OCR MEI Further Mechanics Major 2020 November — Question 7 13 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2020
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeSHM on inclined plane
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem involving elastic strings and SHM on an inclined plane. Part (a) requires setting up forces and applying Hooke's law to derive a differential equation (routine but multi-step), part (b) involves solving the SHM equation and finding when extension becomes zero (standard technique), and part (c) uses SHM formulas for velocity. While it requires careful bookkeeping of equilibrium position, extensions, and signs, all techniques are standard for Further Maths students who have studied elastic strings and SHM. The 13 marks reflect length rather than exceptional difficulty.
Spec3.02f Non-uniform acceleration: using differentiation and integration4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings
\includegraphics{figure_7} A particle P of mass \(m\) is attached to one end of a light elastic string of natural length \(6a\) and modulus of elasticity \(3mg\). The other end of the string is fixed to a point O on a smooth plane, which is inclined at an angle of \(30°\) to the horizontal. The string lies along a line of greatest slope of the plane and P rests in equilibrium on the inclined plane at a point A, as shown in Fig. 7. P is now pulled a further distance \(2a\) down the line of greatest slope through A and released from rest. At time \(t\) later, the displacement of P from A is \(x\), where the positive direction of \(x\) is down the plane.
  1. Show that, until the string slackens, \(x\) satisfies the differential equation $$\frac{d^2x}{dt^2} + \frac{gx}{2a} = 0.$$ [6]
  2. Determine, in terms of \(a\) and \(g\), the time at which the string slackens. [5]
  3. Find, in terms of \(a\) and \(g\), the speed of P when the string slackens. [2]
Question 7:
AnswerMarks Guidance
7(a) 3mge
Hooke’s law: T =
AnswerMarks Guidance
6aB1 1.2
extension of the string
when P is in
equilibrium
AnswerMarks Guidance
T =mgsin30M1 3.3
equilibriumAllow sin/cos
confusion
mge mg
= ⇒e=a
AnswerMarks Guidance
2a 2A1 2.2a
d2x
mgsin30−T =m
AnswerMarks Guidance
dt2M1* 3.3
number of terms)
3mg d2x
mgsin30− (a+x)=m
AnswerMarks Guidance
6a dt2M1dep* 3.4
extension
d2x mg 3mga 3mgx d2x gx
m − + + =0⇒ + =0
AnswerMarks Guidance
dt2 2 6a 6a dt2 2aA1 2.2a
as answer givn
[6]
AnswerMarks Guidance
7(b) g g
x= Acos t+Bsin t
AnswerMarks Guidance
2a 2aB1 1.2a
g
point of the motion x= Acos t
AnswerMarks
2aA and B are arbitrary
constants
t =0,x=2a⇒ A=2a
AnswerMarks Guidance
t =0,x=0⇒B=0M1 3.4
just A)
g
x=2acos t
AnswerMarks Guidance
2aA1 1.1
 g  1
Slack when x=−a⇒cost =−
 
AnswerMarks Guidance
 2a  2M1 3.1b
for x
g 2π 2π 2a
t = ⇒t =
AnswerMarks Guidance
2a 3 3 gA1 2.2a
[5]
AnswerMarks Guidance
7(c) g ( )
v2 = (2a)2 −(−a)2
AnswerMarks Guidance
2aM1 3.4
v2 =ω2 A2 −x2
Use of with their
values or differentiation of their x
3ga
v=
AnswerMarks Guidance
2A1 1.1
[2]
Question 7:
7 | (a) | 3mge
Hooke’s law: T =
6a | B1 | 1.2 | Where e is the
extension of the string
when P is in
equilibrium
T =mgsin30 | M1 | 3.3 | Resolving parallel to the plane with P in
equilibrium | Allow sin/cos
confusion
mge mg
= ⇒e=a
2a 2 | A1 | 2.2a
d2x
mgsin30−T =m
dt2 | M1* | 3.3 | NII parallel to the plane (with correct
number of terms)
3mg d2x
mgsin30− (a+x)=m
6a dt2 | M1dep* | 3.4 | Use of Hooke’s law in NII with correct
extension
d2x mg 3mga 3mgx d2x gx
m − + + =0⇒ + =0
dt2 2 6a 6a dt2 2a | A1 | 2.2a | AG – sufficient working must be shown
as answer givn
[6]
7 | (b) | g g
x= Acos t+Bsin t
2a 2a | B1 | 1.2a | Or as the motion starts at the extreme
g
point of the motion x= Acos t
2a | A and B are arbitrary
constants
t =0,x=2a⇒ A=2a
t =0,x=0⇒B=0 | M1 | 3.4 | Use initial conditions to find A and B (or
just A)
g
x=2acos t
2a | A1 | 1.1
 g  1
Slack when x=−a⇒cost =−
 
 2a  2 | M1 | 3.1b | Substituting x=−a into their equation
for x
g 2π 2π 2a
t = ⇒t =
2a 3 3 g | A1 | 2.2a
[5]
7 | (c) | g ( )
v2 = (2a)2 −(−a)2
2a | M1 | 3.4 | ( )
v2 =ω2 A2 −x2
Use of with their
values or differentiation of their x
3ga
v=
2 | A1 | 1.1
[2]
\includegraphics{figure_7}

A particle P of mass $m$ is attached to one end of a light elastic string of natural length $6a$ and modulus of elasticity $3mg$. The other end of the string is fixed to a point O on a smooth plane, which is inclined at an angle of $30°$ to the horizontal. The string lies along a line of greatest slope of the plane and P rests in equilibrium on the inclined plane at a point A, as shown in Fig. 7.

P is now pulled a further distance $2a$ down the line of greatest slope through A and released from rest. At time $t$ later, the displacement of P from A is $x$, where the positive direction of $x$ is down the plane.

\begin{enumerate}[label=(\alph*)]
\item Show that, until the string slackens, $x$ satisfies the differential equation
$$\frac{d^2x}{dt^2} + \frac{gx}{2a} = 0.$$ [6]
\item Determine, in terms of $a$ and $g$, the time at which the string slackens. [5]
\item Find, in terms of $a$ and $g$, the speed of P when the string slackens. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q7 [13]}}
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