AQA M3 (Mechanics 3) 2015 June

1 A formula for calculating the lift force acting on the wings of an aircraft moving through the air is of the form $$F = k v ^ { \alpha } A ^ { \beta } \rho ^ { \gamma }$$ where \(F\) is the lift force in newtons,
\(k\) is a dimensionless constant,
\(v\) is the air velocity in \(\mathrm { m } \mathrm { s } ^ { - 1 }\),
\(A\) is the surface area of the aircraft's wings in \(\mathrm { m } ^ { 2 }\), and
\(\rho\) is the density of the air in \(\mathrm { kg } \mathrm { m } ^ { - 3 }\).
By using dimensional analysis, find the values of the constants \(\alpha , \beta\) and \(\gamma\).
[0pt] [6 marks]

2 A projectile is launched from a point \(O\) on top of a cliff with initial velocity \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation \(\alpha\) and moves in a vertical plane. During the motion, the position vector of the projectile relative to the point \(O\) is \(( x \mathbf { i } + y \mathbf { j } )\) metres where \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal and vertical unit vectors respectively.

1. Show that, during the motion, the equation of the trajectory of the projectile is given by $$y = x \tan \alpha - \frac { 4.9 x ^ { 2 } } { u ^ { 2 } \cos ^ { 2 } \alpha }$$
2. When \(u = 21\) and \(\alpha = 55 ^ { \circ }\), the projectile hits a small buoy \(B\). The buoy is at a distance \(s\) metres vertically below \(O\) and at a distance \(s\) metres horizontally from \(O\), as shown in the diagram.
   \includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-04_601_935_964_548}
   1. Find the value of \(s\).
   2. Find the acute angle between the velocity of the projectile and the horizontal just before the projectile hits \(B\), giving your answer to the nearest degree.
      [0pt] [5 marks]

3 A disc of mass 0.5 kg is moving with speed \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on a smooth horizontal surface when it receives a horizontal impulse in a direction perpendicular to its direction of motion. Immediately after the impulse, the disc has speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Find the magnitude of the impulse received by the disc.
2. Before the impulse, the disc is moving parallel to a smooth vertical wall, as shown in the diagram. \section*{11/1/1/1/1/1/1/1/1/1/1/1/ Wall} $$\overbrace { 3 \mathrm {~ms} ^ { - 1 } } ^ { \underset { < } { \bigcirc } } \text { Disc }$$ After the impulse, the disc hits the wall and rebounds with speed \(3 \sqrt { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
   Find the coefficient of restitution between the disc and the wall.
   [0pt] [4 marks]

4 Three uniform smooth spheres, \(A , B\) and \(C\), have equal radii and masses \(m , 2 m\) and \(6 m\) respectively. The spheres lie at rest in a straight line on a smooth horizontal surface with \(B\) between \(A\) and \(C\). The sphere \(A\) is projected with speed \(u\) directly towards \(B\) and collides with it.
\includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-10_218_1164_500_438} The coefficient of restitution between \(A\) and \(B\) is \(\frac { 2 } { 3 }\).

1. 1. Show that the speed of \(B\) immediately after the collision is \(\frac { 5 } { 9 } u\).
   2. Find, in terms of \(u\), the speed of \(A\) immediately after the collision.
2. Subsequently, \(B\) collides with \(C\). The coefficient of restitution between \(B\) and \(C\) is \(e\). Show that \(B\) will collide with \(A\) again if \(e > k\), where \(k\) is a constant to be determined.
3. Explain why it is not necessary to model the spheres as particles in this question.
   [0pt] [2 marks]

5 Two smooth spheres, \(A\) and \(B\), have equal radii and masses 2 kg and 1 kg respectively. The spheres move on a smooth horizontal surface and collide. As they collide, \(A\) has velocity \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction inclined at an angle \(\alpha\) to the line of centres of the spheres, and \(B\) has velocity \(2.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in a direction inclined at an angle \(\beta\) to the line of centres, as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-14_458_1068_541_625} The coefficient of restitution between \(A\) and \(B\) is \(\frac { 4 } { 7 }\).
Given that \(\sin \alpha = \frac { 4 } { 5 }\) and \(\sin \beta = \frac { 12 } { 13 }\), find the speeds of \(A\) and \(B\) immediately after the collision.
[0pt] [11 marks]

6 A ship and a navy frigate are a distance of 8 km apart, with the frigate on a bearing of \(120 ^ { \circ }\) from the ship, as shown in the diagram.
\includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-16_451_549_411_760} The ship travels due east at a constant speed of \(50 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). The frigate travels at a constant speed of \(35 \mathrm {~km} \mathrm {~h} ^ { - 1 }\).

1. 1. Find the bearings, to the nearest degree, of the two possible directions in which the frigate can travel to intercept the ship.
      [0pt] [5 marks]
   2. Hence find the shorter of the two possible times for the frigate to intercept the ship.
      [0pt] [5 marks]
2. The captain of the frigate would like the frigate to travel at less than \(35 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). Find the minimum speed at which the frigate can travel to intercept the ship.
   [0pt] [3 marks] \(7 \quad\) A particle is projected from a point \(O\) on a plane which is inclined at an angle \(\theta\) to the horizontal. The particle is projected up the plane with velocity \(u\) at an angle \(\alpha\) above the horizontal. The particle strikes the plane for the first time at a point \(A\). The motion of the particle is in a vertical plane which contains the line \(O A\).
   \includegraphics[max width=\textwidth, alt={}, center]{bcd20c69-cace-408c-8961-169c19ff0231-20_469_624_502_685}
3. Find, in terms of \(u , \theta , \alpha\) and \(g\), the time taken by the particle to travel from \(O\) to \(A\).
4. The particle is moving horizontally when it strikes the plane at \(A\). By using the identity \(\sin ( P - Q ) = \sin P \cos Q - \cos P \sin Q\), or otherwise, show that $$\tan \alpha = k \tan \theta$$ where \(k\) is a constant to be determined.
   [0pt] [5 marks]
   \includegraphics[max width=\textwidth, alt={}]{bcd20c69-cace-408c-8961-169c19ff0231-24_2488_1728_219_141}