Standard +0.3 This is a straightforward energy conservation problem with elastic strings. Students need to equate gravitational PE lost to elastic PE gained at the lowest point, then solve a quadratic equation. The setup is standard and the method is well-practiced in Further Mechanics, requiring only routine application of Hooke's law and energy principles with no novel insight.
A particle P of mass \(0.5\) kg is attached to a fixed point O by a light elastic string of natural length \(3\) m and modulus of elasticity \(75\) N.
P is released from rest at O and is allowed to fall freely.
Determine the length of the string when P is at its lowest point in the subsequent motion. [5]
(Provided the string has not been stretched beyond its
Answer
Marks
Guidance
elastic limit the) length of string is 4.30 (m)
A1
2.2b
[5]
Question 1:
1 | (±)0.5g(3+x)
At lowest point, PE = | B1 | 1.1 | Correct use of PE | x is the extension in
the string
75x2
EPE =
2(3) | B1 | 1.1 | λx2
Correct use of
2a
75x2
−0.5g(3+x)=0
6 | M1 | 3.3 | Use of conservation of energy (correct
number of terms)
75x2 −29.4x−88.2=0
x=1.29... (because x≠−0.90...) as x > 0 | A1 | 2.3 | BC – as a minimum must state x > 0 if no
explicit rejection of –0.906… seen | x = 1.298005445…
or x = –0.906005…
(Provided the string has not been stretched beyond its
elastic limit the) length of string is 4.30 (m) | A1 | 2.2b | Awrt 4.30
[5]
A particle P of mass $0.5$ kg is attached to a fixed point O by a light elastic string of natural length $3$ m and modulus of elasticity $75$ N.
P is released from rest at O and is allowed to fall freely.
Determine the length of the string when P is at its lowest point in the subsequent motion. [5]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q1 [5]}}