| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with standard mechanics techniques. Part (i) uses basic elastic PE = KE equation, part (ii) is a verification requiring GPE + KE = elastic PE calculation, and part (iii) adds a resistance force but follows the same energy method. All are routine applications of work-energy principles with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{2} \times 5000x^2 = \frac{1}{2} \times 400 \times 3^2\) | M1, A1 | Equation involving EE and KE |
| Compression is \(0.849\) m | A1 | Accept \(\frac{3\sqrt{2}}{5}\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Change in PE is \(400 \times 9.8 \times (7.35 + 1.4)\sin\theta\) | M1 | Or \(400 \times 9.8 \times 1.4\sin\theta\) and \(\frac{1}{2} \times 400 \times 4.54^2\) |
| \(= 400 \times 9.8 \times 8.75 \times \frac{1}{7}\) | ||
| \(= 4900\) J | A1 | Or \(784 + 4116\) |
| Change in EE is \(\frac{1}{2} \times 5000 \times 1.4^2 = 4900\) J | M1 | M1M1A1 can also be given for correct equation in \(x\) (compression): \(2500x^2 - 560x - 4116 = 0\) |
| Since Loss of PE = Gain of EE, car will be at rest | E1 | Conclusion required, or solving equation to obtain \(x = 1.4\) |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| WD against resistance is \(7560(24+x)\) | B1 | \((= 181440 + 7560x)\) |
| Change in EE is \(\frac{1}{2} \times 5000x^2\) | B1 | \((= 2500x^2)\) |
| Change in KE is \(\frac{1}{2} \times 400 \times 30^2\) | B1 | \((= 180000)\) |
| Change in PE is \(400 \times 9.8 \times (24+x) \times \frac{1}{7}\) | B1 | \((= 13440 + 560x)\) |
| OR Speed \(7.75\text{ ms}^{-1}\) when it hits buffer, then WD against resistance is \(7560x\) | B1 | |
| Change in EE is \(\frac{1}{2} \times 5000x^2\) | B1 | \((= 2500x^2)\) |
| Change in KE is \(\frac{1}{2} \times 400 \times 7.75^2\) | B1 | \((= 12000)\) |
| Change in PE is \(400 \times 9.8 \times x \times \frac{1}{7}\) | B1 | \((= 560x)\) |
| \(-7560(24+x) = \frac{1}{2} \times 5000x^2 - \frac{1}{2} \times 400 \times 30^2 - 400 \times 9.8 \times (24+x) \times \frac{1}{7}\) | M1 | Equation involving WD, EE, KE, PE |
| F1 | ||
| \(-7560(24+x) = 2500x^2 - 180000 - 560(24+x)\) | ||
| \(-3.024(24+x) = x^2 - 72 - 0.224(24+x)\) | ||
| \(x^2 + 2.8x - 4.8 = 0\) | M1, A1 | Simplification to three term quadratic |
| \(x = \frac{-2.8 + \sqrt{2.8^2 + 19.2}}{2}\) | M1 | |
| \(= 1.2\) | A1 | |
| Total: 10 |
# Question 2:
## Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{2} \times 5000x^2 = \frac{1}{2} \times 400 \times 3^2$ | M1, A1 | Equation involving EE and KE |
| Compression is $0.849$ m | A1 | Accept $\frac{3\sqrt{2}}{5}$ |
| | **Total: 3** | |
## Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| Change in PE is $400 \times 9.8 \times (7.35 + 1.4)\sin\theta$ | M1 | Or $400 \times 9.8 \times 1.4\sin\theta$ and $\frac{1}{2} \times 400 \times 4.54^2$ |
| $= 400 \times 9.8 \times 8.75 \times \frac{1}{7}$ | | |
| $= 4900$ J | A1 | Or $784 + 4116$ |
| Change in EE is $\frac{1}{2} \times 5000 \times 1.4^2 = 4900$ J | M1 | M1M1A1 can also be given for correct equation in $x$ (compression): $2500x^2 - 560x - 4116 = 0$ |
| Since Loss of PE = Gain of EE, car will be at rest | E1 | Conclusion required, or solving equation to obtain $x = 1.4$ |
