3.03a Force: vector nature and diagrams

6 A block of mass 50 kg is pulled up a straight hill and passes through points \(A\) and \(B\) with speeds \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) respectively. The distance \(A B\) is 200 m and \(B\) is 15 m higher than \(A\). For the motion of the block from \(A\) to \(B\), find

1. the loss in kinetic energy of the block,
2. the gain in potential energy of the block. The resistance to motion of the block has magnitude 7.5 N.
3. Find the work done by the pulling force acting on the block. The pulling force acting on the block has constant magnitude 45 N and acts at an angle \(\alpha ^ { \circ }\) upwards from the hill.
4. Find the value of \(\alpha\).

1 Two small smooth spheres \(A\) and \(B\) have equal radii and have masses \(m\) and \(k m\) respectively. They are moving in a straight line in the same direction on a smooth horizontal table. The speed of \(A\) is \(u\) and the speed of \(B\) is \(\frac { 2 } { 3 } u\). Sphere \(A\) collides directly with sphere \(B\). The coefficient of restitution between the spheres is \(\frac { 4 } { 5 }\).

1. Show that the speed of \(A\) after the collision is \(\frac { u ( 2 k + 5 ) } { 5 ( k + 1 ) }\).
2. Given that the magnitude of the impulse experienced by \(B\) during the collision is \(\frac { 2 } { 5 } m u\), find the value of \(k\).

3 hours
Additional Materials:
Answer Booklet/Paper
Graph Paper
List of Formulae (MF10) \section*{READ THESE INSTRUCTIONS FIRST} If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet.
Write your Centre number, candidate number and name on all the work you hand in.
Write in dark blue or black pen.
You may use an HB pencil for any diagrams or graphs.
Do not use staples, paper clips, glue or correction fluid.
DO NOT WRITE IN ANY BARCODES. Answer all the questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
Where a numerical value is necessary, take the acceleration due to gravity to be \(10 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
The use of a calculator is expected, where appropriate.
Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit.
You are reminded of the need for clear presentation in your answers.
At the end of the examination, fasten all your work securely together.
[0pt] The number of marks is given in brackets [ ] at the end of each question or part question.

4 \includegraphics[max width=\textwidth, alt={}, center]{f8961f84-c080-4407-a178-45b76f200111-3_561_606_260_767} A uniform rod \(A B\) has mass \(m\) and length \(2 d\). The rod rests in equilibrium on a smooth peg \(C\), with the end \(A\) resting on a rough horizontal plane. The distance \(A C\) is \(2 a\) and the angle between \(A B\) and the horizontal is \(\alpha\), where \(\cos \alpha = \frac { 3 } { 5 }\). A particle of mass \(\frac { 1 } { 2 } m\) is attached to the rod at \(B\) (see diagram). Find the normal reaction at \(A\) and deduce that \(d < \frac { 25 } { 6 } a\). The coefficient of friction between the rod and the plane is \(\mu\). Show that \(\mu \geqslant \frac { 8 d } { 25 a - 6 d }\).

5 \includegraphics[max width=\textwidth, alt={}, center]{f8961f84-c080-4407-a178-45b76f200111-3_533_698_1343_721} A uniform rectangular lamina \(A B C D\), in which \(A B = 8 a\) and \(B C = 6 a\), has mass \(M\). A uniform circular lamina of radius \(\frac { 5 } { 2 } a\) has mass \(\frac { 1 } { 3 } M\). The two laminas are fixed together in the same plane with their centres coinciding at the point \(O\) (see diagram). A particle \(P\) of mass \(\frac { 1 } { 2 } M\) is attached at \(C\). The system is free to rotate about a fixed smooth horizontal axis through \(A\) and perpendicular to the plane \(A B C D\). Show that the moment of inertia of the system about this axis is \(\frac { 2225 } { 24 } M a ^ { 2 }\). The system is released from rest with \(A C\) horizontal and \(D\) below \(A C\). Find, in the form \(k \sqrt { } \left( \frac { g } { a } \right)\), the greatest angular speed in the subsequent motion, giving the value of \(k\) correct to 3 decimal places.
[0pt] [4]

