Vertical elastic string: general symbolic/proof questions

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.

6 One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(k\), is attached to a particle \(P\) of mass \(m\). The other end of the string is attached to a fixed point \(Q\). The particle \(P\) is projected vertically upwards from \(Q\). When \(P\) is moving upwards and at a distance \(\frac { 4 } { 3 } a\) directly above \(Q\), it has a speed \(\sqrt { 2 g a }\). At this point, its acceleration is \(\frac { 7 } { 3 } g\) downwards. Show that \(\mathrm { k } = 4 \mathrm { mg }\) and find in terms of \(a\) the greatest height above \(Q\) reached by \(P\).

1 One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(3 m g\), is attached to a fixed point \(O\). The other end of the string is attached to a particle \(P\) of mass \(m\). The string hangs with \(P\) vertically below \(O\). The particle \(P\) is pulled vertically downwards so that the extension of the string is \(2 a\). The particle \(P\) is then released from rest.

1. Find the speed of \(P\) when it is at a distance \(\frac { 3 } { 4 } a\) below \(O\).
2. Find the initial acceleration of \(P\) when it is released from rest. \includegraphics[max width=\textwidth, alt={}, center]{454be64a-204f-4fa4-a5fc-72fd88e1289f-03_741_473_269_836} A particle \(P\) of mass \(m\) is moving with speed \(u\) on a fixed smooth horizontal surface. It collides at an angle \(\alpha\) with a fixed smooth vertical barrier. After the collision, \(P\) moves at an angle \(\theta\) with the barrier, where \(\tan \theta = \frac { 1 } { 2 }\) (see diagram). The coefficient of restitution between \(P\) and the barrier is \(e\). The particle \(P\) loses 20\% of its kinetic energy as a result of the collision. Find the value of \(e\).

3 A light elastic string has natural length \(a\) and modulus of elasticity 12 mg . One end of the string is attached to a fixed point \(O\). The other end of the string is attached to a particle of mass \(m\). The particle hangs in equilibrium vertically below \(O\). The particle is pulled vertically down and released from rest with the extension of the string equal to \(e\), where \(\mathrm { e } > \frac { 1 } { 3 } \mathrm { a }\). In the subsequent motion the particle has speed \(\sqrt { 2 \mathrm { ga } }\) when it has ascended a distance \(\frac { 1 } { 3 } a\). Find \(e\) in terms of \(a\). \includegraphics[max width=\textwidth, alt={}, center]{b10c65ef-dafd-4746-be5b-789130b7d030-06_488_496_269_781} A uniform lamina \(A E C F\) is formed by removing two identical triangles \(B C E\) and \(C D F\) from a square lamina \(A B C D\). The square has side \(3 a\) and \(E B = D F = h\) (see diagram).

1. Find the distance of the centre of mass of the lamina \(A E C F\) from \(A D\) and from \(A B\), giving your answers in terms of \(a\) and \(h\).
   The lamina \(A E C F\) is placed vertically on its edge \(A E\) on a horizontal plane.
2. Find, in terms of \(a\), the set of values of \(h\) for which the lamina remains in equilibrium.

7. A light elastic string has natural length \(a\) and modulus of elasticity \(\frac { 3 } { 2 } m g\). A particle \(P\) of mass \(m\) is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). The particle is released from rest at \(A\) and falls vertically. When \(P\) has fallen a distance \(a + x\), where \(x > 0\), the speed of \(P\) is \(v\).

1. Show that \(v ^ { 2 } = 2 g ( a + x ) - \frac { 3 g x ^ { 2 } } { 2 a }\).
2. Find the greatest speed attained by \(P\) as it falls. After release, \(P\) next comes to instantaneous rest at a point \(D\).
3. Find the magnitude of the acceleration of \(P\) at \(D\).

A light elastic string has modulus of elasticity \(\frac { 3 } { 2 } m g\) and natural length \(a\). A particle of mass \(m\) is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). The particle is released from rest at \(A\). Show that when the particle has fallen a distance \(k a\) from \(A\), where \(k > 1\), its kinetic energy is $$\frac { 1 } { 4 } m g a \left( 10 k - 3 - 3 k ^ { 2 } \right) .$$ Show that the particle first comes to instantaneous rest at the point \(B\) which is at a distance \(3 a\) vertically below \(A\). Show that the time taken by the particle to travel from \(A\) to \(B\) is $$\sqrt { } \left( \frac { 2 a } { g } \right) + \frac { 2 \pi } { 3 } \sqrt { } \left( \frac { 2 a } { 3 g } \right)$$

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
   1. the maximum extension of the string,
   2. the maximum speed of the particle,
   3. the maximum magnitude of the acceleration of the particle.

1. A particle \(P\), of mass \(m\), is attached to one end of a light elastic spring of natural length \(a\) and modulus of elasticity kmg.

The other end of the spring is attached to a fixed point \(O\) on a ceiling.
The point \(A\) is vertically below \(O\) such that \(O A = 3 a\) The point \(B\) is vertically below \(O\) such that \(O B = \frac { 1 } { 2 } a\) The particle is held at rest at \(A\), then released and first comes to instantaneous rest at the point \(B\).

1. Show that \(k = \frac { 4 } { 3 }\)
2. Find, in terms of \(g\), the acceleration of \(P\) immediately after it is released from rest at \(A\).
3. Find, in terms of \(g\) and \(a\), the maximum speed attained by \(P\) as it moves from \(A\) to \(B\).

1. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(3 m g\).

The other end of the string is attached to a fixed point \(O\) on a ceiling.
The particle hangs freely in equilibrium at a distance \(d\) vertically below \(O\).

1. Show that \(d = \frac { 4 } { 3 } a\). The point \(A\) is vertically below \(O\) such that \(O A = 2 a\).
   The particle is held at rest at \(A\), then released and first comes to instantaneous rest at the point \(B\).
2. Find, in terms of \(g\), the acceleration of \(P\) immediately after it is released from rest.
3. Find, in terms of \(g\) and \(a\), the maximum speed attained by \(P\) as it moves from \(A\) to \(B\).
4. Find, in terms of \(a\), the distance \(O B\).

One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(kmg\), is attached to a fixed point A. The other end of the string is attached to a particle \(P\) of mass \(4m\). The particle \(P\) hangs in equilibrium a distance \(x\) vertically below A.

1. Show that \(k = \frac{4a}{x-a}\). [1]

An additional particle, of mass \(2m\), is now attached to \(P\) and the combined particle is released from rest at the original equilibrium position of \(P\). When the combined particle has descended a distance \(\frac{3}{4}a\), its speed is \(\frac{1}{2}\sqrt{ga}\).

1. Find \(x\) in terms of \(a\). [6]

One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(kmg\), is attached to a fixed point A. The other end of the string is attached to a particle \(P\) of mass \(4m\). The particle \(P\) hangs in equilibrium a distance \(x\) vertically below A.

1. Show that \(k = \frac{4a}{x-a}\). [1]

An additional particle, of mass \(2m\), is now attached to \(P\) and the combined particle is released from rest at the original equilibrium position of \(P\). When the combined particle has descended a distance \(\frac{3}{4}a\), its speed is \(\frac{1}{3}\sqrt{ga}\).

1. Find \(x\) in terms of \(a\). [6]