Question 3 - A-Level Maths

OCR MEI Further Mechanics A AS 2018 June — Question 3 9 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2018
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Dimensional Analysis
Type	Find exponents with all unknowns
Difficulty	Moderate -0.8 This is a straightforward dimensional analysis problem with standard mechanics quantities. Students set up three simultaneous equations from matching dimensions of T, M, and L, then solve algebraically. The method is routine for Further Mechanics students, though it requires careful bookkeeping. Part (iii) involves simple substitution, and part (iv) tests conceptual understanding that period is independent of mass—a well-known result.
Spec	6.01a Dimensions: M, L, T notation 6.01d Unknown indices: using dimensions

3 Jodie is doing an experiment involving a simple pendulum. The pendulum consists of a small object tied to one end of a piece of string. The other end of the string is attached to a fixed point O and the object is allowed to swing between two fixed points A and B and back again, as shown in Fig. 3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-3_328_350_584_886} \captionsetup{labelformat=empty} \caption{Fig. 3}

\end{figure} Jodie thinks that \(P\), the time the pendulum takes to swing from A to B and back again, depends on the mass, \(m\), of the small object, the length, \(l\), of the piece of string, and the acceleration due to gravity \(g\). She proposes the formula \(P = k m ^ { \alpha } l ^ { \beta } g ^ { \gamma }\).

What is the significance of \(k\) in Jodie's formula?
Use dimensional analysis to determine the values of \(\alpha , \beta\) and \(\gamma\). Jodie finds that when the mass of the object is 1.5 kg and the length of the string is 80 cm the time taken for the pendulum to swing from A to B and back again is 1.8 seconds.
Use Jodie's formula and your answers to part (ii) to find each of the following.
(A) The value of \(k\) (B) The time taken for the pendulum to swing from A to B and back again when the mass of the object is 0.9 kg and the length of the string is 1.4 m
Comment on the assumption made by Jodie that the formula for the time taken for the pendulum to swing from A to B and back again is dependent on \(m , l\) and \(g\).

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks	Guidance
3	(i)	(k is a dimensionless) constant
[1]	1.2	Ignore extra comments
(ii)	M: 0 = 𝛼

L: 0 = 𝛽 +𝛾

T: 1 = −2𝛾

1 1

𝛼 = 0 𝛽 = 𝛾 = −

Answer	Marks
2 2	M1

A1

Answer	Marks
[3]	1.1

1.1

Answer	Marks
1.1	Correct method; can be implied by

T = M𝛼L𝛽(LT−2)𝛾

3 correct equations soi

(iii)

Answer	Marks
(A)	0.8

1.8 = 𝑘√

𝑔

Answer	Marks
𝑘 = 6.3	M1

A1

Answer	Marks
[2]	3.4
1.1	Ft attempt to use their values (but

not all 0)

www

(iii)

Answer	Marks
(B)	1.4

𝑡 = 6.3×√

𝑔

9√7

Time = 2.38 (seconds) or

Answer	Marks
10	M1

A1

Answer	Marks
[2]	3.4
1.1	Ft attempt to use their values (not all

0) and their k

Answer	Marks
www 2.381176	SC: Use of 80 cm in (i) and

140 cm in (ii) leads to k =

0.63 (M1A0) and time =

2.38 (M1A1)

Answer	Marks
(iv)	(a) It turns out that the formula is in fact

independent of the mass OR (b) it is quite

possible that it could have been dependent on

Answer	Marks	Guidance
another unknown quantity.	B1
[1]	2.2b	Or any valid equivalent statement

(a) depends on α = 0 in (ii)

Question 3:
3 | (i) | (k is a dimensionless) constant | B1
[1] | 1.2 | Ignore extra comments
(ii) | M: 0 = 𝛼
L: 0 = 𝛽 +𝛾
T: 1 = −2𝛾
1 1
𝛼 = 0 𝛽 = 𝛾 = −
2 2 | M1
A1
A1
[3] | 1.1
1.1
1.1 | Correct method; can be implied by
T = M𝛼L𝛽(LT−2)𝛾
3 correct equations soi
(iii)
(A) | 0.8
1.8 = 𝑘√
𝑔
𝑘 = 6.3 | M1
A1
[2] | 3.4
1.1 | Ft attempt to use their values (but
not all 0)
www
(iii)
(B) | 1.4
𝑡 = 6.3×√
𝑔
9√7
Time = 2.38 (seconds) or
10 | M1
A1
[2] | 3.4
1.1 | Ft attempt to use their values (not all
0) and their k
www 2.381176 | SC: Use of 80 cm in (i) and
140 cm in (ii) leads to k =
0.63 (M1A0) and time =
2.38 (M1A1)
(iv) | (a) It turns out that the formula is in fact
independent of the mass OR (b) it is quite
possible that it could have been dependent on
another unknown quantity. | B1
[1] | 2.2b | Or any valid equivalent statement
(a) depends on α = 0 in (ii)

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3 Jodie is doing an experiment involving a simple pendulum. The pendulum consists of a small object tied to one end of a piece of string. The other end of the string is attached to a fixed point O and the object is allowed to swing between two fixed points A and B and back again, as shown in Fig. 3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-3_328_350_584_886}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

Jodie thinks that $P$, the time the pendulum takes to swing from A to B and back again, depends on the mass, $m$, of the small object, the length, $l$, of the piece of string, and the acceleration due to gravity $g$. She proposes the formula $P = k m ^ { \alpha } l ^ { \beta } g ^ { \gamma }$.
\begin{enumerate}[label=(\roman*)]
\item What is the significance of $k$ in Jodie's formula?
\item Use dimensional analysis to determine the values of $\alpha , \beta$ and $\gamma$.

Jodie finds that when the mass of the object is 1.5 kg and the length of the string is 80 cm the time taken for the pendulum to swing from A to B and back again is 1.8 seconds.
\item Use Jodie's formula and your answers to part (ii) to find each of the following.\\
(A) The value of $k$\\
(B) The time taken for the pendulum to swing from A to B and back again when the mass of the object is 0.9 kg and the length of the string is 1.4 m
\item Comment on the assumption made by Jodie that the formula for the time taken for the pendulum to swing from A to B and back again is dependent on $m , l$ and $g$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2018 Q3 [9]}}

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