| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Find exponents with all unknowns |
| Difficulty | Standard +0.3 Part (a) is a standard dimensional analysis exercise requiring systematic application of dimensional consistency - finding dimensions of λ and solving simultaneous equations for exponents α, β, γ. Part (b) is a routine energy conservation problem with elastic rope. All techniques are standard M3 material with no novel insights required, making this slightly easier than average. |
| Spec | 6.01a Dimensions: M, L, T notation6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EE is \(\frac{1}{2} \times \frac{2060}{24} \times 6^2 = 1545\) | B1 | |
| (PE gained) = (EE lost) + (KE lost) | M1 | Equation involving PE, EE and KE; can be awarded from start to point where string becomes slack *or* any complete method (e.g. SHM) for finding \(v^2\) at natural length; if B0, give A1 for \(v^2 = 88.2\) correctly obtained |
| \(50 \times 9.8 \times h = 1545 + \frac{1}{2} \times 50 \times 12^2\) | F1 | *or* \(0 = 88.2 - 2 \times 9.8 \times s \quad (s=4.5)\) |
| \(490h = 1545 + 3600\) | *Notes: \(\frac{1}{2} \times \frac{2060}{24} \times 6\) used as EE can earn B0M1F1A0; \(\frac{2060}{24} \times 6\) used as EE gets B0M0* | |
| \(h = 10.5\) | ||
| \(OA = 30 - h = 19.5\) m | A1 | 4 marks total |
# Question 1(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| EE is $\frac{1}{2} \times \frac{2060}{24} \times 6^2 = 1545$ | B1 | |
| (PE gained) = (EE lost) + (KE lost) | M1 | Equation involving PE, EE and KE; can be awarded from start to point where string becomes slack *or* any complete method (e.g. SHM) for finding $v^2$ at natural length; if B0, give A1 for $v^2 = 88.2$ correctly obtained |
| $50 \times 9.8 \times h = 1545 + \frac{1}{2} \times 50 \times 12^2$ | F1 | *or* $0 = 88.2 - 2 \times 9.8 \times s \quad (s=4.5)$ |
| $490h = 1545 + 3600$ | | *Notes: $\frac{1}{2} \times \frac{2060}{24} \times 6$ used as EE can earn B0M1F1A0; $\frac{2060}{24} \times 6$ used as EE gets B0M0* |
| $h = 10.5$ | | |
| $OA = 30 - h = 19.5$ m | A1 | **4 marks total** |
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\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the dimensions of velocity, acceleration and force.
A ball of mass $m$ is thrown vertically upwards with initial velocity $U$. When the velocity of the ball is $v$, it experiences a force $\lambda v ^ { 2 }$ due to air resistance where $\lambda$ is a constant.
\item Find the dimensions of $\lambda$.
A formula approximating the greatest height $H$ reached by the ball is
$$H \approx \frac { U ^ { 2 } } { 2 g } - \frac { \lambda U ^ { 4 } } { 4 m g ^ { 2 } }$$
where $g$ is the acceleration due to gravity.
\item Show that this formula is dimensionally consistent.
A better approximation has the form $H \approx \frac { U ^ { 2 } } { 2 g } - \frac { \lambda U ^ { 4 } } { 4 m g ^ { 2 } } + \frac { 1 } { 6 } \lambda ^ { 2 } U ^ { \alpha } m ^ { \beta } g ^ { \gamma }$.
\item Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
\end{enumerate}\item A girl of mass 50 kg is practising for a bungee jump. She is connected to a fixed point O by a light elastic rope with natural length 24 m and modulus of elasticity 2060 N . At one instant she is 30 m vertically below O and is moving vertically upwards with speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. She comes to rest instantaneously, with the rope slack, at the point A . Find the distance OA .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2008 Q1 [18]}}