E = 5 a$ and $O E$ is perpendicular to the edge of the desk. The particle is held at the point $A$, vertically above $O$, with the string taut.
The particle is projected horizontally from $A$ with speed $\sqrt { 8 a g }$ in the direction $O E$, as shown in Figure 5.
When the particle is above the level of $O E$ the particle is moving in a vertical circle with radius $8 a$.
Given that, when the string makes an angle $\theta$ with the upward vertical through $O$, the tension in the string is $T$,
\begin{enumerate}[label=(\alph*)]
\item show that $T = 3 m g ( 1 - \cos \theta )$
At the instant when the string is horizontal, the particle passes through the point $B$.
\item Find the instantaneous change in the tension in the string as the particle passes through $B$.
The particle hits the vertical side $E F$ of the desk and rebounds. As a result of the impact, the particle loses one third of the kinetic energy it had immediately before the impact.
In the subsequent motion the string becomes slack when it makes an angle $\alpha$ with the upward vertical through $O$.
\item Show that $\cos \alpha = \frac { 7 } { 12 }$
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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Q7 \\
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\hfill \mbox{\textit{Edexcel M3 2020 Q7 [17]}}