Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and the condition for slack string (tension = 0). Part (i) involves setting up energy and centripetal force equations at the slack point, requiring careful angle work with the given tan α = 15/8. Part (ii) requires finding where maximum tension occurs (lowest point) and applying T = mg + mv²/r. The multi-step reasoning, coordinate geometry with the specific angle, and integration of multiple mechanics principles place this well above average difficulty, though it follows a recognizable framework for conical pendulum problems.
\includegraphics{figure_3}
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 point \(A\) is such that \(OA = a\) and \(OA\) makes an angle \(\alpha\) with the upward vertical, where \(\tan \alpha = \frac{15}{8}\). The particle is projected downwards from \(A\) with speed \(u\) perpendicular to the string and moves in a vertical plane (see diagram). The string becomes slack after the string has rotated through \(270°\) from its initial position, with the particle now at the point \(B\).
\begin{enumerate}[label=(\roman*)]
\item Show that \(u^2 = 2ag\).
[5]
\item Find the maximum tension in the string as the particle moves from \(A\) to \(B\).
[4]
\end{enumerate]
Find speed v at B by conservation of energy (A0 if no m)
B
[sin α = 12/13, cos α = 5/13]
[T =] mv 2/a – mg sin α = 0 [v 2 = (12/13) ag]
Answer
Marks
Guidance
B B B
B1
Equate tension T at B to zero by using F = ma radially
B
Answer
Marks
Guidance
u2 = (3 sin α – 2 cos α) ag
M1
Combine to verify u2
= (36/13 – 10/13) ag = 2 ag AG
A1
5
Answer
Marks
Guidance
Question
Answer
Marks
Answer
Marks
3(ii)
EITHER: ½mv 2 = ½mu2 + mga (1 + cos α)
C
or ½mv 2 + mga (1 + sin α) [v 2 = 62ag/13]
Answer
Marks
Guidance
B C
M1
Find v 2 at lowest point C by conservation of energy
C
T = mv 2/a + mg
Answer
Marks
Guidance
max C
B1
Find tension T at lowest point from F = ma radially
max
Answer
Marks
Guidance
= 62mg/13 + mg
M1
Combine to find T
max
Answer
Marks
= 75mg/13 or 5.77 mg or 57.7 m
A1
OR: ½mV2 = ½mu2 + mga (cos α + cos θ )
Answer
Marks
Guidance
[V2 = 2ag (18/13 + cos θ )]
(M1
Find V2 at general point by conservation of energy
(where OP is e.g. at θ to downward vertical)
Answer
Marks
Guidance
T = mV2/a + mg cos θ = (36/13 + 3 cos θ ) mg
B1
Find tension T at general point from F = ma radially
T = (36/13 + 3) mg
Answer
Marks
Guidance
max
M1
Combine to find T at lowest point where θ = 0
max
Answer
Marks
= 75mg/13 or 5.77 mg or 57.7 m
A1)
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 3:
--- 3(i) ---
3(i) | ½mv 2 = ½mu2 – mga (sin α – cos α)
B
[v 2 = u2 – (14/13) ag]
B | M1A1 | Find speed v at B by conservation of energy (A0 if no m)
B
[sin α = 12/13, cos α = 5/13]
[T =] mv 2/a – mg sin α = 0 [v 2 = (12/13) ag]
B B B | B1 | Equate tension T at B to zero by using F = ma radially
B
u2 = (3 sin α – 2 cos α) ag | M1 | Combine to verify u2
= (36/13 – 10/13) ag = 2 ag AG | A1
5
Question | Answer | Marks | Guidance
--- 3(ii) ---
3(ii) | EITHER: ½mv 2 = ½mu2 + mga (1 + cos α)
C
or ½mv 2 + mga (1 + sin α) [v 2 = 62ag/13]
B C | M1 | Find v 2 at lowest point C by conservation of energy
C
T = mv 2/a + mg
max C | B1 | Find tension T at lowest point from F = ma radially
max
= 62mg/13 + mg | M1 | Combine to find T
max
= 75mg/13 or 5.77 mg or 57.7 m | A1
OR: ½mV2 = ½mu2 + mga (cos α + cos θ )
[V2 = 2ag (18/13 + cos θ )] | (M1 | Find V2 at general point by conservation of energy
(where OP is e.g. at θ to downward vertical)
T = mV2/a + mg cos θ = (36/13 + 3 cos θ ) mg | B1 | Find tension T at general point from F = ma radially
T = (36/13 + 3) mg
max | M1 | Combine to find T at lowest point where θ = 0
max
= 75mg/13 or 5.77 mg or 57.7 m | A1)
4
Question | Answer | Marks | Guidance
\includegraphics{figure_3}
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 point $A$ is such that $OA = a$ and $OA$ makes an angle $\alpha$ with the upward vertical, where $\tan \alpha = \frac{15}{8}$. The particle is projected downwards from $A$ with speed $u$ perpendicular to the string and moves in a vertical plane (see diagram). The string becomes slack after the string has rotated through $270°$ from its initial position, with the particle now at the point $B$.
\begin{enumerate}[label=(\roman*)]
\item Show that $u^2 = 2ag$.
[5]
\item Find the maximum tension in the string as the particle moves from $A$ to $B$.
[4]
\end{enumerate]
\hfill \mbox{\textit{CAIE FP2 2018 Q3 [9]}}