3.03d Newton's second law: 2D vectors

2 The acceleration of a particle of mass 4 kg is given by \(\mathbf { a } = ( 9 \mathbf { i } - 4 t \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { 2 }\), where \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors and \(t\) is the time in seconds.

1. Find the acceleration of the particle when \(t = 0\) and also when \(t = 3\).
2. Calculate the force acting on the particle when \(t = 3\). The particle has velocity \(( 4 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } { } ^ { 1 }\) when \(t = 1\).
3. Find an expression for the velocity of the particle at time \(t\).

3 The position vector, \(r\), of a particle of mass 4 kg at time \(t\) is given by $$\mathbf { r } = t ^ { 2 } \mathbf { i } + \left( 5 t - 2 t ^ { 2 } \right) \mathbf { j }$$ where \(\mathbf { i }\) and \(\mathbf { j }\) are the standard unit vectors, lengths are in metres and time is in seconds.

1. Find an expression for the acceleration of the particle. The particle is subject to a force \(\mathbf { F }\) and a force \(12 \mathbf { j } \mathbf { N }\).
2. Find \(\mathbf { F }\).

1 Force \(\mathbf { F } _ { 1 }\) is \(\binom { 6 } { 13 } \mathrm {~N}\) and force \(\mathbf { F } _ { 2 }\) is \(\binom { 3 } { 5 }\), where \({ } _ { 0 }\) and \(\binom { 0 } { 1 }\) are vectors east and north respectively.

1. Calculate the magnitude of \(\mathbf { F } _ { 1 }\), correct to three significant figures.
2. Calculate the direction of the force \(\mathbf { F } _ { 1 } - \mathbf { F } _ { 2 }\) as a bearing. Force \(\mathbf { F } _ { 2 }\) is the resultant of all the forces acting on an object of mass 5 kg .
3. Calculate the acceleration of the object and the change in its velocity after 10 seconds.

2 A train consists of a locomotive pulling 17 identical trucks.
The mass of the locomotive is 120 tonnes and the mass of each truck is 40 tonnes. The locomotive gives a driving force of 121000 N . The resistance to motion on each truck is \(R \mathrm {~N}\) and the resistance on the locomotive is \(5 R \mathrm {~N}\).
Initially the train is travelling on a straight horizontal track and its acceleration is \(0.11 \mathrm {~ms} ^ { - 2 }\).

1. Show that \(R = 1500\).
2. Find the tensions in the couplings between
   (A) the last two trucks,
   (B) the locomotive and the first truck. The train now comes to a place where the track goes up a straight, uniform slope at an angle \(\alpha\) with the horizontal, where \(\sin \alpha = \frac { 1 } { 80 }\). The driving force and the resistance forces remain the same as before.
3. Find the magnitude and direction of the acceleration of the train. The train then comes to a straight uniform downward slope at an angle \(\beta\) to the horizontal.
   The driver of the train reduces the driving force to zero and the resistance forces remain the same as before. The train then travels at a constant speed down the slope.
4. Find the value of \(\beta\).

2 The battery on Carol and Martin's car is flat so the car will not start. They hope to be able to "bump start" the car by letting it run down a hill and engaging the engine when the car is going fast enough. Fig. 6.1 shows the road leading away from their house, which is at A . The road is straight, and at all times the car is steered directly along it.

• From A to B the road is horizontal.
• Between B and C , it goes up a hill with a uniform slope of \(1.5 ^ { \circ }\) to the horizontal.
• Between C and D the road goes down a hill with a uniform slope of \(3 ^ { \circ }\) to the horizontal. CD is 100 m . (This is the part of the road where they hope to get the car started.)
• From D to E the road is again horizontal.

\begin{figure}[h] \end{figure} The mass of the car is 750 kg , Carol's mass is 50 kg and Martin's mass is 80 kg .
Throughout the rest of this question, whenever Martin pushes the car, he exerts a force of 300 N along the line of the car.

