OCR MEI M1 (Mechanics 1)

1 Fig. 7 shows the trajectory of an object which is projected from a point O on horizontal ground. Its initial velocity is \(40 \mathrm {~ms} ^ { - 1 }\) at an angle of \(\alpha\) to the horizontal. \begin{figure}[h] \end{figure}

1. Show that, according to the standard projectile model in which air resistance is neglected, the flight time, \(T \mathrm {~s}\), and the range, \(R \mathrm {~m}\), are given by $$T = \frac { 80 \sin \alpha } { g } \text { and } R = \frac { 3200 \sin \alpha \cos \alpha } { g }$$ A company is designing a new type of ball and wants to model its flight.
2. Initially the company uses the standard projectile model. Use this model to show that when \(\alpha = 30 ^ { \circ }\) and the initial speed is \(40 \mathrm {~ms} ^ { - 1 } , T\) is approximately 4.08 and \(R\) is approximately 141.4 . Find the values of \(T\) and \(R\) when \(\alpha = 45 ^ { \circ }\). The company tests the ball using a machine that projects it from ground level across horizontal ground. The speed of projection is set at \(40 \mathrm {~ms} ^ { - 1 }\). When the angle of projection is set at \(30 ^ { \circ }\), the range is found to be 125 m .
3. Comment briefly on the accuracy of the standard projectile model in this situation. The company refines the model by assuming that the ball has a constant deceleration of \(2 \mathrm {~ms} ^ { - 2 }\) in the horizontal direction. In this new model, the resistance to the vertical motion is still neglected and so the flight time is still 4.08 s when the angle of projection is \(30 ^ { \circ }\).
4. Using the new model, with \(\alpha = 30 ^ { \circ }\), show that the horizontal displacement from the point of projection, \(x \mathrm {~m}\) at time \(t \mathrm {~s}\), is given by $$x = 40 t \cos 30 ^ { \circ } - t ^ { 2 }$$ Find the range and hence show that this new model is reasonably accurate in this case. The company then sets the angle of projection to \(45 ^ { \circ }\) while retaining a projection speed of \(40 \mathrm {~ms} ^ { - 1 }\). With this setting the range of the ball is found to be 135 m .
5. Investigate whether the new model is also accurate for this angle of projection.
6. Make one suggestion as to how the model could be further refined. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{7bcde451-5c86-4ed6-b6f5-62c1ad77618c-2_722_1311_192_453} \captionsetup{labelformat=empty} \caption{Fig. 7}
   \end{figure} Fig. 7 shows a platform 10 m long and 2 m high standing on horizontal ground. A small ball projected from the surface of the platform at one end, O , just misses the other end, P . The ball is projected at \(68.5 ^ { \circ }\) to the horizontal with a speed of \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Air resistance may be neglected. At time \(t\) seconds after projection, the horizontal and vertical displacements of the ball from O are \(x \mathrm {~m}\) and \(y \mathrm {~m}\).

1. Obtain expressions, in terms of \(U\) and \(t\), for
   (A) \(x\),
   (B) \(y\).
2. The ball takes \(T\) s to travel from O to P . Show that \(T = \frac { U \sin 68.5 ^ { \circ } } { 4.9 }\) and write down a second equation connecting \(U\) and \(T\).
3. Hence show that \(U = 12.0\) (correct to three significant figures).
4. Calculate the horizontal distance of the ball from the platform when the ball lands on the ground.
5. Use the expressions you found in part (i) to show that the cartesian equation of the trajectory of the ball in terms of \(U\) is $$y = x \tan 68.5 ^ { \circ } - \frac { 4.9 x ^ { 2 } } { U ^ { 2 } \left( \cos 68.5 ^ { \circ } \right) ^ { 2 } }$$ Use this equation to show again that \(U = 12.0\) (correct to three significant figures). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{7bcde451-5c86-4ed6-b6f5-62c1ad77618c-3_391_1480_248_364} \captionsetup{labelformat=empty} \caption{Fig. 7}
   \end{figure} Fig. 7 shows the graph of \(y = \frac { 1 } { 100 } \left( 100 + 15 x - x ^ { 2 } \right)\).
   For \(0 \leqslant x \leqslant 20\), this graph shows the trajectory of a small stone projected from the point Q where \(y \mathrm {~m}\) is the height of the stone above horizontal ground and \(x \mathrm {~m}\) is the horizontal displacement of the stone from O . The stone hits the ground at the point R .
6. Write down the height of Q above the ground.
7. Find the horizontal distance from O of the highest point of the trajectory and show that this point is 1.5625 m above the ground.
8. Show that the time taken for the stone to fall from its highest point to the ground is 0.565 seconds, correct to 3 significant figures.
9. Show that the horizontal component of the velocity of the stone is \(22.1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), correct to 3 significant figures. Deduce the time of flight from Q to R .
10. Calculate the speed at which the stone hits the ground.

5 Small stones A and B are initially in the positions shown in Fig. 6 with B a height \(H \mathrm {~m}\) directly above A. \begin{figure}[h] \end{figure} At the instant when B is released from rest, A is projected vertically upwards with a speed of \(29.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Air resistance may be neglected. The stones collide \(T\) seconds after they begin to move. At this instant they have the same speed, \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and A is still rising. By considering when the speed of A upwards is the same as the speed of B downwards, or otherwise, show that \(T = 1.5\) and find the values of \(V\) and \(H\).