OCR MEI M1 (Mechanics 1) 2007 January

1 Fig. 1 is the velocity-time graph for the motion of a body. The velocity of the body is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t\) seconds. \begin{figure}[h] \end{figure} The displacement of the body from \(t = 0\) to \(t = 100\) is 1400 m . Find the value of \(V\).

2 A particle moves along a straight line containing a point O . Its displacement, \(x \mathrm {~m}\), from O at time \(t\) seconds is given by $$x = 12 t - t ^ { 3 } , \text { where } - 10 \leqslant t \leqslant 10$$ Find the values of \(x\) for which the velocity of the particle is zero.

3 A box of mass 5 kg is at rest on a rough horizontal floor.

1. Find the value of the normal reaction of the floor on the box. The box remains at rest on the floor when a force of 10 N is applied to it at an angle of \(40 ^ { \circ }\) to the upward vertical, as shown in Fig. 3. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{52d6c914-b204-4587-a82e-fbab6693fcf8-2_293_472_2131_794} \captionsetup{labelformat=empty} \caption{Fig. 3}
   \end{figure}
2. Draw a diagram showing all the forces acting on the box.
3. Calculate the new value of the normal reaction of the floor on the box and also the frictional force.

4 Fig. 4 shows forces of magnitudes 20 N and 16 N inclined at \(60 ^ { \circ }\). \begin{figure}[h] \end{figure}

1. Calculate the component of the resultant of these two forces in the direction of the 20 N force.
2. Calculate the magnitude of the resultant of these two forces. These are the only forces acting on a particle of mass 2 kg .
3. Find the magnitude of the acceleration of the particle and the angle the acceleration makes with the 20 N force.

5 A block of mass 4 kg slides on a horizontal plane against a constant resistance of 14.8 N . A light, inextensible string is attached to the block and, after passing over a smooth pulley, is attached to a freely hanging sphere of mass 2 kg . The part of the string between the block and the pulley is horizontal. This situation is shown in Fig. 5. \begin{figure}[h] \end{figure} The tension in the string is \(T \mathrm {~N}\) and the acceleration of the block and of the sphere is \(a \mathrm {~ms} ^ { - 2 }\).

1. Write down the equation of motion of the block and also the equation of motion of the sphere, each in terms of \(T\) and \(a\).
2. Find the values of \(T\) and \(a\).

6 The velocity of a model boat, \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), is given by $$\mathbf { v } = \binom { - 5 } { 10 } + t \binom { 6 } { - 8 }$$ where \(t\) is the time in seconds and the vectors \(\binom { 1 } { 0 }\) and \(\binom { 0 } { 1 }\) are east and north respectively.

1. Show that when \(t = 2.5\) the boat is travelling south-east (i.e. on a bearing of \(135 ^ { \circ }\) ). Calculate its speed at this time. The boat is at a point O when \(t = 0\).
2. Calculate the bearing of the boat from O when \(t = 2.5\).

7 A horizontal force of 24 N acts on a block of mass 12 kg on a horizontal plane. The block is initially at rest. This situation is first modelled assuming the plane is smooth.

1. Write down the acceleration of the block according to this model. The situation is now modelled assuming a constant resistance to motion of 15 N .
2. Calculate the acceleration of the block according to this new model. How much less distance does the new model predict that the block will travel in the first 4 seconds? The 24 N force is removed and the block slides down a slope at \(5 ^ { \circ }\) to the horizontal. The speed of the block at the top of the slope is \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), as shown in Fig. 7. The answers to parts (iii) and (iv) should be found using the assumption that the resistance to the motion of the block is still a constant 15 N . \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{52d6c914-b204-4587-a82e-fbab6693fcf8-5_255_901_1128_575} \captionsetup{labelformat=empty} \caption{Fig. 7}
   \end{figure}
3. Calculate the acceleration of the block in the direction of its motion.
4. For how much time does the block slide down the slope before coming to rest and how far does it slide in that time? Measurements show that the block actually comes to rest in 3.5 seconds.
5. Assuming that the error in the prediction is due only to the value of the resistance, calculate the true value of the resistance.

8 In this question the value of \(\boldsymbol { g \) should be taken as \(\mathbf { 1 0 } \mathbf { m ~ s } ^ { \mathbf { - 2 } }\).} As shown in Fig. 8, particles A and B are projected towards one another. Each particle has an initial speed of \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) vertically and \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) horizontally. Initially A and B are 70 m apart horizontally and B is 15 m higher than A . Both particles are projected over horizontal ground. \begin{figure}[h] \end{figure}

1. Show that, \(t\) seconds after projection, the height in metres of each particle above its point of projection is \(10 t - 5 t ^ { 2 }\).
2. Calculate the horizontal range of A . Deduce that A hits the horizontal ground between the initial positions of A and B .
3. Calculate the horizontal distance travelled by B before reaching the ground.
4. Show that the paths of the particles cross but that the particles do not collide if they are projected at the same time. In fact, particle A is projected 2 seconds after particle B .
5. Verify that the particles collide 0.75 seconds after A is projected.