OCR FM1 AS (Further Mechanics 1 AS) 2021 June

Question 1 7 marks

1 A particle \(A\) of mass 3.6 kg is attached by a light inextensible string to a particle \(B\) of mass 2.4 kg . \(A\) and \(B\) are initially at rest, with the string slack, on a smooth horizontal surface. \(A\) is projected directly away from \(B\) with a speed of \(7.2 \mathrm {~ms} ^ { - 1 }\).

Calculate the speed of \(A\) after the string becomes taut.
Find the impulse exerted on \(A\) at the instant that the string becomes taut.
Find the loss in kinetic energy as a result of the string becoming taut.

Question 2 11 marks

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2 A car of mass 1500 kg has an engine with maximum power 60 kW . When the car is travelling at \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along a straight horizontal road using maximum power, its acceleration is \(3.3 \mathrm {~ms} ^ { - 2 }\). In an initial model of the motion of the car it is assumed that the resistance to motion is constant.

Using this initial model, find the greatest possible steady speed of the car along the road. In a refined model the resistance to motion is assumed to be proportional to the speed of the car.
Using this refined model, find the greatest possible steady speed of the car along the road. The greatest possible steady speed of the car on the road is measured and found to be \(21.6 \mathrm {~ms} ^ { - 1 }\).
Explain what this value means about the models used in parts (a) and (b). \includegraphics[max width=\textwidth, alt={}, center]{aa25b8a6-9a5a-4de2-9534-18db8a175c34-03_583_378_169_255} As shown in the diagram, \(A B\) is a long thin rod which is fixed vertically with \(A\) above \(B\). One end of a light inextensible string of length 1 m is attached to \(A\) and the other end is attached to a particle \(P\) of mass \(m _ { 1 } \mathrm {~kg}\). One end of another light inextensible string of length 1 m is also attached to \(P\). Its other end is attached to a small smooth ring \(R\), of mass \(m _ { 2 } \mathrm {~kg}\), which is free to move on \(A B\). Initially, \(P\) moves in a horizontal circle of radius 0.6 m with constant angular velocity \(\omega \mathrm { rads } ^ { - 1 }\). The magnitude of the tension in string \(A P\) is denoted by \(T _ { 1 } \mathrm {~N}\) while that in string \(P R\) is denoted by \(T _ { 2 } \mathrm {~N}\).
1. By considering forces on \(R\), express \(T _ { 2 }\) in terms of \(m _ { 2 }\).
2. Show that
  1. \(T _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right)\),
  2. \(\omega ^ { 2 } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } }\).
  3. Deduce that, in the case where \(m _ { 1 }\) is much bigger than \(m _ { 2 } , \omega \approx 3.5\). In a different case, where \(m _ { 1 } = 2.5\) and \(m _ { 2 } = 2.8 , P\) slows down. Eventually the system comes to rest with \(P\) and \(R\) hanging in equilibrium.
3. Find the total energy lost by \(P\) and \(R\) as the angular velocity of \(P\) changes from the initial value of \(\omega \mathrm { rads } ^ { - 1 }\) to zero.

Question 4 14 marks

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4
2.4
&

B1 for each of two correct statements about the models.

If commenting on the accuracy of (a), must emphasise that (a) is very inaccurate or at least quite inaccurate

Do not allow e.g.

- model (a) is not very effective

- Neither model is accurate

- (a) and (b) are not very accurate

Clear comparison between the accuracy of the two models (must emphasise that (b) is fairly accurate or considerably more accurate than (a)), or other suitable distinct second comment

Do not allow e.g.

- model (b) is more accurate than model (a)

- (b) is not accurate

Do not allow statement claiming that resistance is proportional to speed, or to speed \({ } ^ { 2 }\)

Suitable comments for (a):

- is very inaccurate

- predicted speed is nearly three times the actual value

- constant resistance is not a suitable model

- both models underestimate the resistance (as top speed is lower than expected)

For the linear model (b)

- is fairly accurate (but probably underestimates the resistance at higher speeds)

- resistance is not proportional to speed but is a much better model than constant resistance

(a)

\(T _ { 2 } \cos \theta = m _ { 2 } g\) \(T _ { 2 } = \frac { m _ { 2 } \times 9.8 } { 0.8 } = 12.25 m _ { 2 }\)

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1.1a

1.1

Resolving \(T _ { 2 }\) vertically and balancing forces on \(R\)

Do not allow extra forces present

Allow use of g, e.g. \(\frac { 5 } { 4 } g m _ { 2 }\)

In this solution \(\theta\) is the angle between \(R P\) and \(R A\) Sin may be seen instead if \(\theta\) is measured horizontally.

