OCR M3 (Mechanics 3) 2009 January

1
\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-2_385_741_269_701} A particle \(P\) of mass 0.5 kg is moving in a straight line with speed \(6.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). An impulse of magnitude 2.6 N s applied to \(P\) deflects its direction of motion through an angle \(\theta\), and reduces its speed to \(2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). By considering an impulse-momentum triangle, or otherwise,

1. show that \(\cos \theta = 0.6\),
2. find the angle that the impulse makes with the original direction of motion of \(P\).

2 \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Two uniform rods \(A B\) and \(B C\), of weights 70 N and 110 N respectively, are freely jointed at \(B\). The rods are in equilibrium in a vertical plane with \(A\) and \(C\) at the same horizontal level and \(A C = 2 \mathrm {~m}\). The \(\operatorname { rod } A B\) is freely jointed to a fixed point at \(A\) and the rod \(B C\) is freely jointed to a fixed point at \(C\). The horizontal distance between \(B\) and \(A\) is 4 m and \(B\) is 4 m below \(A C\); angle \(B A C\) is obtuse (see Fig. 1). The force exerted on the \(\operatorname { rod } A B\) at \(B\), by the \(\operatorname { rod } B C\), has horizontal and vertical components as shown in Fig. 2.

1. By taking moments about \(A\) for the \(\operatorname { rod } A B\) find the value of \(X - Y\).
2. By taking moments about \(C\) for the rod \(B C\) show that \(2 X - 3 Y + 165 = 0\).
3. Find the magnitude of the force acting between \(A B\) and \(B C\) at \(B\).

3
\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-3_387_181_274_982}
\(A\) and \(B\) are fixed points with \(B\) at a distance of 1.8 m vertically below \(A\). One end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N is attached to \(A\), and one end of an identical elastic string is attached to \(B\). A particle \(P\) of weight 12 N is attached to the other ends of the strings (see diagram).

1. Verify that \(P\) is in equilibrium when it is at a distance of 1.05 m vertically below \(A\).
   \(P\) is released from rest at the point 1.2 m vertically below \(A\) and begins to move.
2. Show that, when \(P\) is \(x \mathrm {~m}\) below its equilibrium position, the tensions in \(P A\) and \(P B\) are \(( 18 + 40 x ) \mathrm { N }\) and \(( 6 - 40 x ) \mathrm { N }\) respectively.
3. Show that \(P\) moves with simple harmonic motion of period 0.777 s , correct to 3 significant figures.
4. Find the speed with which \(P\) passes through the equilibrium position.
   \includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-3_540_655_1564_744} One end of a light inextensible string of length 0.5 m is attached to a fixed point \(O\). A particle \(P\) of mass 0.2 kg is attached to the other end of the string. With the string taut and horizontal, \(P\) is projected with a velocity of \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) vertically downward. \(P\) begins to move in a vertical circle with centre \(O\). While the string remains taut the angular displacement of \(O P\) is \(\theta\) radians from its initial position, and the speed of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram).

1. Show that \(v ^ { 2 } = 9 + 9.8 \sin \theta\).
2. Find, in terms of \(\theta\), the radial and tangential components of the acceleration of \(P\).
3. Show that the tension in the string is \(( 3.6 + 5.88 \sin \theta ) \mathrm { N }\) and hence find the value of \(\theta\) at the instant when the string becomes slack, giving your answer correct to 1 decimal place.

5
\includegraphics[max width=\textwidth, alt={}, center]{14403602-94a6-4441-a673-65f9b98180e5-4_369_953_269_596} Two smooth uniform spheres \(A\) and \(B\), of equal radius, have masses 3 kg and 4 kg respectively. They are moving on a horizontal surface, each with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), when they collide. The directions of motion of \(A\) and \(B\) make angles \(\alpha\) and \(\beta\) respectively with the line of centres of the spheres, where \(\sin \alpha = \cos \beta = 0.6\) (see diagram). The coefficient of restitution between the spheres is 0.75 . Find the angle that the velocity of \(A\) makes, immediately after impact, with the line of centres of the spheres.
[0pt] [10]

6 A stone of mass 0.125 kg falls freely under gravity, from rest, until it has travelled a distance of 10 m . The stone then continues to fall in a medium which exerts an upward resisting force of \(0.025 v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the stone \(t \mathrm {~s}\) after the instant that it enters the resisting medium.

1. Show by integration that \(v = 49 - 35 \mathrm { e } ^ { - 0.2 t }\).
2. Find how far the stone travels during the first 3 seconds in the medium.

7 A particle of mass 0.8 kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 20 N . The other end of the string is attached to a fixed point \(O\). The particle is held at rest at \(O\) and then released. When the extension of the string is \(x \mathrm {~m}\), the particle is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. By considering energy show that \(v ^ { 2 } = 39.2 + 19.6 x - 12.5 x ^ { 2 }\).
2. Hence find
   (a) the maximum extension of the string,
   (b) the maximum speed of the particle,
   (c) the maximum magnitude of the acceleration of the particle. \footnotetext{OCR
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