OCR MEI Further Mechanics Major 2019 June — Question 10 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeInverse power force - non-gravitational context
DifficultyChallenging +1.2 This is a Further Maths mechanics question involving inverse-square forces and friction. Part (a) requires applying Newton's second law with the chain rule (a standard FM technique), part (b) is verification by differentiation (routine), and part (c) requires analyzing equilibrium conditions. While it involves multiple concepts, the techniques are standard for Further Mechanics and the question provides significant scaffolding through its structure.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)3.03r Friction: concept and vector form6.06a Variable force: dv/dt or v*dv/dx methods
A particle P, of mass \(m\), moves on a rough horizontal table. P is attracted towards a fixed point O on the table by a force of magnitude \(\frac{kmg}{x^2}\), where \(x\) is the distance OP. The coefficient of friction between P and the table is \(\mu\). P is initially projected in a direction directly away from O. The velocity of P is first zero at a point A which is a distance \(a\) from O.
  1. Show that the velocity \(v\) of P, when P is moving away from O, satisfies the differential equation $$\frac{\mathrm{d}}{\mathrm{d}x}(v^2) + \frac{2kg}{x^2} + 2\mu g = 0.$$ [3]
  2. Verify that $$v^2 = 2gk\left(\frac{1}{x} - \frac{1}{a}\right) + 2\mu g(a-x).$$ [3]
  3. Find, in terms of \(k\) and \(a\), the range of values of \(\mu\) for which P remains at A. [2]
Question 10:
AnswerMarks Guidance
10(a) kmg
mx F
x2
kmg
mx mg
x2
dv kg 1 d   kg
v  g0 v2  g0
dx x2 2dx x2
d   2kg
 v2  2g0
AnswerMarks
dx x2M1*
M1dep*
A1
AnswerMarks
[3]3.3
3.4
AnswerMarks
2.2aN2L with correct number of terms –
accept any form for a – M0 if
1mxkmg Fseen
2 x2
Use of F R and substitute in their
application of N2L
dv
Use of av to get to AG
dx
Note that d  1mv2  kmg Fis
dx 2 x2
equivalent to the first M mark (work-
AnswerMarks
energy principle for a variable force)Allow 1 d v2 or for
a
2dx
acceleration allow any of
adv vdv d2xbut not
dt dx dt2
if circular motion
implied
Allow any correct form
for a
 
Must see the 1 d v2
2dx
term before AG
AnswerMarks Guidance
10(b) 1 1
Whenxa,v2 2gk  2gaa0
a a
therefore v = 0 at x = a
d  1 1  2gk
2gk  2gax  2g
  
dx x a  x2
d   2kg
v2  2g
dx x2
2gk 2kg
 2g 2g 0
AnswerMarks
x2 x2B1
M1
A1
AnswerMarks
[3]3.5a
2.1
AnswerMarks
2.2aMust explicitly show that when x = a,
v = 0
Differentiate v2 to get two terms of
the form  
x2
Correctly shown – could re-arrange
their derivative for v2 to the AG e.g.
d   2gk
v2  2gtherefore
dx x2
d   2gk
v2  2g 0
AnswerMarks
dx x2Allow v2 = 0
Allow for
differentiating v
Must be = 0
AnswerMarks Guidance
10(b) ALT
2kg 2kg
v2  2gdxv2  2gxc
AnswerMarks Guidance
x2 xM1* Separating and integrating to the form
kg
v2  gx
AnswerMarks
x+c not required
2kg
v = 0 at x = a c2ga
AnswerMarks Guidance
aM1dep* Using given conditions to find c
2kg 2kg
v2  2gx2ga
x a
1 1
v2 2gk  2gax
 
AnswerMarks Guidance
x aA1 AG – so sufficient working must be
shown
AnswerMarks Guidance
10(c) kmg
Remain at A if mg
a2
k

AnswerMarks
a2M1
A1
AnswerMarks
[2]3.1b
1.1Considering relationship between
attractive force and friction (accept
any inequality or equals)
AnswerMarks
caoAllow x for a for the M
mark
Question 10:
10 | (a) | kmg
mx F
x2
kmg
mx mg
x2
dv kg 1 d   kg
v  g0 v2  g0
dx x2 2dx x2
d   2kg
 v2  2g0
dx x2 | M1*
M1dep*
A1
[3] | 3.3
3.4
2.2a | N2L with correct number of terms –
accept any form for a – M0 if
1mxkmg Fseen
2 x2
Use of F R and substitute in their
application of N2L
dv
Use of av to get to AG
dx
Note that d  1mv2  kmg Fis
dx 2 x2
equivalent to the first M mark (work-
energy principle for a variable force) | Allow 1 d v2 or for
a
2dx
acceleration allow any of
adv vdv d2xbut not
dt dx dt2
if circular motion
implied
Allow any correct form
for a
 
Must see the 1 d v2
2dx
term before AG
10 | (b) | 1 1
Whenxa,v2 2gk  2gaa0
a a
therefore v = 0 at x = a
d  1 1  2gk
2gk  2gax  2g
  
dx x a  x2
d   2kg
v2  2g
dx x2
2gk 2kg
 2g 2g 0
x2 x2 | B1
M1
A1
[3] | 3.5a
2.1
2.2a | Must explicitly show that when x = a,
v = 0
Differentiate v2 to get two terms of

the form  
x2
Correctly shown – could re-arrange
their derivative for v2 to the AG e.g.
d   2gk
v2  2gtherefore
dx x2
d   2gk
v2  2g 0
dx x2 | Allow v2 = 0
Allow for
differentiating v
Must be = 0
10 | (b) | ALT | Solving differential equation
2kg 2kg
v2  2gdxv2  2gxc
x2 x | M1* | Separating and integrating to the form
kg
v2  gx
x | +c not required
2kg
v = 0 at x = a c2ga
a | M1dep* | Using given conditions to find c
2kg 2kg
v2  2gx2ga
x a
1 1
v2 2gk  2gax
 
x a | A1 | AG – so sufficient working must be
shown
10 | (c) | kmg
Remain at A if mg
a2
k

a2 | M1
A1
[2] | 3.1b
1.1 | Considering relationship between
attractive force and friction (accept
any inequality or equals)
cao | Allow x for a for the M
mark
A particle P, of mass $m$, moves on a rough horizontal table. P is attracted towards a fixed point O on the table by a force of magnitude $\frac{kmg}{x^2}$, where $x$ is the distance OP.

The coefficient of friction between P and the table is $\mu$.

P is initially projected in a direction directly away from O. The velocity of P is first zero at a point A which is a distance $a$ from O.

\begin{enumerate}[label=(\alph*)]
\item Show that the velocity $v$ of P, when P is moving away from O, satisfies the differential equation
$$\frac{\mathrm{d}}{\mathrm{d}x}(v^2) + \frac{2kg}{x^2} + 2\mu g = 0.$$ [3]

\item Verify that
$$v^2 = 2gk\left(\frac{1}{x} - \frac{1}{a}\right) + 2\mu g(a-x).$$ [3]

\item Find, in terms of $k$ and $a$, the range of values of $\mu$ for which P remains at A. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q10 [8]}}
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