# Question 4:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Mass of elemental disc $\delta m = k\left(2 + \frac{x}{a}\right)\pi a^2 \delta x$ | B1 | Or $\delta m = \rho\pi a^2 \delta x$; condone lack of $\delta x$ |
| About diameter of elemental disc $\delta I = \frac{1}{4}\left(k\left(2+\frac{x}{a}\right)\pi a^2\delta x\right)a^2$ | B1 | Or $\delta I = \frac{1}{4}(\rho\pi a^2\delta x)a^2$; condone lack of $\delta x$ |
| By parallel axes theorem, about the given axis: $\delta I = \frac{1}{4}\left(k\left(2+\frac{x}{a}\right)\pi a^2\delta x\right)a^2 + k\left(2+\frac{x}{a}\right)\pi a^2\delta x \cdot x^2$ | M1* | |
| $= k\left(2+\frac{x}{a}\right)\left(\frac{1}{4}a^2 + x^2\right)\pi a^2\,\delta x$ | A1 | Must be of form $\rho\left(\lambda a^2 + \mu x^2\right)a^2$; condone lack of $\delta x$ |
| $I = k\pi a^2\int_0^{3a}\left(2+\frac{x}{a}\right)\left(\frac{1}{4}a^2 + x^2\right)dx = k\pi a^2\int_0^{3a}\frac{1}{2}a^2 + \frac{1}{4}ax + 2x^2 + \frac{1}{a}x^3\,dx$ | M1dep* A1 | Condone lack of limits for both M1 A1; A1 correct simplified integral |
| $= k\pi a^2\left[\frac{1}{2}a^2x + \frac{1}{8}ax^2 + \frac{2}{3}x^3 + \frac{1}{4a}x^4\right]_0^{3a}$ | M1 | Integrating and using correct limits; dependent on both previous M marks |
| $= \frac{327}{8}k\pi a^5$ | A1 | |
| $= \frac{327}{8}\cdot\frac{2M}{21\pi a^3}\cdot\pi a^5 = \frac{109}{28}Ma^2$ | E1 | If $\delta x$ omitted throughout then withhold final mark |
| **[9]** | | |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $M\bar{x} = \int x\,dm = \int_0^{3a} k\left(2+\frac{x}{a}\right)\pi a^2 x\,dx$ | B1 | Condone lack of limits |
| $= k\pi a^2\int_0^{3a} 2x + \frac{1}{a}x^2\,dx = k\pi a^2\left[x^2 + \frac{1}{3a}x^3\right]_0^{3a}$ | M1 | Integrating and using correct limits |
| $= 18k\pi a^4$ | A1 | |
| $= 18\cdot\frac{2M}{21\pi a^3}\cdot\pi a^4 = \frac{12}{7}Ma \Rightarrow \bar{x} = \frac{12}{7}a$ | E1 | |
| **[4]** | | |

## Question 4(iii):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $I_B = I_G + Md_{BG}^2$ and $I_A = I_G + Md_{AG}^2$ | M1 | Attempt at use of parallel axis theorem (for both). Note: $d_{BG} = \frac{9}{7}a$, $d_{AG} = \frac{12}{7}a$, $I_A = \frac{109}{28}Ma^2$ |
| $I_B = (I_A - Md_{AG}^2) + Md_{BG}^2$ | A1 | Eliminate or evaluate $I_G$ ($I_G = \frac{187}{196}Ma^2$) |
| $= \frac{109}{28}Ma^2 + M\left(\frac{81}{49} - \frac{144}{49}\right)a^2$ | A1 | |
| $= \frac{73}{28}Ma^2$ | E1 | |
| **[4]** | | |

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## Question 4(iv):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Moment of momentum of object about axis before collision is $0.2 \times 20 \times 2.1 = 8.4$ | M1 | $0.2 \times 20 \times (2.1$ or $7/\sqrt{10})$ |
| $0.2 \times 20 \times 2.1 = 8.4$ | A1 | M1 A1 for 8.4 seen (www) |
| MI of object about axis after coalesces is $0.2\left(\dfrac{7}{\sqrt{10}}\right)^2 = 0.98$ | B1 | |
| MI of cylinder about the axis is $5.11$ | B1 | |
| MI of combined object about axis is $6.09$ | B1 | 5.992 with no working scores B1 only |
| Conservation of angular momentum $8.4 = 6.09\dot{\theta}$ | M1 | Use of conservation of angular momentum |
| $\dot{\theta} = 1.379\ldots$ | A1 | Cao |
| **[7]** | | |