One end of a light elastic string of natural length \(2.1\) m and modulus of elasticity \(4.8\) N is attached to a particle, \(P\), of mass \(1.75\) kg. The other end of the string is attached to a fixed point, \(O\), which is on a rough inclined plane. The angle between the plane and the horizontal is \(\theta\) where \(\sin\theta = \frac{3}{5}\). The coefficient of friction between \(P\) and the plane is \(0.732\).
Particle \(P\) is placed on the plane at \(O\) and then projected down a line of greatest slope of the plane with an initial speed of \(2.4\) m s\(^{-1}\).
Determine the distance that \(P\) has travelled from \(O\) at the instant when it first comes to rest. You can assume that during its motion \(P\) does not reach the bottom of the inclined plane. [8]
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Question 8:
Answer Marks
Guidance
8 Initial KE = ½1.752.42 J
B1
Suppose the distance is d m.
Answer Marks
Guidance
Initial PE = 1.75gdsin J M1
3.1b
use mgh with h different from d
4 3
𝑐𝑜𝑠𝜃 = ,𝑠𝑖𝑛𝜃 =
Answer Marks
5 5 d is the distance moved along the
plane. h is the vertical distance
between O and the point where P
stops.
EPE/Total energy when P stops =
Answer Marks
Guidance
4.8(d – 2.1)2/(22.1) J M1
3.4
Correct use of (but could be
2 l
Answer Marks
in terms of x rather than d). 8(d – 2.1)2 / 7 or 8x2 / 7
C = 1.75gcos => F = 0.7321.75gcos
Answer Marks
Guidance
r B1
1.1
an expression for the frictional
force
WD against friction = F d J
Answer Marks
Guidance
r M1
3.4
expression for the energy lost 1.0248gd or 10.04304d
Do not allow sin cos interchange
Answer Marks
Guidance
5.04 + 10.29d = 8(d – 2.1)2/7 + 10.04304d M1
3.4
involving initial energy, final
energy and energy loss on the
correct side, signs correct, in
terms of one correct unknown
distance. Do not allow sin/cos
Answer Marks
interchange Equation could be in terms of
extension, x:
5.04 + 10.29(x + 2.1)= 8x2/7 +
10.04304(x +2.1)
ALT method
Candidates may consider interim
energy at 2.1m (giving
𝑣2 =6.3527). Can award B1 B1 for
the initial KE and frictional force as
per the main scheme. All method
marks awarded as per the main
scheme for energy consideration
involving a correct algebraic
distance by considering motion
from 2.1m to the point the particle
comes to rest.
KE (d=2.1) +PE(d=2.1) = EPE
stored + WD against Fr
5.558616 + 10.29x= 8x2/7 +
10.04304x
Or
5.558616 = 8x2/7 – 0.24696x
d(d – 4.41609) = 0 => (d = 0) or d = awrt 4.42
Answer Marks
Guidance
so distance travelled must be awrt 4.42 m A1
1.1
correct root seen) 8x2 – 1.72872x – 38.910312 = 0
(x + 2.1)(8x – 18.52872) = 0
x = –2.1 or awrt 2.32
So: awrt 2.1 + 2.32 = 4.42 m
Answer Marks
Guidance
Rejection of d = 0 (as start point) A1
3.2a
for correct roots only But x > 0 do reject x =-2.1m
[8]
PMT
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Copy
Question 8:
8 | Initial KE = ½1.752.42 J | B1 | 3.4 | 5.04 J
Suppose the distance is d m.
Initial PE = 1.75gdsin J | M1 | 3.1b | 1.05gd or 10.29d. Attempt to
use mgh with h different from d
4 3
𝑐𝑜𝑠𝜃 = ,𝑠𝑖𝑛𝜃 =
5 5 | d is the distance moved along the
plane. h is the vertical distance
between O and the point where P
stops.
EPE/Total energy when P stops =
4.8(d – 2.1)2/(22.1) J | M1 | 3.4 | 2 x
Correct use of (but could be
2 l
in terms of x rather than d). | 8(d – 2.1)2 / 7 or 8x2 / 7
C = 1.75gcos => F = 0.7321.75gcos
r | B1 | 1.1 | Using law of friction to derive
an expression for the frictional
force
WD against friction = F d J
r | M1 | 3.4 | Using WD = Fd to find an
expression for the energy lost | 1.0248gd or 10.04304d
Do not allow sin cos interchange
5.04 + 10.29d = 8(d – 2.1)2/7 + 10.04304d | M1 | 3.4 | Energy budget equation
involving initial energy, final
energy and energy loss on the
correct side, signs correct, in
terms of one correct unknown
distance. Do not allow sin/cos
interchange | Equation could be in terms of
extension, x:
5.04 + 10.29(x + 2.1)= 8x2/7 +
10.04304(x +2.1)
ALT method
Candidates may consider interim
energy at 2.1m (giving
𝑣2 =6.3527). Can award B1 B1 for
the initial KE and frictional force as
per the main scheme. All method
marks awarded as per the main
scheme for energy consideration
involving a correct algebraic
distance by considering motion
from 2.1m to the point the particle
comes to rest.
KE (d=2.1) +PE(d=2.1) = EPE
stored + WD against Fr
5.558616 + 10.29x= 8x2/7 +
10.04304x
Or
5.558616 = 8x2/7 – 0.24696x
d(d – 4.41609) = 0 => (d = 0) or d = awrt 4.42
so distance travelled must be awrt 4.42 m | A1 | 1.1 | Could be BC (at least one
correct root seen) | 8x2 – 1.72872x – 38.910312 = 0
(x + 2.1)(8x – 18.52872) = 0
x = –2.1 or awrt 2.32
So: awrt 2.1 + 2.32 = 4.42 m
Rejection of d = 0 (as start point) | A1 | 3.2a | explicit rejection must be seen,
for correct roots only | But x > 0 do reject x =-2.1m
[8]
PMT
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support@ocr.org.uk
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ocr.org.uk
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/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
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One end of a light elastic string of natural length $2.1$ m and modulus of elasticity $4.8$ N is attached to a particle, $P$, of mass $1.75$ kg. The other end of the string is attached to a fixed point, $O$, which is on a rough inclined plane. The angle between the plane and the horizontal is $\theta$ where $\sin\theta = \frac{3}{5}$. The coefficient of friction between $P$ and the plane is $0.732$.
Particle $P$ is placed on the plane at $O$ and then projected down a line of greatest slope of the plane with an initial speed of $2.4$ m s$^{-1}$.
Determine the distance that $P$ has travelled from $O$ at the instant when it first comes to rest. You can assume that during its motion $P$ does not reach the bottom of the inclined plane. [8]
\hfill \mbox{\textit{OCR Further Mechanics 2023 Q8 [8]}}