1. Between A and B, Martin pushes the car and Carol sits inside to steer it. The car has an acceleration of \(0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Show that the resistance to the car's motion is 100 N . Throughout the rest of this question you should assume that the resistance to motion is constant at 100 N .
2. They stop at B and then Martin tries to push the car up the hill BC. Show that Martin cannot push the car up the hill with Carol inside it but can if she gets out.
   Find the acceleration of the car when Martin is pushing it and Carol is standing outside.
3. While between B and C, Carol opens the window of the car and pushes it from outside while steering with one hand. Carol is able to exert a force of 150 N parallel to the surface of the road but at an angle of \(30 ^ { \circ }\) to the line of the car. This is illustrated in Fig. 6.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9bff41e0-7be0-4932-ae50-a612abb3fe19-2_216_425_1964_870} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{figure} Find the acceleration of the car.
4. At C, both Martin and Carol get in the car and, starting from rest, let it run down the hill under gravity. If the car reaches a speed of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) they can get the engine to start.

3 Fig. 7 illustrates a train with a locomotive, L, pulling two trucks, A and B. The locomotive has mass 90 tonnes and is subject to a resistance force of 2000 N .
Each of the trucks \(A\) and \(B\) has mass 30 tonnes and is subject to a resistance force of \(500 N\). \begin{figure}[h] \end{figure} Initially the train is travelling along a straight horizontal track. The locomotive is exerting a driving force of 12000 N .

1. Find the acceleration of the train.
2. Find the tension in the coupling between trucks A and B . When the train is travelling at \(10 \mathrm {~ms} ^ { - 1 }\), a fault occurs with truck A and the resistance to its motion changes from 500 N to 5000 N . The driver reduces the driving force to zero and allows the train to slow down under the resistance forces and come to a stop.
3. Find the distance the train travels while slowing down and coming to a stop. Find also the force in the coupling between trucks A and B while the train is slowing down, and state whether it is a tension or a thrust. The fault in truck A is repaired so that the resistance to its motion is again 500 N . The train continues and comes to a place where the track goes up a uniform slope at an angle of \(\alpha ^ { \circ }\) to the horizontal.
4. When the train is on the slope, it travels at uniform speed. The driving force remains at 12000 N . Find the value of \(\alpha\).
5. Show that the force in the coupling between trucks A and B has the same value that it had in part (ii).

4 Fig. 5 shows blocks of mass 4 kg and 6 kg on a smooth horizontal table. They are connected by a light, inextensible string. As shown, a horizontal force \(F \mathrm {~N}\) acts on the 4 kg block and a horizontal force of 30 N acts on the 6 kg block. The magnitude of the acceleration of the system is \(2 \mathrm {~ms} ^ { - 2 }\). \begin{figure}[h] \end{figure}

1. Find the two possible values of \(F\).
2. Find the tension in the string in each case.

5 Fig. 8.1 shows a sledge of mass 40 kg . It is being pulled across a horizontal surface of deep snow by a light horizontal rope. There is a constant resistance to its motion. The tension in the rope is 120 N . \begin{figure}[h] \end{figure} The sledge is initially at rest. After 10 seconds its speed is \(5 \mathrm {~ms} ^ { - 1 }\).

1. Show that the resistance to motion is 100 N . When the speed of the sledge is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the rope breaks.
   The resistance to motion remains 100 N .
2. Find the speed of the sledge
   (A) 1.6 seconds after the rope breaks,
   (B) 6 seconds after the rope breaks. The sledge is then pushed to the bottom of a ski slope. This is a plane at an angle of \(15 ^ { \circ }\) to the horizontal. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9bff41e0-7be0-4932-ae50-a612abb3fe19-5_263_854_1391_637} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure} The sledge is attached by a light rope to a winch at the top of the slope. The rope is parallel to the slope and has a constant tension of 200 N . Fig. 8.2 shows the situation when the sledge is part of the way up the slope. The ski slope is smooth.
3. Show that when the sledge has moved from being at rest at the bottom of the slope to the point when its speed is \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), it has travelled a distance of 13.0 m (to 3 significant figures). When the speed of the sledge is \(8 \mathrm {~ms} ^ { - 1 }\), this rope also breaks.
4. Find the time between the rope breaking and the sledge reaching the bottom of the slope.