6 \includegraphics[max width=\textwidth, alt={}, center]{d0fa61eb-f320-427e-8883-de224d293933-4_474_831_269_657} Forces of magnitudes \(P \mathrm {~N}\) and 25 N act at right angles to each other. The resultant of the two forces has magnitude \(R \mathrm {~N}\) and makes an angle of \(\theta ^ { \circ }\) with the \(x\)-axis (see diagram). The force of magnitude \(P \mathrm {~N}\) has components - 2.8 N and 9.6 N in the \(x\)-direction and the \(y\)-direction respectively, and makes an angle of \(\alpha ^ { \circ }\) with the negative \(x\)-axis.

1. Find the values of \(P\) and \(R\).
2. Find the value of \(\alpha\), and hence find the components of the force of magnitude 25 N in
   1. the \(x\)-direction,
   2. the \(y\)-direction.
   3. Find the value of \(\theta\).

1 \includegraphics[max width=\textwidth, alt={}, center]{a4cb105b-55d2-4793-95d2-3d791990a1f6-2_341_929_269_609} Forces of magnitudes 10 N and 8 N act in directions as shown in the diagram.

1. Write down in terms of \(\theta\) the component of the resultant of the two forces
   1. parallel to the force of magnitude 10 N ,
   2. perpendicular to the force of magnitude 10 N .
   3. The resultant of the two forces has magnitude 8 N . Show that \(\cos \theta = \frac { 5 } { 8 }\).

3 \includegraphics[max width=\textwidth, alt={}, center]{a9f3480e-7a8a-497d-a26a-b2aba9b05512-2_462_721_1672_712} Two forces have magnitudes \(P \mathrm {~N}\) and \(Q \mathrm {~N}\). The resultant of the two forces has magnitude 12 N and acts in a direction \(40 ^ { \circ }\) clockwise from the force of magnitude \(P \mathrm {~N}\) and \(80 ^ { \circ }\) anticlockwise from the force of magnitude \(Q \mathrm {~N}\) (see diagram). Find the value of \(Q\).

5 A force of magnitude \(F \mathrm {~N}\) acts in a horizontal plane and has components 27.5 N and - 24 N in the \(x\)-direction and the \(y\)-direction respectively. The force acts at an angle of \(\alpha ^ { \circ }\) below the \(x\)-axis.

1. Find the values of \(F\) and \(\alpha\). A second force, of magnitude 87.6 N , acts in the same plane at \(90 ^ { \circ }\) anticlockwise from the force of magnitude \(F \mathrm {~N}\). The resultant of the two forces has magnitude \(R \mathrm {~N}\) and makes an angle of \(\theta ^ { \circ }\) with the positive \(x\)-axis.
2. Find the values of \(R\) and \(\theta\).

4 A particle \(P\) starts from rest and moves in a straight line for 18 seconds. For the first 8 seconds of the motion \(P\) has constant acceleration \(0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Subsequently \(P\) 's velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t\) seconds after the motion started, is given by $$v = - 0.1 t ^ { 2 } + 2.4 t - k ,$$ where \(8 \leqslant t \leqslant 18\) and \(k\) is a constant.

1. Find the value of \(v\) when \(t = 8\) and hence find the value of \(k\).
2. Find the maximum velocity of \(P\).
3. Find the displacement of \(P\) from its initial position when \(t = 18\).

