| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Ratio of tensions/forces |
| Difficulty | Standard +0.8 This is a standard Further Maths circular motion problem requiring energy conservation and Newton's second law in two positions, followed by finding minimum tension. While it involves multiple steps and careful algebraic manipulation with the tension condition, the techniques are well-established for FM students. The given cos α = 4/5 simplifies calculations. More routine than proof-based FM questions but requires solid understanding of vertical circular motion. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks |
|---|---|
| 5(i) | ½mv2 = ½mu2 + mga (cos α + sin α) |
| Answer | Marks | Guidance |
|---|---|---|
| = u2 + 14 ag/5 AG | M1 A1 | Verify v by conservation of energy (A0 if no m) |
| Total: | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | T = mu2/a – mg cos α | |
| P | B1 | Find tension T at P by using F = ma |
| Answer | Marks | Guidance |
|---|---|---|
| Q | B1 | Find tension T at Q by using F = ma |
| Answer | Marks | Guidance |
|---|---|---|
| v2 = 2u2 – ag(2 cos α + sin α) = 2u 2 – 11ag/5 (AEF) | M1 A1 | Relate u2, v2 using T = 2T (A0 if reqd. eqn omitted) |
| Answer | Marks | Guidance |
|---|---|---|
| u = √(5ag) or 2⋅24√(ag) [v2 = 39ag/5] | A1 | Eliminate v2 to find u |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(iii) | T + mg = mV 2/a | |
| min | B1 | Find tension T at top from F = ma radially |
| Answer | Marks | Guidance |
|---|---|---|
| or ½mv2 – mga (1 + sin α) | M1 | Find V 2 at top by conservation of energy (A0 if no m) |
| Answer | Marks | Guidance |
|---|---|---|
| min | A1 | Combine to find T |
| Answer | Marks |
|---|---|
| Total: | 3 |
Question 5:
--- 5(i) ---
5(i) | ½mv2 = ½mu2 + mga (cos α + sin α)
v2 = u2 + 2 ag (4/5 + 3/5)
= u2 + 14 ag/5 AG | M1 A1 | Verify v by conservation of energy (A0 if no m)
Total: | 2
--- 5(ii) ---
5(ii) | T = mu2/a – mg cos α
P | B1 | Find tension T at P by using F = ma
P
T = mv2/a + mg sin α
Q | B1 | Find tension T at Q by using F = ma
Q
v2 = 2u2 – ag(2 cos α + sin α) = 2u 2 – 11ag/5 (AEF) | M1 A1 | Relate u2, v2 using T = 2T (A0 if reqd. eqn omitted)
Q P
u = √(5ag) or 2⋅24√(ag) [v2 = 39ag/5] | A1 | Eliminate v2 to find u
Total: | 5
--- 5(iii) ---
5(iii) | T + mg = mV 2/a
min | B1 | Find tension T at top from F = ma radially
min
½mV2 = ½mu 2 – mga (1 – cos α)
or ½mv2 – mga (1 + sin α) | M1 | Find V 2 at top by conservation of energy (A0 if no m)
[V2 = (5 – 2/5)ag = (39/5 – 16/5)ag = 23ag/5]
T = 23mg/5 – mg = 18mg/5 or 3⋅6mg
min | A1 | Combine to find T
min
Total: | 3\includegraphics{figure_5}
A particle of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle is moving in complete vertical circles with the string taut. When the particle is at the point $P$, where $OP$ makes an angle $\alpha$ with the upward vertical through $O$, its speed is $u$. When the particle is at the point $Q$, where angle $QOP = 90°$, its speed is $v$ (see diagram). It is given that $\cos \alpha = \frac{4}{5}$.
\begin{enumerate}[label=(\roman*)]
\item Show that $v^2 = u^2 + \frac{14}{5}ag$. [2]
\end{enumerate}
The tension in the string when the particle is at $Q$ is twice the tension in the string when the particle is at $P$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Obtain another equation relating $u^2$, $v^2$, $a$ and $g$, and hence find $u$ in terms of $a$ and $g$. [5]
\item Find the least tension in the string during the motion. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q5 [10]}}