| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse on inclined plane |
| Difficulty | Moderate -0.3 This is a straightforward multi-part mechanics question requiring standard application of impulse-momentum theorem, work-energy principle, and equations of motion. Part (i) is a 'show that' verification requiring one calculation. Parts (ii)-(iv) follow a clear sequential path with no novel insight needed—just systematic application of SUVAT and energy methods to an inclined plane with friction. The numerical values are designed to work out cleanly, and the question structure guides students through each step. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(33.6 = 3.5 \times v - 3.5 \times (-2.1)\) | M1 | Use of Impulse = change in momentum. Condone sign error. |
| \(v = (+/-)7.5\) | A1 (AG) | Sign must be consistent with changed direction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial energy \(= \frac{1}{2} \times 3.5 \times 7.5^2\) | M1 | Can be implied by awrt 98.4 |
| Final energy \(= \frac{1}{2} \times 3.5 \times 1.5^2 + 3.5g \times 4.2\sin 20\) | M1 | Can be implied by awrt 53.2 |
| Work done against resistance = Initial energy \(-\) final energy | M1 | Allow 1 slip in energy equations. NB do not allow use of \(\pm 2.1\) for velocity |
| Or: Loss of \(KE = 94.5\text{J}\) | M1 | |
| Gain in \(GPE = 49.27\text{J}\) | M1 | |
| Work done against resistance = loss of \(KE\) - gain in \(GPE\) | M1 | Allow wrong sign at this stage |
| Or: \(1.5^2 = 7.5^2 - 2(a)(4.2)\), (so \(a = (-)6.43...\)) | M1 | |
| \(Fr + 3.5g \times \sin 20 = -ma\) (so \(Fr = 10.76...\)) | M1 | |
| \(Fr \times 4.2\) | M1 | |
| \(45.2\) | A1 | Do not allow negative values even if corrected, unless justified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(45.2 / 4.2\) | M1FT | Their 45.2 (must be \(> 0\)). Must come from 3 term equation in (ii) involving initial energy \(-\) final energy or \(\Delta KE - GPE\) |
| \(10.8\) | A1 | May have been found in (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Final energy \(= 3.5g \times x\sin 20\) | M1 | |
| \(\frac{1}{2} \times 3.5 \times 7.5^2 - 3.5g \times x\sin 20 = 10.8x\) | M1 | Setting up equation for \(x\) using energy balance, using their friction. Allow 1 slip. Could use energy at B and add answer to 4.2 |
| Or: \(Fr + 3.5g \times \sin 20 = 22.5\text{N}\) | M1 | Overall force down the slope. Must be from 2 terms. NB could use \(F = ma\) and uvast to find \(x\) at this stage |
| \(22.5x = \frac{1}{2} \times 3.5 \times 7.5^2\) | M1 | Do not accept use of \(Fr = 22.5\) if seen in part (iii) |
| Or: \(1.5^2 = 7.5^2 - 2(a)(4.2)\), (so \(a = -6.43...\)) | M1 | Using uvast/suvat. Could use uvast from B |
| \(0 = 7.5^2 - 2("-6.43...")(x)\) | M1 | |
| \(x = 4.4\) (= AC) | A1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $33.6 = 3.5 \times v - 3.5 \times (-2.1)$ | M1 | Use of Impulse = change in momentum. Condone sign error. |
| $v = (+/-)7.5$ | A1 (AG) | Sign must be consistent with changed direction |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial energy $= \frac{1}{2} \times 3.5 \times 7.5^2$ | M1 | Can be implied by awrt 98.4 |
| Final energy $= \frac{1}{2} \times 3.5 \times 1.5^2 + 3.5g \times 4.2\sin 20$ | M1 | Can be implied by awrt 53.2 |
| Work done against resistance = Initial energy $-$ final energy | M1 | Allow 1 slip in energy equations. NB do not allow use of $\pm 2.1$ for velocity |
| **Or:** Loss of $KE = 94.5\text{J}$ | M1 | |
| Gain in $GPE = 49.27\text{J}$ | M1 | |
| Work done against resistance = loss of $KE$ - gain in $GPE$ | M1 | Allow wrong sign at this stage |
| **Or:** $1.5^2 = 7.5^2 - 2(a)(4.2)$, (so $a = (-)6.43...$) | M1 | |
| $Fr + 3.5g \times \sin 20 = -ma$ (so $Fr = 10.76...$) | M1 | |
| $Fr \times 4.2$ | M1 | |
| $45.2$ | A1 | Do not allow negative values even if corrected, unless justified |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $45.2 / 4.2$ | M1FT | Their 45.2 (must be $> 0$). Must come from 3 term equation in (ii) involving initial energy $-$ final energy or $\Delta KE - GPE$ |
| $10.8$ | A1 | May have been found in (ii) |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Final energy $= 3.5g \times x\sin 20$ | M1 | |
| $\frac{1}{2} \times 3.5 \times 7.5^2 - 3.5g \times x\sin 20 = 10.8x$ | M1 | Setting up equation for $x$ using energy balance, using their friction. Allow 1 slip. Could use energy at B and add answer to 4.2 |
| **Or:** $Fr + 3.5g \times \sin 20 = 22.5\text{N}$ | M1 | Overall force down the slope. Must be from 2 terms. NB could use $F = ma$ and uvast to find $x$ at this stage |
| $22.5x = \frac{1}{2} \times 3.5 \times 7.5^2$ | M1 | Do not accept use of $Fr = 22.5$ if seen in part (iii) |
| **Or:** $1.5^2 = 7.5^2 - 2(a)(4.2)$, (so $a = -6.43...$) | M1 | Using uvast/suvat. Could use uvast from B |
| $0 = 7.5^2 - 2("-6.43...")(x)$ | M1 | |
| $x = 4.4$ (= AC) | A1 | |
---2 A particle $P$ of mass 3.5 kg is moving down a line of greatest slope of a rough inclined plane. At the instant that its speed is $2.1 \mathrm {~ms} ^ { - 1 } P$ is at a point $A$ on the plane. At that instant an impulse of magnitude 33.6 Ns , directed up the line of greatest slope, acts on $P$.\\
(i) Show that as a result of the impulse $P$ starts moving up the plane with a speed of $7.5 \mathrm {~ms} ^ { - 1 }$.
While still moving up the plane, $P$ has speed $1.5 \mathrm {~ms} ^ { - 1 }$ at a point $B$ where $A B = 4.2 \mathrm {~m}$. The plane is inclined at an angle of $20 ^ { \circ }$ to the horizontal. The frictional force exerted by the plane on $P$ is modelled as constant.\\
(ii) Calculate the work done against friction as $P$ moves from $A$ to $B$.\\
(iii) Hence find the magnitude of the frictional force acting on $P$.\\
$P$ first comes to instantaneous rest at point $C$ on the plane.\\
(iv) Calculate $A C$.
\hfill \mbox{\textit{OCR Further Mechanics AS 2018 Q2 [11]}}