| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Particle in circular tube or on wire |
| Difficulty | Challenging +1.2 This is a multi-stage mechanics problem requiring circular motion analysis (normal reaction with centripetal force) and work-energy principles across collision and resistance phases. Part (a) is standard circular motion; part (b) requires systematic application of coefficient of restitution, work-energy theorem, and algebraic manipulation to reach a given result. While involving several steps, the techniques are standard Further Mechanics fare without requiring novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (a) | 1 |
| Answer | Marks |
|---|---|
| | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Use of conservation of energy – correct |
| Answer | Marks |
|---|---|
| Substitute an expression for v 2 | Note that the reference |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (b) | 1 1 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| Answer | Marks |
|---|---|
| [6] | 3.4 |
| Answer | Marks |
|---|---|
| 2.2a | π |
| Answer | Marks |
|---|---|
| k need not be stated explicitly | Where v is the speed |
Question 10:
10 | (a) | 1
[At B,] KE = mu2 , PE = 0
2
1
[ At θ , ] KE = mv2 , PE = mga (1− cosθ )
2
1 1
mu2 = mv2 + mga (1 − cosθ )
2 2
mv2
R − mg cosθ =
a
R − mg cosθ = m ( u2 − 2ga (1 − cosθ ))
a
u2
R = m 3g cosθ − 2g +
a
| B1
B1
M1*
A1
M1*
M1dep*
A1
[7] | 1.1
1.1
3.3
1.1
3.3
3.4
1.1 | Use of conservation of energy – correct
number of terms
cao
N2L radially with correct number of
terms and weight resolved
Substitute an expression for v 2 | Note that the reference
level for zero GPE
might be taken at C
10 | (b) | 1 1
Before collision at C, mu2 = mv2 + mga
2 2
After collision at C, speed of P is e u2 − 2ga
( )
2
1 B 1 2
mv 2 = mga + m e u − 2ga
2 2
( )
v 2 = 2ga + e2 u2 − 2ga
B
1 1
mv 2 − mv 2 = Fb
2
B 2 A
( ( ))
m 2ga + e2 u2 − 2ga − 2bF ≥ 0
1
Fb ≤ mga + me2u2 − me2ga
2
1 ( )
⇒ Fb ≤ m e2u2 + 2 1 − e2 ga, so k = 2
2 | M1
A1
M1
M1
M1
A1
[6] | 3.4
1.1
3.1b
3.1b
2.5
2.2a | π
Substituting θ = into their
2
conservation of energy equation from (a)
Conservation of energy to find an
expression for the speed of P at B
Work-energy principle for motion
between B and A
Set v ≥ 0 and substitute for v 2
A B
k need not be stated explicitly | Where v is the speed
B
of P at B
\begin{enumerate}[label=(\alph*)]
\item Determine the magnitude of the normal reaction of the wire on P in terms of $m , g , a , u$ and $\theta$, when P is between B and C .
P collides with a fixed barrier at C . The coefficient of restitution between P and the fixed barrier is $e$. After this collision P moves back towards B .
On the straight portion BA , the motion of P is resisted by a constant horizontal force $F$.
\item Show that P will reach A if
$$F b \leqslant \frac { 1 } { 2 } m \left[ e ^ { 2 } u ^ { 2 } + k \left( 1 - e ^ { 2 } \right) g a \right] ,$$
where $k$ is an integer to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q10 [13]}}