6 \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Particles \(P\) and \(Q\) have a total mass of 1 kg . The particles are attached to opposite ends of a light inextensible string which passes over a smooth fixed pulley. \(P\) is held at rest and \(Q\) hangs freely, with both straight parts of the string vertical. Both particles are at a height of \(h \mathrm {~m}\) above the floor (see Fig. 1). \(P\) is released from rest and the particles start to move with the string taut. Fig. 2 shows the velocity-time graphs for \(P\) 's motion and for \(Q\) 's motion, where the positive direction for velocity is vertically upwards. Find

1. the magnitude of the acceleration with which the particles start to move and the mass of each of the particles,
2. the value of \(h\),
3. the greatest height above the floor reached by particle \(P\).

7 \includegraphics[max width=\textwidth, alt={}, center]{9c7e8624-c4cd-4a8e-83d9-f92d0bd6f95b-4_668_848_260_653} A small block of mass 3 kg is initially at rest at the bottom \(O\) of a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = 0.6\) and \(\cos \alpha = 0.8\). A force of magnitude 35 N acts on the block at an angle \(\beta\) above the plane, where \(\sin \beta = 0.28\) and \(\cos \beta = 0.96\). The block starts to move up a line of greatest slope of the plane and passes through a point \(A\) with speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The distance \(O A\) is 12.5 m (see diagram).

1. For the motion of the block from \(O\) to \(A\), find the work done against the frictional force acting on the block.
2. Find the coefficient of friction between the block and the plane. At the instant that the block passes through \(A\) the force of magnitude 35 N ceases to act.
3. Find the distance the block travels up the plane after passing through \(A\). \end{document}

3. Three forces \(\mathbf { F } _ { 1 } , \mathbf { F } _ { 2 }\) and \(\mathbf { F } _ { 3 }\) acting on a particle \(P\) are given by $$\begin{aligned} & \mathbf { F } _ { 1 } = ( 7 \mathbf { i } - 9 \mathbf { j } ) \mathrm { N } \\ & \mathbf { F } _ { 2 } = ( 5 \mathbf { i } + 6 \mathbf { j } ) \mathrm { N } \\ & \mathbf { F } _ { 3 } = ( p \mathbf { i } + q \mathbf { j } ) \mathrm { N } \end{aligned}$$ where \(p\) and \(q\) are constants.
Given that \(P\) is in equilibrium,

1. find the value of \(p\) and the value of \(q\). The force \(\mathbf { F } _ { 3 }\) is now removed. The resultant of \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) is \(\mathbf { R }\). Find
2. the magnitude of \(\mathbf { R }\),
3. the angle, to the nearest degree, that the direction of \(\mathbf { R }\) makes with \(\mathbf { j }\).

7. Two forces \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) act on a particle \(P\). The force \(\mathbf { F } _ { 1 }\) is given by \(\mathbf { F } _ { 1 } = ( - \mathbf { i } + 2 \mathbf { j } ) \mathrm { N }\) and \(\mathbf { F } _ { 2 }\) acts in the direction of the vector \(( \mathbf { i } + \mathbf { j } )\).
Given that the resultant of \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) acts in the direction of the vector ( \(\mathbf { i } + 3 \mathbf { j }\) ),

1. find \(\mathbf { F } _ { 2 }\) (7) The acceleration of \(P\) is \(( 3 \mathbf { i } + 9 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }\). At time \(t = 0\), the velocity of \(P\) is \(( 3 \mathbf { i } - 22 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\)
2. Find the speed of \(P\) when \(t = 3\) seconds.
   

6. [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors due east and due north respectively] Two forces \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) act on a particle \(P\) of mass 0.5 kg . \(\mathbf { F } _ { 1 } = ( 4 \mathbf { i } - 6 \mathbf { j } ) \mathrm { N }\) and \(\mathbf { F } _ { 2 } = ( p \mathbf { i } + q \mathbf { j } ) \mathrm { N }\).
Given that the resultant force of \(\mathbf { F } _ { 1 }\) and \(\mathbf { F } _ { 2 }\) is in the same direction as \(- 2 \mathbf { i } - \mathbf { j }\),

1. show that \(p - 2 q = - 16\) Given that \(q = 3\)
2. find the magnitude of the acceleration of \(P\),
3. find the direction of the acceleration of \(P\), giving your answer as a bearing to the nearest degree. XXXXXXXXXXIXITEINTIIS AREA XX女X女X女X女X DO NOT WIRIE IN THS AREA.