Do not allow incomplete expressions e.g. \(\frac { m _ { 2 } g } { \sin 53.13 }\)

(b)

(i)

\(\begin{aligned}

T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta

T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } =

\qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\)

[2]

3.1b

2.1

Vertical forces on \(P\); 3 terms including resolving of \(T _ { 1 }\); allow sign error

AG Dividing by \(\cos \theta ( = 0.8 )\), substituting their \(T _ { 2 }\) and rearranging

Allow 12.25 instead of \(\frac { 49 } { 4 }\)

Or \(T _ { 1 } \cos \theta = m _ { 1 } g + m _ { 2 } g\) (equation for the system as a whole)

At least one intermediate step must be seen

(b)

(ii)

\(\begin{aligned}

T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a

12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 }

\omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\)

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3.1b

1.1

2.1

NII horizontally for \(P ; 3\) terms including resolving of tensions; allow sign error

Substituting for \(T _ { 1 }\), their \(T _ { 2 } , \sin \theta\) and \(\alpha\)

AG Must see an intermediate step

Could see \(a\) or \(0.6 \omega ^ { 2 }\) or \(\frac { v ^ { 2 } } { 0.6 }\) or \(\omega ^ { 2 } r\) or \(\frac { v ^ { 2 } } { r } \sin \theta = 0.6\)

must be \(a = 0.6 \omega ^ { 2 }\)

(c)

\(\begin{aligned}

\text { E.g } m _ { 1 } \gg m _ { 2 } \Rightarrow \frac { 2 m _ { 2 } } { m _ { 1 } } \approx 0 \text { or } \frac { 49 m _ { 2 } } { 4 m _ { 1 } } \approx 0

\omega \approx \sqrt { \frac { 49 m } { 4 m } } = 3.5 \end{aligned}\)

M1 A1

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1.1

Allow argument such as if \(m _ { 1 } \gg m _ { 2 }\) then \(m _ { 1 } + 2 m _ { 2 } \approx m _ { 1 }\)

AG \(m\) may be missing

SC1 for result following argument that \(m _ { 2 }\) is negligible (by comparison with \(m _ { 1 }\) ) without justification, or using trial values of \(m _ { 1 }\) and \(m _ { 2 }\) with \(m _ { 1 } \gg m _ { 2 }\).

Do not allow the assumption that \(m _ { 2 } = 0\)

If using trial values, \(m _ { 1 }\) must be at least \(70 \times m _ { 2 }\) to give \(\omega = 3.5\) to 1 dp .

\multirow{3}{*}{(d)}

\(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\)

Final energy \(= 2.5 \times g \times 1\) \(\text { Initial } \mathrm { KE } = \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\)

Initial PE \(= 2.5 \times g \times 1.2 + 2.8 \times g \times 0.4\)

Energy loss \(= 17.8605 + 40.376 - 24.5 = 33.7365\)

1.2

1.1

3.2a

Use of \(v = r \omega\) with values for \(m _ { 1 }\) and \(m _ { 2 }\)

(Assuming zero PE level at 2 m below \(A\); other values possible)

Do not allow use of \(\omega = 3.5\)

oe with different zero PE level awrt 33.7

( \(v = 3.78 , v ^ { 2 } = 14.2884\) )

NB \(\omega = 6.3\) (24.5)

(17.8605)

(40.376)

Alternate method \(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\)

Initial KE \(= \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\)

\(\triangle P E\) for \(m _ { 1 } = \pm 2.5 \times 9.8 \times ( 0.8 - 1 )\)

\(\triangle P E\) for \(m _ { 2 } = \pm 2.8 \times 9.8 ( 1.6 - 2 )\)

Energy loss \(= 17.8605 + 4.9 + 10.976\)

Use of \(v = r \omega\) with values for \(m _ { 1 }\) and \(m _ { 2 }\)

Or \(- \triangle P E\) \(= 2.5 \times 9.8 \times 0.2 + 2.8 \times 9.8 \times 0.4\)

awrt 33.7

( \(v = 3.78 , v ^ { 2 } = 14.2884\) ) \(\mathrm { NB } \omega = 6.3\)

(17.8605)

\(( \pm 4.9 )\)

\(( \pm 10.976 )\)

\(( \pm 15.876 )\)

Or 15.876 + 17.8605

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