| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Work-energy with multiple stages |
| Difficulty | Challenging +1.2 This is a multi-step M3 question combining elastic strings, inclined planes, energy conservation, and momentum conservation. While it requires several techniques (equilibrium to find initial extension, energy for Q's speed, momentum for collision, then energy again for final position), each step follows standard M3 procedures with clear guidance from the question structure. The algebra is moderately involved but straightforward, making this somewhat harder than average but well within typical M3 scope. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| \(112e/4 = 3.5 \times 9.8 \times \frac{40}{49}\) | M1 | For using \(mg\sin\theta\) and \(\lambda e/L\) |
| \(V^2 = 2x8x(4+1)\) | M1 | For using \(s = 4 + e\) and \(a = 8\) in \(v^2 = 2as\), or by energy |
| \(V^2 = 80\) | A1 | |
| \(0.5\sqrt{80} = (0.5 + 3.5)u\) | M1 | For using the principle of conservation of momentum |
| Initial speed of combined particles is \(\frac{1}{2}\sqrt{5}\text{ ms}^{-1}\) | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Gain in \(EE = (112/(2\times4))((X+1)^2 - 1^2)\) | M1 | For using \(EE = \lambda x^2/2L\) |
| Loss of \(KE = \frac{1}{2}(0.5 + 3.5) \times 5/4\) | A1 | |
| Loss of \(PE = (0.5 + 3.5) \times 9.8 \times \frac{40}{49}X\) | B1 | |
| \(14(X^2 + 2X) = 2.5 + 32X\) | M1 | For using the principle of conservation of energy |
| \(28X^2 - 8X - 5 = 0\) | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - mg\sin\theta = -ma\) | M1 | For use of \(F = ma\); allow one slip for A1 |
| \(\frac{112(x+1)}{4} - 4g\frac{40}{49} = -4a\) | M1 | Using \(a = v\frac{dv}{dx}\) and integrating |
| \(\int(7x - 1)dx = -\int vdv (+c)\) | M1 | |
| \(\frac{7x^2}{2} - x = -\frac{v^2}{2} + c\) | A1 | |
| \(c = \frac{5}{8}\) | A1 | |
| \(28x^2 - 8x - 5 = 0\) | A1 | |
| [6] | AG Convincingly |
**Part i**
$112e/4 = 3.5 \times 9.8 \times \frac{40}{49}$ | M1 | For using $mg\sin\theta$ and $\lambda e/L$
$V^2 = 2x8x(4+1)$ | M1 | For using $s = 4 + e$ and $a = 8$ in $v^2 = 2as$, or by energy
$V^2 = 80$ | A1 |
$0.5\sqrt{80} = (0.5 + 3.5)u$ | M1 | For using the principle of conservation of momentum
Initial speed of combined particles is $\frac{1}{2}\sqrt{5}\text{ ms}^{-1}$ | A1 |
| [6] |
**Part ii**
Gain in $EE = (112/(2\times4))((X+1)^2 - 1^2)$ | M1 | For using $EE = \lambda x^2/2L$
Loss of $KE = \frac{1}{2}(0.5 + 3.5) \times 5/4$ | A1 |
Loss of $PE = (0.5 + 3.5) \times 9.8 \times \frac{40}{49}X$ | B1 |
$14(X^2 + 2X) = 2.5 + 32X$ | M1 | For using the principle of conservation of energy
$28X^2 - 8X - 5 = 0$ | A1 |
| [6] |
**OR alternative approach**
$T - mg\sin\theta = -ma$ | M1 | For use of $F = ma$; allow one slip for A1
$\frac{112(x+1)}{4} - 4g\frac{40}{49} = -4a$ | M1 | Using $a = v\frac{dv}{dx}$ and integrating
$\int(7x - 1)dx = -\int vdv (+c)$ | M1 |
$\frac{7x^2}{2} - x = -\frac{v^2}{2} + c$ | A1 |
$c = \frac{5}{8}$ | A1 |
$28x^2 - 8x - 5 = 0$ | A1 |
| [6] | AG Convincingly6\\
\includegraphics[max width=\textwidth, alt={}, center]{67af8d98-85af-42b1-9e7f-c6380a1f8a3f-4_638_473_260_836}
A particle $P$, of mass 3.5 kg , is in equilibrium suspended from the top $A$ of a smooth slope inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 40 } { 49 }$, by an elastic rope of natural length 4 m and modulus of elasticity 112 N (see diagram). Another particle $Q$, of mass 0.5 kg , is released from rest at $A$ and slides freely downwards until it reaches $P$ and becomes attached to it.\\
(i) Find the value of $V ^ { 2 }$, where $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of $Q$ immediately before it becomes attached to $P$, and show that the speed of the combined particles, immediately after $Q$ becomes attached to $P$, is $\frac { 1 } { 2 } \sqrt { 5 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The combined particles slide downwards for a distance of $X \mathrm {~m}$, before coming instantaneously to rest at $B$.\\
(ii) Show that $28 X ^ { 2 } - 8 X - 5 = 0$.
\hfill \mbox{\textit{OCR M3 2011 Q6 [12]}}