Edexcel FM2 AS (Further Mechanics 2 AS) 2023 June

1. Three particles of masses \(4 m , 2 m\) and \(k m\) are placed at the points with coordinates \(( - 3 , - 1 ) , ( 6,1 )\) and \(( - 1,5 )\) respectively.

Given that the centre of mass of the three particles is at the point with coordinates \(( \bar { x } , \bar { y } )\)

1. show that \(\bar { x } = \frac { - k } { k + 6 }\)
2. find \(\bar { y }\) in terms of \(k\). Given that the centre of mass of the three particles lies on the line with equation \(y = 2 x + 3\)
3. find the value of \(k\). A fourth particle is placed at the point with coordinates \(( \lambda , 4 )\).
   Given that the centre of mass of the four particles also lies on the line with equation \(y = 2 x + 3\)
4. find the value of \(\lambda\).

1. A particle \(P\) is moving along the \(x\)-axis.

At time \(t\) seconds, \(t \geqslant 0 , P\) has acceleration \(a \mathrm {~ms} ^ { - 2 }\) and velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction of \(x\) increasing, where $$v = \mathrm { e } ^ { 2 t } + 6 \mathrm { e } ^ { t } - k t$$ and \(k\) is a positive constant.
When \(t = \ln 2\), \(a = 0\)

1. Find the value of \(k\). When \(t = 0\), the particle passes through the fixed point \(A\).
   When \(t = \ln 2\), the particle is \(d\) metres from \(A\).
2. Showing all stages of your working, find the value of \(d\) correct to 2 significant figures.
   [0pt] [Solutions relying entirely on calculator technology are not acceptable.]

1. A girl is cycling round a circular track.

The girl and her bicycle have a combined mass of 55 kg .
The coefficient of friction between the track surface and the tyres of the bicycle is \(\mu\).
The track is banked at an angle of \(15 ^ { \circ }\) to the horizontal.
The girl and her bicycle are modelled as a particle moving in a horizontal circle of radius 50 m
The minimum speed at which the girl can cycle round this circle without slipping is \(4.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) Using the model, find the value of \(\mu\).

4. \begin{figure}[h] \end{figure} A uniform triangular lamina \(A B C\) is isosceles, with \(A C = B C\). The midpoint of \(A B\) is \(M\). The length of \(A B\) is \(18 a\) and the length of \(C M\) is \(18 a\). The triangular lamina \(C D E\), with \(D E = 6 a\) and \(C D = 12 a\), has \(E D\) parallel to \(A B\) and \(M D C\) is a straight line. Triangle \(C D E\) is removed from triangle \(A B C\) to form the lamina \(L\), shown shaded in Figure 1. The distance of the centre of mass of \(L\) from \(M C\) is \(d\).

1. Show that \(d = \frac { 4 } { 7 } a\) The lamina \(L\) is suspended by two light inextensible strings. One string is attached to \(L\) at \(A\) and the other string is attached to \(L\) at \(B\).
   The lamina hangs in equilibrium in a vertical plane with the strings vertical and \(A B\) horizontal.
   The weight of \(L\) is \(W\)
2. Find, in terms of \(W\), the tension in the string attached to \(L\) at \(B\) The string attached to \(L\) at \(B\) breaks, so that \(L\) is now suspended from \(A\). When \(L\) is hanging in equilibrium in a vertical plane, the angle between \(A B\) and the downward vertical through \(A\) is \(\theta ^ { \circ }\)
3. Find the value of \(\theta\)