# Question 7:

## Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $I = \Delta mv \Rightarrow 44.1 = 3.5u \Rightarrow u = 44.1/3.5 = 12.6\ \text{ms}^{-1}$ | B1 | AO 3.3; use of impulse-momentum principle to find initial speed; FT their $u$ |
| Initial $KE = \frac{1}{2} \times 3.5 \times \text{``}12.6\text{''}^2$ ($= 277.83$ J) | B1FT | AO 1.1 |
| $PE = 3.5g \times 5.4(1 - \cos\theta)$ | M1 | AO 3.4; PE with allowance made for change in height in terms of $\theta$ and use of $mgh$; could see separate PE terms and/or errors in sign(s) of PE term(s); allow sin/cos confusion; their initial KE must be $> 0$ |
| ``$277.83$'' $= 3.5g \times 5.4(1-\cos\theta)$ | M1 | AO 3.4; conservation of energy with final KE set to 0 |
| $\Rightarrow \cos\theta = -\frac{1}{2} \Rightarrow \theta = \frac{2\pi}{3}$ | A1 | AO 1.1; or awrt $2.09$ ($2.09439\ldots$); radians only |

**Alternative Method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $I = \Delta mv \Rightarrow 44.1 = 3.5u \Rightarrow u = 12.6\ \text{ms}^{-1}$ | B1 | |
| Initial $KE = \frac{1}{2} \times 3.5 \times \text{``}12.6\text{''}^2$ ($= 277.83$ J) | B1FT | |
| $PE = 3.5g \times h$ and ``$277.83$'' $= 3.5g \times h$ | M1 | NB $h = 8.1$ m |
| $\cos\theta = (\pm)\frac{5.4 - \text{``}8.1\text{''}}{5.4}$ | M1 | Or $\sin\alpha = \frac{8.1-5.4}{5.4}$ oe (to horizontal); allow sin/cos confusion; NB $\alpha = \frac{\pi}{6}$; probably using a diagram; may be explained by diagram |
| $\theta = \frac{2\pi}{3}$ | A1 | $\theta = \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3}$ |

## Part (b)

| Answer | Marks | Guidance |
|--------|-------|----------|
| ``$277.83$'' $=$ ``$3.5g \times 5.4(1-\cos\theta)$'' $+ 20 \times k\theta \times 5.4$ | M1 | AO 3.4; applying work-energy principle with their PE and KE and an energy loss term; FT their previous values; $k$ may be positive or negative e.g. $\frac{\pi}{180}$; allow use of $h$ instead of $\theta$ |
| $277.83 = 185.22 - 185.22\cos\theta + 108\theta$; $343 + 686\cos\theta = 400\theta$; $343(1 + 2\cos\theta) = 400\theta$ **AG** | A1 | AO 1.1; rearranging convincingly to AG |

## Question (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Energy lost to resistance $= 20 \times 5.4 \times 1.306$ J $= 141.048$ J | M1 | Using "$r\theta$" and 20 to find work done against resistance |
| $141.048 = 3.5 \times 9.8h$ | M1 | Using W-E principle (i.e. lost energy would manifest as more PE) |
| $h = 4.11$ (m) | A1 | $4.112\ldots$ |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $5.4(1 - \cos(\frac{2\pi}{3}))$ $[= \text{"8.1"}]$ | M1 | Finding maximum height predicted by first model (may be seen in part (a)). Award this mark if re-used in part (c). Allow sin/cos confusion. NB $\theta = 2.09$ gives $8.079\ldots$ |
| $5.4(1 - \cos(1.306))$ $[= 3.986\ldots]$ | M1 | Finding maximum height predicted by second model |
| $\text{"8.1"} - 3.986\ldots = 4.11$ m (3 sf) | A1 | $4.09$-$4.11$ (m). Allow FT of their $\theta$ from part (a) |

**[3]**

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## Question (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Take into account the mass and/or weight of the rod (which is unlikely to be negligible) | B1 | Or other sensible improvement, e.g. making resistance to motion dependent on speed or account for rod not being perfectly inextensible etc. Allow reference to size and shape of the particle, but not just a generalisation such as: "do not model as a particle". Allow the idea that tension may cause the rod to extend or affect the friction at the hinge, but not just tension on its own. Allow the idea that there could be some spin in particle P or motion in and out of the plane. Ignore using (light inextensible) string instead of a rod. Ignore gravity as resistive force. Ignore reference to friction at the hinge. Ignore reference to tangential acceleration or speed. Splitting resistance into air resistance and friction must be justified as an improvement. Ignore reference to air resistance not being constant, or reason why friction may not be constant. |

**[1]**