3. \begin{figure}[h] \end{figure} Two forces \(\mathbf { P }\) and \(\mathbf { Q }\) act on a particle at a point \(O\). Force \(\mathbf { P }\) has magnitude 6 N and force \(\mathbf { Q }\) has magnitude 7 N . The angle between the line of action of \(\mathbf { P }\) and the line of action of \(\mathbf { Q }\) is \(120 ^ { \circ }\), as shown in Figure 1. The resultant of \(\mathbf { P }\) and \(\mathbf { Q }\) is \(\mathbf { R }\). Find

1. the magnitude of \(\mathbf { R }\),
2. the angle between the line of action of \(\mathbf { R }\) and the line of action of \(\mathbf { P }\).

6. A force \(\mathbf { F }\) is given by \(\mathbf { F } = ( 10 \mathbf { i } + \mathbf { j } ) \mathrm { N }\).

1. Find the exact value of the magnitude of \(\mathbf { F }\).
2. Find, in degrees, the size of the angle between the direction of \(\mathbf { F }\) and the direction of the vector \(( \mathbf { i } + \mathbf { j } )\). The resultant of the force \(\mathbf { F }\) and the force \(( - 15 \mathbf { i } + a \mathbf { j } ) \mathrm { N }\), where \(a\) is a constant, is parallel to, but in the opposite direction to, the vector \(( 2 \mathbf { i } - 3 \mathbf { j } )\).
3. Find the value of \(a\). \includegraphics[max width=\textwidth, alt={}, center]{916543cb-14f7-486c-ba3c-eda9be134045-19_104_59_2613_1886}

5. A particle is acted upon by two forces \(\mathbf { F }\) and \(\mathbf { G }\). The force \(\mathbf { F }\) has magnitude 8 N and acts in a direction with a bearing of \(240 ^ { \circ }\). The force \(\mathbf { G }\) has magnitude 10 N and acts due South. Given that \(\mathbf { R } = \mathbf { F } + \mathbf { G }\), find

1. the magnitude of \(\mathbf { R }\),
2. the direction of \(\mathbf { R }\), giving your answer as a bearing to the nearest degree. in a direction with a bearing of \(240 ^ { \circ }\). The force \(\mathbf { G }\) has magnitude 10 N and acts due South. Given that \(\mathbf { R } = \mathbf { F } + \mathbf { G }\), find

4. \begin{figure}[h] \end{figure} A particle \(P\) of mass \(m \mathrm {~kg}\) is attached to one end of a light inextensible string of length 2.5 m . The other end of the string is attached to a fixed point \(A\) on a vertical wall. The tension in the string is 16 N . The particle is held in equilibrium by a force of magnitude \(F\) newtons, acting in the vertical plane which is perpendicular to the wall and contains the string. This force acts in a direction perpendicular to the string, as shown in Figure 2. Given that the horizontal distance of \(P\) from the wall is 1.5 m , find

1. the value of \(F\),
2. the value of \(m\).
   \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5f2d38d9-b719-4205-8cb0-caa959afc46f-16_186_830_292_557} \captionsetup{labelformat=empty} \caption{Figure 3}
   \end{figure} Two posts, \(A\) and \(B\), are fixed at the side of a straight horizontal road and are 816 m apart, as shown in Figure 3. A car and a van are at rest side by side on the road and level with \(A\). The car and the van start to move at the same time in the direction \(A B\). The car accelerates from rest with constant acceleration until it reaches a speed of \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The car then moves at a constant speed of \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The van accelerates from rest with constant acceleration for 12 s until it reaches a speed of \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The van then moves at a constant speed of \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When the car has been moving at \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) for 30 s , the van draws level with the car at \(B\), and each vehicle has then travelled a distance of 816 m .
   1. Sketch, on the same diagram, a speed-time graph for the motion of each vehicle from \(A\) to \(B\).
   2. Find the time for which the car is accelerating.
   3. Find the value of \(V\).

