OCR M1 (Mechanics 1) 2005 June

1
\includegraphics[max width=\textwidth, alt={}, center]{99d30766-9c1b-43a8-986a-112b78b08146-2_508_501_274_822} A light inextensible string has its ends attached to two fixed points \(A\) and \(B\). The point \(A\) is vertically above \(B\). A smooth ring \(R\) of mass \(m \mathrm {~kg}\) is threaded on the string and is pulled by a force of magnitude 1.6 N acting upwards at \(45 ^ { \circ }\) to the horizontal. The section \(A R\) of the string makes an angle of \(30 ^ { \circ }\) with the downward vertical and the section \(B R\) is horizontal (see diagram). The ring is in equilibrium with the string taut.

1. Give a reason why the tension in the part \(A R\) of the string is the same as that in the part \(B R\).
2. Show that the tension in the string is 0.754 N , correct to 3 significant figures.
3. Find the value of \(m\).

2
\includegraphics[max width=\textwidth, alt={}, center]{99d30766-9c1b-43a8-986a-112b78b08146-2_643_289_1475_927} Particles \(A\) and \(B\), of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of a light inextensible string. Particle \(A\) is held at rest at a fixed point and \(B\) hangs vertically below \(A\). Particle \(A\) is now released. As the particles fall the air resistance acting on \(A\) is 0.4 N and the air resistance acting on \(B\) is 0.25 N (see diagram). The downward acceleration of each of the particles is \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\) and the tension in the string is \(T \mathrm {~N}\).

1. Write down two equations in \(a\) and \(T\) obtained by applying Newton's second law to \(A\) and to \(B\).
2. Find the values of \(a\) and \(T\).

3 Two small spheres \(P\) and \(Q\) have masses 0.1 kg and 0.2 kg respectively. The spheres are moving directly towards each other on a horizontal plane and collide. Immediately before the collision \(P\) has speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(Q\) has speed \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Immediately after the collision the spheres move away from each other, \(P\) with speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(Q\) with speed \(( 3.5 - u ) \mathrm { m } \mathrm { s } ^ { - 1 }\).

1. Find the value of \(u\). After the collision the spheres both move with deceleration of magnitude \(5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) until they come to rest on the plane.
2. Find the distance \(P Q\) when both \(P\) and \(Q\) are at rest.

4 A particle moves downwards on a smooth plane inclined at an angle \(\alpha\) to the horizontal. The particle passes through the point \(P\) with speed \(u \mathrm {~ms} ^ { - 1 }\). The particle travels 2 m during the first 0.8 s after passing through \(P\), then a further 6 m in the next 1.2 s . Find

1. the value of \(u\) and the acceleration of the particle,
2. the value of \(\alpha\) in degrees.

5
\includegraphics[max width=\textwidth, alt={}, center]{99d30766-9c1b-43a8-986a-112b78b08146-3_697_579_1238_781} Two small rings \(A\) and \(B\) are attached to opposite ends of a light inextensible string. The rings are threaded on a rough wire which is fixed vertically. \(A\) is above \(B\). A horizontal force is applied to a point \(P\) of the string. Both parts \(A P\) and \(B P\) of the string are taut. The system is in equilibrium with angle \(B A P = \alpha\) and angle \(A B P = \beta\) (see diagram). The weight of \(A\) is 2 N and the tensions in the parts \(A P\) and \(B P\) of the string are 7 N and \(T \mathrm {~N}\) respectively. It is given that \(\cos \alpha = 0.28\) and \(\sin \alpha = 0.96\), and that \(A\) is in limiting equilibrium.

1. Find the coefficient of friction between the wire and the ring \(A\).
2. By considering the forces acting at \(P\), show that \(T \cos \beta = 1.96\).
3. Given that there is no frictional force acting on \(B\), find the mass of \(B\).

6 A particle of mass 0.04 kg is acted on by a force of magnitude \(P \mathrm {~N}\) in a direction at an angle \(\alpha\) to the upward vertical.

1. The resultant of the weight of the particle and the force applied to the particle acts horizontally. Given that \(\alpha = 20 ^ { \circ }\) find
   (a) the value of \(P\),
   (b) the magnitude of the resultant,
   (c) the magnitude of the acceleration of the particle.
2. It is given instead that \(P = 0.08\) and \(\alpha = 90 ^ { \circ }\). Find the magnitude and direction of the resultant force on the particle.

7
\includegraphics[max width=\textwidth, alt={}, center]{99d30766-9c1b-43a8-986a-112b78b08146-4_634_1127_934_507} A car \(P\) starts from rest and travels along a straight road for 600 s . The \(( t , v )\) graph for the journey is shown in the diagram. This graph consists of three straight line segments. Find

1. the distance travelled by \(P\),
2. the deceleration of \(P\) during the interval \(500 < t < 600\). Another car \(Q\) starts from rest at the same instant as \(P\) and travels in the same direction along the same road for 600 s . At time \(t \mathrm {~s}\) after starting the velocity of \(Q\) is \(\left( 600 t ^ { 2 } - t ^ { 3 } \right) \times 10 ^ { - 6 } \mathrm {~ms} ^ { - 1 }\).
3. Find an expression in terms of \(t\) for the acceleration of \(Q\).
4. Find how much less \(Q\) 's deceleration is than \(P\) 's when \(t = 550\).
5. Show that \(Q\) has its maximum velocity when \(t = 400\).
6. Find how much further \(Q\) has travelled than \(P\) when \(t = 400\).