Challenging +1.8 This is a challenging multi-part circular motion problem requiring energy conservation, collision mechanics with momentum conservation, and analysis of tension in circular motion. While the individual techniques are A-level standard (energy methods, momentum, resolving forces), the problem requires careful tracking through multiple stages, understanding the critical condition for completing a vertical circle (T≥0 at top), and applying these concepts in sequence. The collision adding complexity mid-problem and the final slack condition requiring solving a non-trivial equation elevate this above typical textbook exercises, though it remains within reach of well-prepared Further Maths students.
A particle \(P\) 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 \(O\). When \(P\) is hanging at rest vertically below \(O\), it is projected horizontally. In the subsequent motion \(P\) completes a vertical circle. The speed of \(P\) when it is at its highest point is \(u\). Show that the least possible value of \(u\) is \(\sqrt{(ag)}\). [2]
It is now given that \(u = \sqrt{(ag)}\). When \(P\) passes through the lowest point of its path, it collides with, and coalesces with, a stationary particle of mass \(\frac{1}{4}m\). Find the speed of the combined particle immediately after the collision. [4]
In the subsequent motion, when \(OP\) makes an angle \(\theta\) with the upward vertical the tension in the string is \(T\). Find an expression for \(T\) in terms of \(m\), \(g\) and \(\theta\). [5]
Find the value of \(\cos \theta\) when the string becomes slack. [2]
EITHER: Find tension at top from F = ma vertically:
T = mu 2 /a – mg B1
OR: Use energy at e.g. θ to upward vertical: ½ mv 2 = ½ mu 2
+ mga (1 – cos θ )
Find tension T by using F = ma radially: T′ = mv 2 /a – mg cos θ
Eliminate v 2 : = mu 2 /a + mg (2 – 3 cos θ )
Find T at top by taking θ = 0: T = mu 2 /a – mg (B1)
Find umin by requiring T ⩾ 0 at top [or T > 0]: u 2 /a – g ⩾ 0 so umin = √ag A.G. B1
Find v at bottom from conservation of energy: ½mv 2 = ½mu 2 + mg × 2a M1
v 2 = ag + 4ag, v = √(5ag) A1
Find new speed V from conservation of momentum:
m′V = mv with m′ = m + ¼m M1
V = 4v/5 = 4√(ag/5)
or (4/5) √(5ag) AEF A1
Find w 2 at angle θ from conservation of energy: ½ m′w 2 = ½ m′V 2
(condone m instead of m′ here since cancels out)
– m′ga (1 + cos θ ) M1 A1
[ w 2 = ag (6/5 – 2 cos θ )]
S.R. Invalid energy method (max 2/5): ½ m′w 2 = ½ mu 2
[gives T ′ = (5mg/4)(2 – 3 cos θ ) ] + mga (1 – cos θ )
– ¼mga (1 + cos θ ) (B1)
Find tension T ′ there by using F = ma radially: T ′ = m′w 2 /a – m′g cos θ B1
Eliminate w 2 : = m′V 2 /a – m′g (2 + 3 cos θ ) A1
Substitute for m′ and V: = (5mg/4)(6/5 – 3 cos θ )
AEF or 3mg/2 – (15/4) mg cos θ A1
Find cos θ when string becomes slack from T ′ = 0: cos θ = ⅓ × 6/5 = 2/5 or 0⋅4 M1 A1
Answer
Marks
S.R. Allow if found from T ′ = mg (6/5 – 3 cos θ )
2
4
5
Answer
Marks
2
13
Question 4:
4 | EITHER: Find tension at top from F = ma vertically:
T = mu 2 /a – mg B1
OR: Use energy at e.g. θ to upward vertical: ½ mv 2 = ½ mu 2
+ mga (1 – cos θ )
Find tension T by using F = ma radially: T′ = mv 2 /a – mg cos θ
Eliminate v 2 : = mu 2 /a + mg (2 – 3 cos θ )
Find T at top by taking θ = 0: T = mu 2 /a – mg (B1)
Find umin by requiring T ⩾ 0 at top [or T > 0]: u 2 /a – g ⩾ 0 so umin = √ag A.G. B1
Find v at bottom from conservation of energy: ½mv 2 = ½mu 2 + mg × 2a M1
v 2 = ag + 4ag, v = √(5ag) A1
Find new speed V from conservation of momentum:
m′V = mv with m′ = m + ¼m M1
V = 4v/5 = 4√(ag/5)
or (4/5) √(5ag) AEF A1
Find w 2 at angle θ from conservation of energy: ½ m′w 2 = ½ m′V 2
(condone m instead of m′ here since cancels out)
– m′ga (1 + cos θ ) M1 A1
[ w 2 = ag (6/5 – 2 cos θ )]
S.R. Invalid energy method (max 2/5): ½ m′w 2 = ½ mu 2
[gives T ′ = (5mg/4)(2 – 3 cos θ ) ] + mga (1 – cos θ )
– ¼mga (1 + cos θ ) (B1)
Find tension T ′ there by using F = ma radially: T ′ = m′w 2 /a – m′g cos θ B1
Eliminate w 2 : = m′V 2 /a – m′g (2 + 3 cos θ ) A1
Substitute for m′ and V: = (5mg/4)(6/5 – 3 cos θ )
AEF or 3mg/2 – (15/4) mg cos θ A1
Find cos θ when string becomes slack from T ′ = 0: cos θ = ⅓ × 6/5 = 2/5 or 0⋅4 M1 A1
S.R. Allow if found from T ′ = mg (6/5 – 3 cos θ ) | 2
4
5
2 | 13
A particle $P$ 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 $O$. When $P$ is hanging at rest vertically below $O$, it is projected horizontally. In the subsequent motion $P$ completes a vertical circle. The speed of $P$ when it is at its highest point is $u$. Show that the least possible value of $u$ is $\sqrt{(ag)}$. [2]
It is now given that $u = \sqrt{(ag)}$. When $P$ passes through the lowest point of its path, it collides with, and coalesces with, a stationary particle of mass $\frac{1}{4}m$. Find the speed of the combined particle immediately after the collision. [4]
In the subsequent motion, when $OP$ makes an angle $\theta$ with the upward vertical the tension in the string is $T$. Find an expression for $T$ in terms of $m$, $g$ and $\theta$. [5]
Find the value of $\cos \theta$ when the string becomes slack. [2]
\hfill \mbox{\textit{CAIE FP2 2015 Q4 [13]}}