SPS SPS ASFM Mechanics (SPS ASFM Mechanics) 2021 May

1. In this question you must show detailed reasoning.

The equation \(x ^ { 3 } + 3 x ^ { 2 } - 2 x + 4 = 0\) has roots \(\alpha , \beta\) and \(\gamma\).

1. Using the identity \(\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } \equiv ( \alpha + \beta + \gamma ) ^ { 3 } - 3 ( \alpha \beta + \beta \gamma + \gamma \alpha ) ( \alpha + \beta + \gamma ) + 3 \alpha \beta \gamma\) find the value of \(\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 }\).
2. Given that \(\alpha ^ { 3 } \beta ^ { 3 } + \beta ^ { 3 } \gamma ^ { 3 } + \gamma ^ { 3 } \alpha ^ { 3 } = 112\) find a cubic equation whose roots are \(\alpha ^ { 3 } , \beta ^ { 3 }\) and \(\gamma ^ { 3 }\).
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2. Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\), $$\mathrm { f } ( n ) = 2 ^ { 2 n - 1 } + 3 ^ { 2 n - 1 } \text { is divisible by } 5 .$$ [BLANK PAGE]

3. The \(2 \times 2\) matrix \(A\) represents a transformation \(T\) which has the following properties.

• The image of the point \(( 0,1 )\) is the point \(( 3,4 )\).
• An object shape whose area is 7 is transformed to an image shape whose area is 35 .
• T has a line of invariant points.
  1. Find a possible matrix for \(\mathbf { A }\).

The transformation \(S\) is represented by the matrix \(\mathbf { B }\) where \(\mathbf { B } = \left( \begin{array} { l l } 3 & 1
2 & 2 \end{array} \right)\).

Find the equation of the line of invariant points of \(S\).

Show that any line of the form \(y = x + c\) is an invariant line of S .
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4.
\includegraphics[max width=\textwidth, alt={}, center]{3ecf08a3-5cf0-400c-ab79-b79a4dc8c9b4-08_663_446_228_164} 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.
4. 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.
   [0pt] [BLANK PAGE] A car of mass 1250 kg experiences a resistance to its motion of magnitude \(k v ^ { 2 } \mathrm {~N}\), where \(k\) is a constant and \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the car's speed. The car travels in a straight line along a horizontal road with its engine working at a constant rate of \(P \mathrm {~W}\). At a point \(A\) on the road the car's speed is \(15 \mathrm {~ms} ^ { - 1 }\) and it has an acceleration of magnitude \(0.54 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). At a point \(B\) on the road the car's speed is \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and it has an acceleration of magnitude \(0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
   1. Find the values of \(k\) and \(P\). The power is increased to 15 kW .
   2. Calculate the maximum steady speed of the car on a straight horizontal road.
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6. At a demolition site, bricks slide down a straight chute into a container. The chute is rough and is inclined at an angle of \(30 ^ { \circ }\) to the horizontal. The distance travelled down the chute by each brick is 8 m . A brick of mass 3 kg is released from rest at the top of the chute. When it reaches the bottom of the chute, its speed is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Find the potential energy lost by the brick in moving down the chute.
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2. By using the work-energy principle, or otherwise, find the constant frictional force acting on the brick as it moves down the chute.
   (5)
3. Hence find the coefficient of friction between the brick and the chute.
   (3) Another brick of mass 3 kg slides down the chute. This brick is given an initial speed of \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at the top of the chute.
4. Find the speed of this brick when it reaches the bottom of the chute.
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