CAIE M2 (Mechanics 2) 2009 November

3 \includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-3_408_1164_248_493} A particle \(P\) is released from rest at a point \(A\) which is 7 m above horizontal ground. At the same instant that \(P\) is released a particle \(Q\) is projected from a point \(O\) on the ground. The horizontal distance of \(O\) from \(A\) is 24 m . Particle \(Q\) moves in the vertical plane containing \(O\) and \(A\), with initial speed \(50 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and initial direction making an angle \(\theta\) above the horizontal, where \(\tan \theta = \frac { 7 } { 24 }\) (see diagram). Show that the particles collide.

4 One end of a light elastic string of natural length 3 m and modulus of elasticity 15 mN is attached to a fixed point \(O\). A particle \(P\) of mass \(m \mathrm {~kg}\) is attached to the other end of the string. \(P\) is released from rest at \(O\) and moves vertically downwards. When the extension of the string is \(x \mathrm {~m}\) the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(v ^ { 2 } = 5 \left( 12 + 4 x - x ^ { 2 } \right)\).
2. Find the magnitude of the acceleration of \(P\) when it is at its lowest point, and state the direction of this acceleration.

5 \includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-3_593_828_1530_660} A horizontal disc of radius 0.5 m is rotating with constant angular speed \(\omega \mathrm { rad } \mathrm { s } ^ { - 1 }\) about a fixed vertical axis through its centre \(O\). One end of a light inextensible string of length 0.8 m is attached to a point \(A\) of the circumference of the disc. A particle \(P\) of mass 0.4 kg is attached to the other end of the string. The string is taut and the system rotates so that the string is always in the same vertical plane as the radius \(O A\) of the disc. The string makes a constant angle \(\theta\) with the vertical (see diagram). The speed of \(P\) is 1.6 times the speed of \(A\).

1. Show that \(\sin \theta = \frac { 3 } { 8 }\).
2. Find the tension in the string.
3. Find the value of \(\omega\).

6 \includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-4_447_736_269_701} \(P\) is the vertex of a uniform solid cone of mass 5 kg , and \(O\) is the centre of its base. Strings are attached to the cone at \(P\) and at \(O\). The cone hangs in equilibrium with \(P O\) horizontal and the strings taut. The strings attached at \(P\) and \(O\) make angles of \(\theta ^ { \circ }\) and \(20 ^ { \circ }\), respectively, with the vertical (see diagram, which shows a cross-section).

1. By taking moments about \(P\) for the cone, find the tension in the string attached at \(O\).
2. Find the value of \(\theta\) and the tension in the string attached at \(P\).

7 A particle \(P\) of mass 0.3 kg is projected vertically upwards from the ground with an initial speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When \(P\) is at height \(x \mathrm {~m}\) above the ground, its upward speed is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that $$3 v - 90 \ln ( v + 30 ) + x = A ,$$ where \(A\) is a constant.

1. Differentiate this equation with respect to \(x\) and hence show that the acceleration of the particle is \(- \frac { 1 } { 3 } ( v + 30 ) \mathrm { m } \mathrm { s } ^ { - 2 }\).
2. Find, in terms of \(v\), the resisting force acting on the particle.
3. Find the time taken for \(P\) to reach its maximum height.