OCR MEI M1 (Mechanics 1) 2014 June

1 Fig. 1 shows the velocity-time graph of a cyclist travelling along a straight horizontal road between two sets of traffic lights. The velocity, \(v\), is measured in metres per second and the time, \(t\), in seconds. The distance travelled, \(s\) metres, is measured from when \(t = 0\). \begin{figure}[h] \end{figure}

1. Find the values of \(s\) when \(t = 4\) and when \(t = 18\).
2. Sketch the graph of \(s\) against \(t\) for \(0 \leqslant t \leqslant 18\).

2 The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) shown in Fig. 2 are in the horizontal and vertically upwards directions. \begin{figure}[h] \end{figure} Forces \(\mathbf { p }\) and \(\mathbf { q }\) are given, in newtons, by \(\mathbf { p } = 12 \mathbf { i } - 5 \mathbf { j }\) and \(\mathbf { q } = 16 \mathbf { i } + 1.5 \mathbf { j }\).

1. Write down the force \(\mathbf { p } + \mathbf { q }\) and show that it is parallel to \(8 \mathbf { i } - \mathbf { j }\).
2. Show that the force \(3 \mathbf { p } + 10 \mathbf { q }\) acts in the horizontal direction.
3. A particle is in equilibrium under forces \(k \mathbf { p } , 3 \mathbf { q }\) and its weight \(\mathbf { w }\). Show that the value of \(k\) must be - 4 and find the mass of the particle.

3 Fig. 3 shows a smooth ball resting in a rack. The angle in the middle of the rack is \(90 ^ { \circ }\). The rack has one edge at angle \(\alpha\) to the horizontal. The weight of the ball is \(W \mathrm {~N}\). The reaction forces of the rack on the ball at the points of contact are \(R \mathrm {~N}\) and \(S \mathrm {~N}\). \begin{figure}[h] \end{figure}

1. Draw a fully labelled triangle of forces to show the forces acting on the ball. Your diagram must indicate which angle is \(\alpha\).
2. Find the values of \(R\) and \(S\) in terms of \(W\) and \(\alpha\).
3. On the same axes draw sketches of \(R\) against \(\alpha\) and \(S\) against \(\alpha\) for \(0 ^ { \circ } \leqslant \alpha \leqslant 90 ^ { \circ }\). For what values of \(\alpha\) is \(R < S\) ?

4 Fig. 4 illustrates a situation in which a film is being made. A cannon is fired from the top of a vertical cliff towards a ship out at sea. The director wants the cannon ball to fall just short of the ship so that it appears to be a near-miss. There are actors on the ship so it is important that it is not hit by mistake. The cannon ball is fired from a height 75 m above the sea with an initial velocity of \(20 \mathrm {~ms} ^ { - 1 }\) at an angle of \(30 ^ { \circ }\) above the horizontal. The ship is 90 m from the bottom of the cliff. \begin{figure}[h] \end{figure}

1. The director calculates where the cannon ball will hit the sea, using the standard projectile model and taking the value of \(g\) to be \(10 \mathrm {~ms} ^ { - 2 }\). Verify that according to this model the cannon ball is in the air for 5 seconds. Show that it hits the water less than 5 m from the ship.
2. Without doing any further calculations state, with a brief reason, whether the cannon ball would be predicted to travel further from the cliff if the value of \(g\) were taken to be \(9.8 \mathrm {~ms} ^ { - 2 }\).

5 In a science fiction story a new type of spaceship travels to the moon. The journey takes place along a straight line. The spaceship starts from rest on the earth and arrives at the moon's surface with zero speed. Its speed, \(v\) kilometres per hour at time \(t\) hours after it has started, is given by $$v = 37500 \left( 4 t - t ^ { 2 } \right) .$$

1. Show that the spaceship takes 4 hours to reach the moon.
2. Find an expression for the distance the spaceship has travelled at time \(t\). Hence find the distance to the moon.
3. Find the spaceship's greatest speed during the journey. Section B (36 marks)

6 In this question the origin is a point on the ground. The directions of the unit vectors \(\left( \begin{array} { l } 1
0
0 \end{array} \right) , \left( \begin{array} { l } 0
1
0 \end{array} \right)\) and \(\left( \begin{array} { l } 0
0
1 \end{array} \right)\) are
\includegraphics[max width=\textwidth, alt={}, center]{63a2dc41-5e8b-4275-8653-ece5067c4306-5_398_689_434_689} Alesha does a sky-dive on a day when there is no wind. The dive starts when she steps out of a moving helicopter. The dive ends when she lands gently on the ground.

• During the dive Alesha can reduce the magnitude of her acceleration in the vertical direction by spreading her arms and increasing air resistance.
• During the dive she can use a power unit strapped to her back to give herself an acceleration in a horizontal direction.
• Alesha's mass, including her equipment, is 100 kg .
• Initially, her position vector is \(\left( \begin{array} { r } - 75
  90
  750 \end{array} \right) \mathrm { m }\) and her velocity is \(\left( \begin{array} { r } - 5
  0
  - 10 \end{array} \right) \mathrm { ms } ^ { - 1 }\).
  1. Calculate Alesha's initial speed, and the initial angle between her motion and the downward vertical.

At a certain time during the dive, forces of \(\left( \begin{array} { r } 0
0
- 980 \end{array} \right) \mathrm { N } , \left( \begin{array} { r } 0
0
880 \end{array} \right) \mathrm { N }\) and \(\left( \begin{array} { r } 50
- 20
0 \end{array} \right) \mathrm { N }\) are acting on Alesha.

Suggest how these forces could arise.

Find Alesha's acceleration at this time, giving your answer in vector form, and show that, correct to 3 significant figures, its magnitude is \(1.14 \mathrm {~ms} ^ { - 2 }\). One suggested model for Alesha's motion is that the forces on her are constant throughout the dive from when she leaves the helicopter until she reaches the ground.

Find expressions for her velocity and position vector at time \(t\) seconds after the start of the dive according to this model. Verify that when \(t = 30\) she is at the origin.

Explain why consideration of Alesha's landing velocity shows this model to be unrealistic.