Collision/impulse during circular motion

4 A particle P of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point O . Particle P is projected so that it moves in complete vertical circles with centre O ; there is no air resistance. A and B are two points on the circle, situated on opposite sides of the vertical through O . The lines OA and OB make angles \(\alpha\) and \(\beta\) with the upward vertical as shown in Fig. 4. \begin{figure}[h] \end{figure} The speed of P at A is \(\sqrt { \frac { 17 a g } { 3 } }\). The speed of P at B is \(\sqrt { 5 a g }\) and \(\cos \beta = \frac { 2 } { 3 }\).

1. Show that \(\cos \alpha = \frac { 1 } { 3 }\). On one occasion, when P is at its lowest point and moving in a clockwise direction, it collides with a stationary particle Q . The two particles coalesce and the combined particle continues to move in the same vertical circle. When this combined particle reaches the point A , the string becomes slack.
2. Show that when the string becomes slack, the speed of the combined particle is \(\sqrt { \frac { a g } { 3 } }\). The mass of the particle Q is \(k m\).
3. Find the value of \(k\).
4. Find, in terms of \(m\) and \(g\), the instantaneous change in the tension in the string as a result of the collision.

Answer only one of the following two alternatives. **EITHER** A particle \(P\), of mass \(m\), is able to move in a vertical circle on the smooth inner surface of a sphere with centre \(O\) and radius \(a\). Points \(A\) and \(B\) are on the inner surface of the sphere and \(AOB\) is a horizontal diameter. Initially, \(P\) is projected vertically downwards with speed \(\sqrt{\left(\frac{21}{2}ag\right)}\) from \(A\) and begins to move in a vertical circle. At the lowest point of its path, vertically below \(O\), the particle \(P\) collides with a stationary particle \(Q\), of mass \(4m\), and rebounds. The speed acquired by \(Q\), as a result of the collision, is just sufficient for it to reach the point \(B\).

1. Find the speed of \(P\) and the speed of \(Q\) immediately after their collision. [7] In its subsequent motion, \(P\) loses contact with the inner surface of the sphere at the point \(D\), where the angle between \(OD\) and the upward vertical through \(O\) is \(\theta\).
2. Find \(\cos \theta\). [5] **OR** A farmer grows two different types of cherries, Type A and Type B. He assumes that the masses of each type are normally distributed. He chooses a random sample of 8 cherries of Type A. He finds that the sample mean mass is 15.1 g and that a 95% confidence interval for the population mean mass, \(\mu\) g, is \(13.5 \leqslant \mu \leqslant 16.7\).
3. Find an unbiased estimate for the population variance of the masses of cherries of Type A. [3] The farmer now chooses a random sample of 6 cherries of Type B and records their masses as follows. $$12.2 \quad 13.3 \quad 16.4 \quad 14.0 \quad 13.9 \quad 15.4$$
4. Test at the 5% significance level whether the mean mass of cherries of Type B is less than the mean mass of cherries of Type A. You should assume that the population variances for the two types of cherry are equal. [9]

\includegraphics{figure_7} A particle \(P\) is attached to a fixed point \(O\) by a light inextensible string of length \(0.7\) m. A particle \(Q\) is in equilibrium suspended from \(O\) by an identical string. With the string \(OP\) taut and horizontal, \(P\) is projected vertically downwards with speed \(6\) m s\(^{-1}\) so that it strikes \(Q\) directly (see diagram). \(P\) is brought to rest by the collision and \(Q\) starts to move with speed \(4.9\) m s\(^{-1}\).

1. Find the speed of \(P\) immediately before the collision. Hence find the coefficient of restitution between \(P\) and \(Q\). [3]
2. Given that the speed of \(Q\) is \(v\) m s\(^{-1}\) when \(OQ\) makes an angle \(\theta\) with the downward vertical, find an expression for \(v^2\) in terms of \(\theta\), and show that the tension in the string \(OQ\) is \(14.7m(1 + 2\cos\theta)\) N, where \(m\) kg is the mass of \(Q\). [6]
3. Find the radial and transverse components of the acceleration of \(Q\) at the instant that the string \(OQ\) becomes slack. [4]
4. Show that \(V^2 = 0.8575\), where \(V\) m s\(^{-1}\) is the speed of \(Q\) when it reaches its greatest height (after the string \(OQ\) becomes slack). Hence find the greatest height reached by \(Q\) above its initial position. [4]

A uniform rod \(AB\) has mass \(2m\) and length \(4a\).

1. Show by integration that the moment of inertia of the rod about an axis perpendicular to the rod through \(A\) is \(\frac{32}{3}ma^2\). [4]

The rod is initially at rest with \(B\) vertically below \(A\) and it is free to rotate in a vertical plane about a smooth fixed horizontal axis through \(A\). A particle of mass \(m\) is moving horizontally in the plane in which the rod is free to rotate. The particle has speed \(v\), and strikes the rod at \(B\). In the subsequent motion the particle adheres to the rod and the combined rigid body \(Q\), consisting of the rod and the particle, starts to rotate.

1. Find, in terms of \(v\) and \(a\), the initial angular speed of \(Q\). [4]

At time \(t\) seconds the angle between \(Q\) and the downward vertical is \(\theta\) radians.

1. Show that \(\dot{\theta}^2 = k\frac{g}{a}(\cos \theta - 1) + \frac{9v^2}{400a^2}\), stating the value of the constant \(k\). [4]
2. Find, in terms of \(a\) and \(g\), the set of values of \(v^2\) for which \(Q\) makes complete revolutions. [2]

When \(Q\) is horizontal, the force exerted by the axis on \(Q\) has vertically upwards component \(R\).

1. Find \(R\) in terms of \(m\) and \(g\). [4]

A uniform rod \(PQ\), of mass \(m\) and length \(3a\), is free to rotate about a fixed smooth horizontal axis \(L\), which passes through the end \(P\) of the rod and is perpendicular to the rod. The rod hangs at rest in equilibrium with \(Q\) vertically below \(P\). One end of a light inextensible string of length \(2a\) is attached to the rod at \(P\) and the other end is attached to a particle of mass \(3m\). The particle is held with the string taut, and horizontal and perpendicular to \(L\), and is then released. After colliding, the particle sticks to the rod forming a body \(B\).

1. Show that the moment of inertia of \(B\) about \(L\) is \(15ma^2\). [2]

1. Show that \(B\) first comes to instantaneous rest after it has turned through an angle \(\arccos\left(\frac{9}{25}\right)\). [10]

A uniform rod \(PQ\) of mass \(m\) and length \(3a\), is free to rotate about a fixed smooth horizontal axis \(L\), which passes through the end \(P\) of the rod and is perpendicular to the rod. The rod hangs at rest in equilibrium with \(Q\) vertically below \(P\). One end of a light inextensible string of length \(2a\) is attached to the rod at \(P\) and the other end is attached to a particle of mass \(3m\). The particle is held with the string taut, and horizontal and perpendicular to \(L\), and is then released. After colliding, the particle sticks to the rod forming a body \(B\).

1. Show that the moment of inertia of \(B\) about \(L\) is \(15ma^2\). [2]
2. Show that \(B\) first comes to instantaneous rest after it has turned through an angle \(\arccos\left(\frac{9}{25}\right)\). [10]