4 As shown in Fig. 4, boxes P and Q are descending vertically supported by a parachute. Box P has mass 75 kg . Box Q has mass 25 kg and hangs from box P by means of a light vertical wire. Air resistance on the boxes should be neglected. At one stage the boxes are slowing in their descent with the parachute exerting an upward vertical force of 1030 N on box P . The acceleration of the boxes is \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\) upwards and the tension in the wire is \(T \mathrm {~N}\). \begin{figure}[h] \end{figure}

1. Draw a labelled diagram showing all the forces acting on box P and another diagram showing all the forces acting on box Q .
2. Write down separate equations of motion for box P and for box Q .
3. Calculate the tension in the wire.

5 A cylindrical tub of mass 250 kg is on a horizontal floor. Resistance to its motion other than that due to friction is negligible. The first attempt to move the tub is by pulling it with a force of 150 N in the \(\mathbf { i }\) direction, as shown in Fig. 8.1. \begin{figure}[h] \end{figure}

1. Calculate the acceleration of the tub if friction is ignored. In fact, there is friction and the tub does not move.
2. Write down the magnitude and direction of the frictional force opposing the pull. Two more forces are now added to the 150 N force in a second attempt to move the tub, as shown in Fig. 8.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-4_497_927_1350_646} \captionsetup{labelformat=empty} \caption{Fig. 5.2}
   \end{figure} Angle \(\theta\) is acute and chosen so that the resultant of the three forces is in the \(\mathbf { i }\) direction.
3. Determine the value of \(\theta\) and the resultant of the three forces. With this resultant force, the tub moves with constant acceleration and travels 1 metre from rest in 2 seconds.
4. Show that the magnitude of the friction acting on the tub is 661 N , correct to 3 significant figures. When the speed of the tub is \(1.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), it comes to a part of the floor where the friction on the tub is 200 N greater. The pulling forces stay the same.
5. Find the velocity of the tub when it has moved a further 1.65 m .

1 Fig. 1.1 shows a circular cylinder of mass 100 kg being raised by a light, inextensible vertical wire AB . There is negligible air resistance. \begin{figure}[h] \end{figure}

1. Calculate the acceleration of the cylinder when the tension in the wire is 1000 N .
2. Calculate the tension in the wire when the cylinder has an upward acceleration of \(0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). The cylinder is now raised inside a fixed smooth vertical tube that prevents horizontal motion but provides negligible resistance to the upward motion of the cylinder. When the wire is inclined at \(30 ^ { \circ }\) to the vertical, as shown in Fig. 1.2, the cylinder again has an upward acceleration of \(0.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-1_308_490_1230_849} \captionsetup{labelformat=empty} \caption{Fig. 1.2}
   \end{figure}
3. Calculate the new tension in the wire.

2 Boxes A and B slide on a smooth, horizontal plane. Box A has a mass of 4 kg and box B a mass of 5 kg . They are connected by a light, inextensible, horizontal wire. Horizontal forces of 9 N and 135 N act on A and B in the directions shown in Fig. 5. \begin{figure}[h] \end{figure} Calculate the tension in the wire joining the boxes.

4 Two trucks, A and B, each of mass 10000 kg , are pulled along a straight, horizontal track by a constant, horizontal force of \(P \mathrm {~N}\). The coupling between the trucks is light and horizontal. This situation and the resistances to motion of the trucks are shown in Fig. 4. \begin{figure}[h] \end{figure} The acceleration of the system is \(0.2 \mathrm {~ms} ^ { 2 }\) in the direction of the pulling force of magnitude \(P\).

1. Calculate the value of \(P\). Truck A is now subjected to an extra resistive force of 2000 N while \(P\) does not change.
2. Calculate the new acceleration of the trucks.
3. Calculate the force in the coupling between the trucks.

5 A train consists of an engine of mass 10000 kg pulling one truck of mass 4000 kg . The coupling between the engine and the truck is light and parallel to the track. The train is accelerating at \(0.25 \mathrm {~m} \mathrm {~s} ^ { 2 }\) along a straight, level track.

1. What is the resultant force on the train in the direction of its motion? The driving force of the engine is 4000 N .
2. What is the resistance to the motion of the train?
3. If the tension in the coupling is 1150 N , what is the resistance to the motion of the truck? With the same overall resistance to motion, the train now climbs a uniform slope inclined at \(3 ^ { \circ }\) to the horizontal with the same acceleration of \(0.25 \mathrm {~m} \mathrm {~s} ^ { 2 }\).
4. What extra driving force is being applied?

