AS June 2023 Q1
1. Three particles of masses \(4m\), \(2m\) and \(km\) are placed at the points with coordinates \((-3, -1)\), \((6, 1)\) and \((-1, 5)\) respectively.
Given that the centre of mass of the three particles is at the point with coordinates \((\bar{x}, \bar{y})\)
Given that the centre of mass of the three particles lies on the line with equation \(y = 2x + 3\)
A fourth particle is placed at the point with coordinates \((\lambda, 4)\).
Given that the centre of mass of the four particles also lies on the line with equation \(y = 2x + 3\)
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(y\) axis: | M1 | 2.1 |
| \(km \times -1 + 4m \times -3 + 2m \times 6 = (k+6)m \times \bar{x}\) | A1 | 1.1b |
| \(\Rightarrow \bar{x} = \dfrac{-k}{k+6}\) * | A1* | 1.1b |
| (3) |
Notes
M1: Moments about \(y\) axis (or a parallel axis). All terms required. Dimensionally consistent. Condone sign errors.
A1: Correct unsimplified equation
A1*: Obtain given answer from full and correct working
Accept with \(6 + k\)
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(x\) axis: \(\left(km \times 5 + 4m \times -1 + 2m \times 1 = (k+6)m \times \bar{y}\right)\) | M1 | 3.1a |
| \(\Rightarrow \bar{y} = \dfrac{5k-2}{k+6}\) | A1 | 1.1b |
| (2) |
Notes
M1: Moments about \(x\) axis (or a parallel axis). All terms required. Dimensionally consistent. Condone sign errors.
A1: Correct unsimplified expression for \(\bar{y}\).
Any equivalent simplified form
The first 5 marks are available for a combined equation in vector form
| Scheme | Marks | AO |
|---|---|---|
| \(\bar{y} = 2\bar{x} + 3 \Rightarrow \dfrac{5k-2}{k+6} = \dfrac{-2k}{k+6} + 3\) \(\left(5k - 2 = -2k + (3k + 18)\right)\) | M1 | 1.1b |
| \(\Rightarrow k = 5\) | A1 | 1.1b |
| (2) |
Notes
M1: Correct use of their \(\bar{y}\) and given \(\bar{x}\) to find \(k\).
A1: Correct only
| Scheme | Marks | AO |
|---|---|---|
| \(4 = 2\lambda + 3 \Rightarrow\) | M1 | 3.4 |
| \(\lambda = \dfrac{1}{2}\) | A1 | 1.1b |
| (2) | ||
| (9 marks) |
Notes
M1: Use the model to obtain \(\lambda = \ldots\) or \(x = \ldots\)
If working from the beginning and the new particle has mass \(M\) then \(\dfrac{23+4M}{11+M} = 2\left(\dfrac{-5+\lambda M}{11+M}\right) + 3\)
A1: Accept \(x = \dfrac{1}{2}\)