| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Standard +0.3 This is a standard M1 moments question involving two tilting scenarios to find two unknowns (M and d). It requires systematic application of the moment principle about two different pivot points, but the setup is straightforward with clear geometry and no conceptual surprises. Slightly easier than average due to the structured nature and routine algebraic manipulation. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(T_2 = 0\) | M1 | \(T_2 = 0\) seen or implied |
| \(M(C), 64g \times 0.625 = Mg(d-2.5)\) OR e.g. \(M(C), 64g \times 0.625 = Mgx\) | M1 A1 | Correct number of terms, dimensionally correct equation in \(M\) and one unknown length. Allow without \(g\)'s. Omission of a length is M error. A1 for correct equation in \(M\) and \(d\) only or another unknown length |
| Other equations listed: \(T_1 = Mg + 64g\) | ||
| \(M(A), 64g \times 1.875 + Mgd = 2.5T_1\) | ||
| \(M(G), 64g \times (d-1.875) = T_1 \times (d-2.5)\) | ||
| Use of \(S_1 = 0\) | M1 | \(S_1 = 0\) seen or implied |
| \(M(D), 48g \times 1.5 = Mg(6-d)\) OR e.g. \(M(D), 48g \times 1.5 = Mg(3.5-x)\) | M1 A1 | Correct number of terms, dimensionally correct equation in \(M\) and same unknown length. A1 for correct equation in \(M\) and \(d\) only or same unknown length |
| Other equations listed: \(S_2 = Mg + 48g\) | ||
| \(M(A), 48g \times 7.5 + Mgd = 6S_2\) | ||
| \(M(C), 48g \times 5 + Mg(d-2.5) = 3.5S_2\) | ||
| \(M(G), 48g \times (7.5-d) = S_2 \times (6-d)\) | ||
| \(M(B), 48g \times 0.5 + Mg(8-d) = 2S_2\) | ||
| Solve for \(M\) | DM1 | Dependent on all previous M marks |
| \(M = 32\) exact answer | A1 | Correct exact answer |
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $T_2 = 0$ | M1 | $T_2 = 0$ seen or implied |
| $M(C), 64g \times 0.625 = Mg(d-2.5)$ **OR** e.g. $M(C), 64g \times 0.625 = Mgx$ | M1 A1 | Correct number of terms, dimensionally correct equation in $M$ and one unknown length. Allow without $g$'s. Omission of a length is M error. A1 for correct equation in $M$ and $d$ only or another unknown length |
| Other equations listed: $T_1 = Mg + 64g$ | | |
| $M(A), 64g \times 1.875 + Mgd = 2.5T_1$ | | |
| $M(G), 64g \times (d-1.875) = T_1 \times (d-2.5)$ | | |
| Use of $S_1 = 0$ | M1 | $S_1 = 0$ seen or implied |
| $M(D), 48g \times 1.5 = Mg(6-d)$ **OR** e.g. $M(D), 48g \times 1.5 = Mg(3.5-x)$ | M1 A1 | Correct number of terms, dimensionally correct equation in $M$ and same unknown length. A1 for correct equation in $M$ and $d$ only or same unknown length |
| Other equations listed: $S_2 = Mg + 48g$ | | |
| $M(A), 48g \times 7.5 + Mgd = 6S_2$ | | |
| $M(C), 48g \times 5 + Mg(d-2.5) = 3.5S_2$ | | |
| $M(G), 48g \times (7.5-d) = S_2 \times (6-d)$ | | |
| $M(B), 48g \times 0.5 + Mg(8-d) = 2S_2$ | | |
| Solve for $M$ | DM1 | Dependent on all previous M marks |
| $M = 32$ exact answer | A1 | Correct exact answer |4.
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A non-uniform beam $A B$ has length 8 m and mass $M \mathrm {~kg}$.
The centre of mass of the beam is $d$ metres from $A$.
The beam is supported in equilibrium in a horizontal position by two vertical light ropes. One rope is attached to the beam at $C$, where $A C = 2.5 \mathrm {~m}$ and the other rope is attached to the beam at $D$, where $D B = 2 \mathrm {~m}$, as shown in Figure 2.
A gymnast, of mass 64 kg , stands on the beam at the point $X$, where $A X = 1.875 \mathrm {~m}$, and the beam remains in equilibrium in a horizontal position but is now on the point of tilting about $C$.
The gymnast then dismounts from the beam.
A second gymnast, of mass 48 kg , now stands on the beam at the point $Y$, where $Y B = 0.5 \mathrm {~m}$, and the beam remains in equilibrium in a horizontal position but is now on the point of tilting about $D$.
The beam is modelled as a non-uniform rod and the gymnasts are modelled as particles. Find the value of $M$.
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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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\hfill \mbox{\textit{Edexcel M1 2020 Q4 [8]}}