5 A particle moves such that at time \(t\) seconds its acceleration is given by $$( 2 \cos t \mathbf { i } - 5 \sin t \mathbf { j } ) \mathrm { m } \mathrm {~s} ^ { - 2 }$$

1. The mass of the particle is 6 kg . Find the magnitude of the resultant force on the particle when \(t = 0\).
2. When \(t = 0\), the velocity of the particle is \(( 2 \mathbf { i } + 10 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find an expression for the velocity of the particle at time \(t\).

4 A particle moves in a horizontal plane under the action of a single force, \(\mathbf { F }\) newtons. The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed east and north respectively. At time \(t\) seconds, the position vector, \(\mathbf { r }\) metres, of the particle is given by $$\mathbf { r } = \left( t ^ { 3 } - 3 t ^ { 2 } + 4 \right) \mathbf { i } + \left( 4 t + t ^ { 2 } \right) \mathbf { j }$$

1. Find an expression for the velocity of the particle at time \(t\).
2. The mass of the particle is 3 kg .
   1. Find an expression for \(\mathbf { F }\) at time \(t\).
   2. Find the magnitude of \(\mathbf { F }\) when \(t = 3\).
3. Find the value of \(t\) when \(\mathbf { F }\) acts due north.

1 The velocity of a particle at time \(t\) seconds is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), where $$\mathbf { v } = \left( 4 + 3 t ^ { 2 } \right) \mathbf { i } + ( 12 - 8 t ) \mathbf { j }$$

1. When \(t = 0\), the particle is at the point with position vector \(( 5 \mathbf { i } - 7 \mathbf { j } ) \mathrm { m }\). Find the position vector, \(\mathbf { r }\) metres, of the particle at time \(t\).
2. Find the acceleration of the particle at time \(t\).
3. The particle has mass 2 kg . Find the magnitude of the force acting on the particle when \(t = 1\).

2 A particle, of mass 50 kg , moves on a smooth horizontal plane. A single horizontal force $$\left[ \left( 300 t - 60 t ^ { 2 } \right) \mathbf { i } + 100 \mathrm { e } ^ { - 2 t } \mathbf { j } \right] \text { newtons }$$ acts on the particle at time \(t\) seconds.
The vectors \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors.

1. Find the acceleration of the particle at time \(t\).
2. When \(t = 0\), the velocity of the particle is \(( 7 \mathbf { i } - 4 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\). Find the velocity of the particle at time \(t\).
3. Calculate the speed of the particle when \(t = 1\).

6 Alice places a toy, of mass 0.4 kg , on a slope. The toy is set in motion with an initial velocity of \(1 \mathrm {~ms} ^ { - 1 }\) down the slope. The resultant force acting on the toy is \(( 2 - 4 v )\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the toy's velocity at time \(t\) seconds after it is set in motion.

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 10 ( v - 0.5 )\).
2. By using \(\int \frac { 1 } { v - 0.5 } \mathrm {~d} v = - \int 10 \mathrm {~d} t\), find \(v\) in terms of \(t\).
3. Find the time taken for the toy's velocity to reduce to \(0.55 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \(7 \quad\) A small bead, of mass \(m\), is suspended from a fixed point \(O\) by a light inextensible string of length \(a\). With the string taut, the bead is at the point \(B\), vertically below \(O\), when it is set into vertical circular motion with an initial horizontal velocity \(u\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{06c3e260-8167-4616-97d4-0f360a376a0f-5_616_613_520_733} The string does not become slack in the subsequent motion. The velocity of the bead at the point \(A\), where \(A\) is vertically above \(O\), is \(v\).

2 A particle moves in a horizontal plane. The vectors \(\mathbf { i }\) and \(\mathbf { j }\) are perpendicular unit vectors in the horizontal plane. At time \(t\) seconds, the velocity of the particle, \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), is given by $$\mathbf { v } = 12 \cos \left( \frac { \pi } { 3 } t \right) \mathbf { i } - 9 t ^ { 2 } \mathbf { j }$$

1. Find an expression for the acceleration of the particle at time \(t\).
2. The particle, which has mass 4 kg , moves under the action of a single force, \(\mathbf { F }\) newtons.
   1. Find an expression for the force \(\mathbf { F }\) in terms of \(t\).
   2. Find the magnitude of \(\mathbf { F }\) when \(t = 3\).
3. When \(t = 3\), the particle is at the point with position vector \(( 4 \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }\). Find the position vector, \(\mathbf { r }\) metres, of the particle at time \(t\).

