Edexcel M2 (Mechanics 2) 2005 January

Question 4
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4. A particle \(P\) of mass 0.4 kg is moving under the action of a single force \(\mathbf { F }\) newtons. At time \(t\) seconds, the velocity of \(P , \mathbf { v } \mathrm {~m} \mathrm {~s} ^ { - 1 }\), is given by $$\mathbf { v } = ( 6 t + 4 ) \mathbf { i } + \left( t ^ { 2 } + 3 t \right) \mathbf { j } .$$ When \(t = 0 , P\) is at the point with position vector \(( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m }\). When \(t = 4 , P\) is at the point \(S\).
  1. Calculate the magnitude of \(\mathbf { F }\) when \(t = 4\).
  2. Calculate the distance \(O S\).
Question 5
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5. A car of mass 1000 kg is towing a trailer of mass 1500 kg along a straight horizontal road. The tow-bar joining the car to the trailer is modelled as a light rod parallel to the road. The total resistance to motion of the car is modelled as having constant magnitude 750 N . The total resistance to motion of the trailer is modelled as of magnitude \(R\) newtons, where \(R\) is a constant. When the engine of the car is working at a rate of 50 kW , the car and the trailer travel at a constant speed of \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Show that \(R = 1250\). When travelling at \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) the driver of the car disengages the engine and applies the brakes. The brakes provide a constant braking force of magnitude 1500 N to the car. The resisting forces of magnitude 750 N and 1250 N are assumed to remain unchanged. Calculate
  2. the deceleration of the car while braking,
  3. the thrust in the tow-bar while braking,
  4. the work done, in kJ , by the braking force in bringing the car and the trailer to rest.
  5. Suggest how the modelling assumption that the resistances to motion are constant could be refined to be more realistic.
Question 6
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6. A particle \(P\) of mass \(3 m\) is moving with speed \(2 u\) in a straight line on a smooth horizontal table. The particle \(P\) collides with a particle \(Q\) of mass \(2 m\) moving with speed \(u\) in the opposite direction to \(P\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
  1. Show that the speed of \(Q\) after the collision is \(\frac { 1 } { 5 } u ( 9 e + 4 )\). As a result of the collision, the direction of motion of \(P\) is reversed.
  2. Find the range of possible values of \(e\). Given that the magnitude of the impulse of \(P\) on \(Q\) is \(\frac { 32 } { 5 } m u\),
  3. find the value of \(e\).
    (4)
Question 7
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7. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 4} \includegraphics[alt={},max width=\textwidth]{a9e00b5b-3804-4f8d-9cc8-7d1170027726-6_568_1582_360_239}
\end{figure} A particle \(P\) is projected from a point \(A\) with speed \(32 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\alpha\), where \(\sin \alpha = \frac { 3 } { 5 }\). The point \(O\) is on horizontal ground, with \(O\) vertically below \(A\) and \(O A = 20 \mathrm {~m}\). The particle \(P\) moves freely under gravity and passes through a point \(B\), which is 16 m above ground, before reaching the ground at the point \(C\), as shown in Figure 4. Calculate
  1. the time of the flight from \(A\) to \(C\),
  2. the distance \(O C\),
  3. the speed of \(P\) at \(B\),
  4. the angle that the velocity of \(P\) at \(B\) makes with the horizontal.