Challenging +1.2 This is a standard energy conservation problem in mechanics involving elastic potential energy and gravitational potential energy on an inclined plane. While it requires careful bookkeeping of three energy equations at different positions and algebraic manipulation to eliminate V and solve for k, the approach is straightforward and follows a well-practiced method. The inclusion of sin θ = 1/4 simplifies calculations. This is moderately above average difficulty due to the multi-step algebra and need to track extensions/compressions carefully, but remains a standard Further Maths mechanics question without requiring novel insight.
\includegraphics{figure_4}
A light spring of natural length \(a\) and modulus of elasticity \(kmg\) is attached to a fixed point \(O\) on a smooth plane inclined to the horizontal at an angle \(\theta\), where \(\sin\theta = \frac{1}{4}\). A particle of mass \(m\) is attached to the lower end of the spring and is held at the point \(A\) on the plane, where \(OA = 2a\) and \(OA\) is along a line of greatest slope of the plane (see diagram).
The particle is released from rest and is moving with speed \(V\) when it passes through the point \(B\) on the plane, where \(OB = \frac{3}{2}a\). The speed of the particle is \(\frac{1}{3}V\) when it passes through the point \(C\) on the plane, where \(OC = \frac{3}{4}a\).
Find the value of \(k\). [7]
Eliminate V2 from two energy equations to obtain expression involving only k, a
Answer
Marks
Guidance
and possibly sin
M1
At least one of the energy equations must have
scored M1.
9
k
Answer
Marks
4
A1
7
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | Consider one situation:
kmg 1 2 3
A to B: Loss in EPE = a2 a = kmga
2a 2 8
| B1 | Accept unsimplified.
1 mga 3 3
Energy: mV2 sin kmga ( V2 gak1
2 2 8 4 | M1A1 | KE, GPE, EPE terms present.
Must be dimensionally correct.
Must include sin or cos in GPE.
Consider a second situation:
kmg a 2 15
A to C: Loss in EPE = a2 = kmga
2a 4 32
| B1 | Accept unsimplified.
1 1 2 mg5a 15 15
Energy: m V sin kmga V2 agk2
2 2 4 32 4 | M1 | KE, GPE, EPE terms present.
Must be dimensionally correct.
Must include sin or cos in GPE.
Third possible situation:
kmg a 2 a 2 3
B to C : Loss in EPE = = kmga
2a 2 4 32
| (B1) | This may be used in combination with either of
the first two situations.
Mark to the candidate’s benefit,
1 1 2 1 1 2 mg3a 3 1
Energy: m V m V sin kmga V2 ag6k
2 2 2 4 4 32 4 | (M1) | KE, GPE, EPE terms present.
Must be dimensionally correct .
Must include sin or cos in GPE.
Eliminate V2 from two energy equations to obtain expression involving only k, a
and possibly sin | M1 | At least one of the energy equations must have
scored M1.
9
k
4 | A1
7
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
A light spring of natural length $a$ and modulus of elasticity $kmg$ is attached to a fixed point $O$ on a smooth plane inclined to the horizontal at an angle $\theta$, where $\sin\theta = \frac{1}{4}$. A particle of mass $m$ is attached to the lower end of the spring and is held at the point $A$ on the plane, where $OA = 2a$ and $OA$ is along a line of greatest slope of the plane (see diagram).
The particle is released from rest and is moving with speed $V$ when it passes through the point $B$ on the plane, where $OB = \frac{3}{2}a$. The speed of the particle is $\frac{1}{3}V$ when it passes through the point $C$ on the plane, where $OC = \frac{3}{4}a$.
Find the value of $k$. [7]
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q4 [7]}}