| | **Total: 4** | |
## Part (iii):
| Working | Marks | Guidance |
|---------|-------|----------|
| WD against resistance is $7560(24+x)$ | B1 | $(= 181440 + 7560x)$ |
| Change in EE is $\frac{1}{2} \times 5000x^2$ | B1 | $(= 2500x^2)$ |
| Change in KE is $\frac{1}{2} \times 400 \times 30^2$ | B1 | $(= 180000)$ |
| Change in PE is $400 \times 9.8 \times (24+x) \times \frac{1}{7}$ | B1 | $(= 13440 + 560x)$ |
| OR Speed $7.75\text{ ms}^{-1}$ when it hits buffer, then WD against resistance is $7560x$ | B1 | |
| Change in EE is $\frac{1}{2} \times 5000x^2$ | B1 | $(= 2500x^2)$ |
| Change in KE is $\frac{1}{2} \times 400 \times 7.75^2$ | B1 | $(= 12000)$ |
| Change in PE is $400 \times 9.8 \times x \times \frac{1}{7}$ | B1 | $(= 560x)$ |
| $-7560(24+x) = \frac{1}{2} \times 5000x^2 - \frac{1}{2} \times 400 \times 30^2 - 400 \times 9.8 \times (24+x) \times \frac{1}{7}$ | M1 | Equation involving WD, EE, KE, PE |
| | F1 | |
| $-7560(24+x) = 2500x^2 - 180000 - 560(24+x)$ | | |
| $-3.024(24+x) = x^2 - 72 - 0.224(24+x)$ | | |
| $x^2 + 2.8x - 4.8 = 0$ | M1, A1 | Simplification to three term quadratic |
| $x = \frac{-2.8 + \sqrt{2.8^2 + 19.2}}{2}$ | M1 | |
| $= 1.2$ | A1 | |
| | **Total: 10** | |
---2 In trials for a vehicle emergency stopping system, a small car of mass 400 kg is propelled towards a buffer. The buffer is modelled as a light spring of stiffness $5000 \mathrm {~N} \mathrm {~m} ^ { - 1 }$. One end of the spring is fixed, and the other end points directly towards the oncoming car. Throughout this question, there is no driving force acting on the car, and there are no resistances to motion apart from those specifically mentioned.
At first, the buffer is mounted on a horizontal surface, and the car has speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it hits the buffer, as shown in Fig. 2.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3ec81c4e-e0fa-43d9-9c79-ef9df746be8f-3_220_1105_671_520}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
(i) Find the compression of the spring when the car comes (instantaneously) to rest.
The buffer is now mounted on a slope making an angle $\theta$ with the horizontal, where $\sin \theta = \frac { 1 } { 7 }$. The car is released from rest and travels 7.35 m down the slope before hitting the buffer, as shown in Fig. 2.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3ec81c4e-e0fa-43d9-9c79-ef9df746be8f-3_268_1091_1329_529}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
(ii) Verify that the car comes instantaneously to rest when the spring is compressed by 1.4 m .
The surface of the slope (including the section under the buffer) is now covered with gravel which exerts a constant resistive force of 7560 N on the car. The car is moving down the slope, and has speed $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it is 24 m from the buffer, as shown in Fig. 2.3. It comes to rest when the spring has been compressed by $x$ metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3ec81c4e-e0fa-43d9-9c79-ef9df746be8f-3_305_1087_2122_529}
\captionsetup{labelformat=empty}
\caption{Fig. 2.3}
\end{center}
\end{figure}
(iii) By considering work and energy, find the value of $x$.
\hfill \mbox{\textit{OCR MEI M3 2009 Q2 [17]}}