3 A force \(\mathbf { F }\) is given by \(\mathbf { F } = ( 3.5 \mathbf { i } + 12 \mathbf { j } ) \mathrm { N }\), where \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors east and north respectively.

1. Calculate the magnitude of \(\mathbf { F }\) and also its direction as a bearing.
2. \(\mathbf { G }\) is the force \(( 7 \mathbf { i } + 24 \mathbf { j } )\) N. Show that \(\mathbf { G }\) and \(\mathbf { F }\) are in the same direction and compare their magnitudes.
3. Force \(\mathbf { F } _ { 1 }\) is \(( 9 \mathbf { i } - 18 \mathbf { j } ) \mathrm { N }\) and force \(\mathbf { F } _ { 2 }\) is \(( 12 \mathbf { i } + q \mathbf { j } ) \mathrm { N }\). Find \(q\) so that the sum \(\mathbf { F } _ { 1 } + \mathbf { F } _ { 2 }\) is in the direction of \(\mathbf { F }\).

3 In this question, \(\mathbf { i }\) is a horizontal unit vector and \(\mathbf { j }\) is a unit vector pointing vertically upwards.
A force \(\mathbf { F }\) is \(- \mathbf { i } + 5 \mathbf { j }\).

1. Calculate the magnitude of \(\mathbf { F }\). Calculate also the angle between \(\mathbf { F }\) and the upward vertical. Force \(\mathbf { G }\) is \(2 a \mathbf { i } + a \mathbf { j }\) and force \(\mathbf { H }\) is \(- 2 \mathbf { i } + 3 b \mathbf { j }\), where \(a\) and \(b\) are constants. The force \(\mathbf { H }\) is the resultant of forces \(4 \mathbf { F }\) and \(\mathbf { G }\).
2. Find \(\mathbf { G }\) and \(\mathbf { H }\).

2 Fig. 2 shows a small object, P , of weight 20 N , suspended by two light strings. The strings are tied to points A and B on a sloping ceiling which is at an angle of \(60 ^ { \circ }\) to the upward vertical. The string AP is at \(60 ^ { \circ }\) to the downward vertical and the string BP makes an angle of \(30 ^ { \circ }\) with the ceiling. The object is in equilibrium. \begin{figure}[h] \end{figure}

1. Show that \(\angle \mathrm { APB } = 90 ^ { \circ }\).
2. Draw a labelled triangle of forces to represent the three forces acting on P .
3. Hence, or otherwise, find the tensions in the two strings.

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.

1 Fig. 1 shows a block of mass \(M \mathrm {~kg}\) being pushed over level ground by means of a light rod. The force, \(T \mathrm {~N}\), this exerts on the block is along the line of the rod. The ground is rough.
The rod makes an angle \(\alpha\) with the horizontal. \begin{figure}[h] \end{figure}

1. Draw a diagram showing all the forces acting on the block.
2. You are given that \(M = 5 , \alpha = 60 ^ { \circ } , T = 40\) and the acceleration of the block is \(1.5 \mathrm {~ms} ^ { - 2 }\). Find the frictional force.

4. A car of mass 750 kg is moving up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 1 } { 15 }\). The resistance to motion of the car from non-gravitational forces has constant magnitude R newtons. The power developed by the car's engine is 15 kW and the car is moving at a constant speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that \(\mathrm { R } = 260\). The power developed by the car's engine is now increased to 18 kW . The magnitude of the resistance to motion from non-gravitational forces remains at 260 N . At the instant when the car is moving up the road at \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) the car's acceleration is a \(\mathrm { m } \mathrm { s } ^ { - 2 }\).
2. Find the value of a.