Standard +0.3 This is a standard elastic energy conservation problem requiring equilibrium position calculation and energy equation setup. The modulus is given as a simple multiple of mg, making the algebra straightforward. It's slightly above average difficulty due to being Further Maths content and requiring careful energy bookkeeping, but follows a well-practiced method with no novel insights needed.
A particle \(P\) of mass \(m\) is attached to one end of a light elastic spring of natural length \(a\) and modulus of elasticity \(5mg\). The other end of the spring is attached to a fixed point \(O\). The spring hangs vertically with \(P\) below \(O\). The particle \(P\) is pulled down vertically and released from rest when the length of the spring is \(\frac{7}{5}a\).
Find the distance of \(P\) below \(O\) when \(P\) first comes to instantaneous rest. [4]
Extension when P comes to rest is x, EPE loss = −x2
2 a 2
Answer
Marks
Guidance
B1
Both terms seen.
1 5mg a 2 a
Energy: −x2=mg −x
2 a 2 2
Answer
Marks
Guidance
M1
At least one EPE term and a GPE term,
dimensionally correct.
Answer
Marks
Guidance
Solve: 20x2 −8ax−a2 =0
M1
Obtain homogeneous quadratic equation in x and a
Must come from an energy equation involving two
EPE terms.
Note that the correct case simplifies to a linear
5mga
equation +x=mg and this scores M1.
2a 2
a 9
x=− , so distance of P below O is a
Answer
Marks
Guidance
10 10
A1
Question
Answer
Marks
2
Alternative method for question 2
Distance of P below O when it comes to rest comes to rest is h
1 5mg a 2
EPE loss = = −(h−a)2
2 a 2
Answer
Marks
B1
1 5mg a 2 3
Energy: −(h−a)2 =mg a−h
2 a 2 2
Answer
Marks
Guidance
M1
At least one EPE term and a GPE term,
dimensionally correct.
Answer
Marks
Guidance
20h2 −48ah+27a2 =0
M1
Obtain homogeneous quadratic/linear equation in
h and a. Must come from an energy equation
involving two EPE terms.
9
(10h−9a)( 2h−3a)=0, h= a
Answer
Marks
10
A1
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 2:
2 | 1 5mg a 2
Extension when P comes to rest is x, EPE loss = −x2
2 a 2
| B1 | Both terms seen.
1 5mg a 2 a
Energy: −x2=mg −x
2 a 2 2
| M1 | At least one EPE term and a GPE term,
dimensionally correct.
Solve: 20x2 −8ax−a2 =0 | M1 | Obtain homogeneous quadratic equation in x and a
Must come from an energy equation involving two
EPE terms.
Note that the correct case simplifies to a linear
5mga
equation +x=mg and this scores M1.
2a 2
a 9
x=− , so distance of P below O is a
10 10 | A1
Question | Answer | Marks | Guidance
2 | Alternative method for question 2
Distance of P below O when it comes to rest comes to rest is h
1 5mg a 2
EPE loss = = −(h−a)2
2 a 2
| B1
1 5mg a 2 3
Energy: −(h−a)2 =mg a−h
2 a 2 2
| M1 | At least one EPE term and a GPE term,
dimensionally correct.
20h2 −48ah+27a2 =0 | M1 | Obtain homogeneous quadratic/linear equation in
h and a. Must come from an energy equation
involving two EPE terms.
9
(10h−9a)( 2h−3a)=0, h= a
10 | A1
4
Question | Answer | Marks | Guidance
A particle $P$ of mass $m$ is attached to one end of a light elastic spring of natural length $a$ and modulus of elasticity $5mg$. The other end of the spring is attached to a fixed point $O$. The spring hangs vertically with $P$ below $O$. The particle $P$ is pulled down vertically and released from rest when the length of the spring is $\frac{7}{5}a$.
Find the distance of $P$ below $O$ when $P$ first comes to instantaneous rest. [4]
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q2 [4]}}