1 A particle moves in a horizontal plane, in which the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed east and north respectively. At time \(t\) seconds, its position vector, \(\mathbf { r }\) metres, is given by $$\mathbf { r } = \left( 2 t ^ { 3 } - t ^ { 2 } + 6 \right) \mathbf { i } + \left( 8 - 4 t ^ { 3 } + t \right) \mathbf { j }$$

1. Find an expression for the velocity of the particle at time \(t\).
2. 1. Find the velocity of the particle when \(t = \frac { 1 } { 3 }\).
   2. State the direction in which the particle is travelling at this time.
3. Find the acceleration of the particle when \(t = 4\).
4. The mass of the particle is 6 kg . Find the magnitude of the resultant force on the particle when \(t = 4\).

5 The graph shows a model for the resultant horizontal force on a car, which varies as it accelerates from rest for 20 seconds. The mass of the car is 1200 kg . \includegraphics[max width=\textwidth, alt={}, center]{c02cf013-365b-44e2-8c16-aa8209cbe250-4_373_1203_445_390}

1. The acceleration of the car at time \(t\) seconds is \(a \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Show that $$a = \frac { 2 } { 3 } + \frac { t } { 20 } , \text { for } 0 \leqslant t \leqslant 20$$
2. Find an expression for the velocity of the car at time \(t\).
3. Find the distance travelled by the car in the 20 seconds.
4. An alternative model assumes that the resultant force increases uniformly from 900 to 2100 newtons during the 20 seconds. Which term in your expression for the velocity would change as a result of this modification? Explain why.

3 A particle has mass 800 kg . A single force of \(( 2400 \mathbf { i } - 4800 t \mathbf { j } )\) newtons acts on the particle at time \(t\) seconds. No other forces act on the particle.

1. Find the acceleration of the particle at time \(t\).
2. At time \(t = 0\), the velocity of the particle is \(( 6 \mathbf { i } + 30 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The velocity of the particle at time \(t\) is \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). Show that $$\mathbf { v } = ( 6 + 3 t ) \mathbf { i } + \left( 30 - 3 t ^ { 2 } \right) \mathbf { j }$$
3. Initially, the particle is at the point with position vector \(( 2 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }\). Find the position vector, \(\mathbf { r }\) metres, of the particle at time \(t\).

4 A particle has mass 200 kg and moves on a smooth horizontal plane. A single horizontal force, \(\left( 400 \cos \left( \frac { \pi } { 2 } t \right) \mathbf { i } + 600 t ^ { 2 } \mathbf { j } \right)\) newtons, acts on the particle at time \(t\) seconds. The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed east and north respectively.

1. Find the acceleration of the particle at time \(t\).
2. When \(t = 4\), the velocity of the particle is \(( - 3 \mathbf { i } + 56 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Find the velocity of the particle at time \(t\).
3. Find \(t\) when the particle is moving due west.
4. Find the speed of the particle when it is moving due west.
   
   \includegraphics[max width=\textwidth, alt={}]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-09_2484_1709_223_153}

3 A particle moves in a horizontal plane under the action of a single force, \(\mathbf { F }\) newtons. The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are directed east and north respectively. At time \(t\) seconds, the velocity of the particle, \(\mathbf { v } \mathrm { ms } ^ { - 1 }\), is given by $$\mathbf { v } = 4 \mathrm { e } ^ { - 2 t } \mathbf { i } + \left( 6 t - 3 t ^ { 2 } \right) \mathbf { j }$$

1. Find an expression for the acceleration of the particle at time \(t\).
2. The mass of the particle is 5 kg .
   1. Find an expression for the force \(\mathbf { F }\) acting on the particle at time \(t\).
   2. Find the magnitude of \(\mathbf { F }\) when \(t = 0\).
3. Find the value of \(t\) when \(\mathbf { F }\) acts due west.
4. When \(t = 0\), the particle is at the point with position vector \(( 6 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }\). Find the position vector, \(\mathbf { r }\) metres, of the particle at time \(t\).