3.03d Newton's second law: 2D vectors

\includegraphics{figure_2} Two particles \(P\) and \(Q\) have masses 0.3 kg and \(m\) kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a fixed rough plane. The plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac{3}{4}\). The coefficient of friction between \(P\) and the plane is \(\frac{1}{2}\). The string lies in a vertical plane through a line of greatest slope of the inclined plane. The particle \(P\) is held at rest on the inclined plane and the particle \(Q\) hangs freely below the pulley with the string taut, as shown in Figure 2. The system is released from rest and \(Q\) accelerates vertically downwards at 1.4 m s\(^{-2}\). Find

1. the magnitude of the normal reaction of the inclined plane on \(P\), [2]
2. the value of \(m\). [8]

When the particles have been moving for 0.5 s, the string breaks. Assuming that \(P\) does not reach the pulley,

1. find the further time that elapses until \(P\) comes to instantaneous rest. [6]

\includegraphics{figure_2} A fixed rough plane is inclined at 30° to the horizontal. A small smooth pulley \(P\) is fixed at the top of the plane. Two particles \(A\) and \(B\), of mass 2 kg and 4 kg respectively, are attached to the ends of a light inextensible string which passes over the pulley \(P\). The part of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane and \(B\) hangs freely below \(P\), as shown in Figure 2. The coefficient of friction between \(A\) and the plane is \(\frac{1}{\sqrt{3}}\). Initially \(A\) is held at rest on the plane. The particles are released from rest with the string taut and \(A\) moves up the plane. Find the tension in the string immediately after the particles are released. [9]

\includegraphics{figure_4} A truck of mass 1750 kg is towing a car of mass 750 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is inclined at an angle \(\theta\) to the road, as shown in Figure 4. The vehicles are travelling at 20 m s\(^{-1}\) as they enter a zone where the speed limit is 14 m s\(^{-1}\). The truck's brakes are applied to give a constant braking force on the truck. The distance travelled between the instant when the brakes are applied and the instant when the speed of each vehicle is 14 m s\(^{-1}\) is 100 m.

1. Find the deceleration of the truck and the car. [3]

The constant braking force on the truck has magnitude \(R\) newtons. The truck and the car also experience constant resistances to motion of 500 N and 300 N respectively. Given that cos \(\theta = 0.9\), find

1. the force in the towbar, [4]
2. the value of \(R\). [4]

A woman travels in a lift. The mass of the woman is 50 kg and the mass of the lift is 950 kg. The lift is being raised vertically by a vertical cable which is attached to the top of the lift. The lift is moving upwards and has constant deceleration of \(2 \text{ m s}^{-2}\). By modelling the cable as being light and inextensible, find

1. the tension in the cable; [3]
2. the magnitude of the force exerted on the woman by the floor of the lift. [3]

\includegraphics{figure_2} Two particles \(A\) and \(B\) have masses \(2m\) and \(3m\) respectively. The particles are attached to the ends of a light inextensible string. Particle \(A\) is held at rest on a smooth horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table. Particle \(B\) hangs at rest vertically below the pulley with the string taut, as shown in Figure 2. Particle \(A\) is released from rest. Assuming that \(A\) has not reached the pulley, find

1. the acceleration of \(B\), [5]
2. the tension in the string, [1]
3. the magnitude and direction of the force exerted on the pulley by the string. [4]

A particle \(P\) of mass 3 kg is moving under the action of a constant force \(\mathbf{F}\) newtons. At \(t = 0\), \(P\) has velocity \((3\mathbf{i} - 5\mathbf{j})\) m s\(^{-1}\). At \(t = 4\) s, the velocity of \(P\) is \((-5\mathbf{i} + 11\mathbf{j})\) m s\(^{-1}\). Find

1. the acceleration of \(P\), in terms of \(\mathbf{i}\) and \(\mathbf{j}\). [2]
2. the magnitude of \(\mathbf{F}\). [4]

At \(t = 6\) s, \(P\) is at the point \(A\) with position vector \((6\mathbf{i} - 29\mathbf{j})\) m relative to a fixed origin \(O\). At this instant the force \(\mathbf{F}\) newtons is removed and \(P\) then moves with constant velocity. Three seconds after the force has been removed, \(P\) is at the point \(B\).

1. Calculate the distance of \(B\) from \(O\). [6]

A particle \(P\) moves in a horizontal plane. The acceleration of \(P\) is \((-\mathbf{i} + 2\mathbf{j}) \text{ m s}^{-2}\). At time \(t = 0\), the velocity of \(P\) is \((2\mathbf{i} - 3\mathbf{j}) \text{ m s}^{-1}\).

1. Find, to the nearest degree, the angle between the vector \(\mathbf{j}\) and the direction of motion of \(P\) when \(t = 0\). [3]

At time \(t\) seconds, the velocity of \(P\) is \(\mathbf{v} \text{ m s}^{-1}\). Find

1. an expression for \(\mathbf{v}\) in terms of \(t\), in the form \(a\mathbf{i} + b\mathbf{j}\), [2]
2. the speed of \(P\) when \(t = 3\), [3]
3. the time when \(P\) is moving parallel to \(\mathbf{i}\). [2]

A van of mass 1500 kg is driving up a straight road inclined at an angle \(α\) to the horizontal, where \(\sin α = \frac{1}{16}\). The resistance to motion due to non-gravitational forces is modelled as a constant force of magnitude 1000 N. Given that initially the speed of the van is 30 m s\(^{-1}\) and that the van's engine is operating at a rate of 60 kW,

1. calculate the magnitude of the initial deceleration of the van. [4]

When travelling up the same hill, the rate of working of the van's engine is increased to 80 kW. Using the same model for the resistance due to non-gravitational forces,

1. calculate in m s\(^{-1}\) the constant speed which can be sustained by the van at this rate of working. [4]

1. Give one reason why the use of this model for resistance may mean that your answer to part (b) is too high. [1]

A particle \(P\) of mass \(0.3\) kg is moving under the action of a single force \(F\) newtons. At time \(t\) seconds the velocity of \(P\), v m s\(^{-1}\), is given by $$\mathbf{v} = 3t^2\mathbf{i} + (6t - 4)\mathbf{j}.$$

1. Calculate, to 3 significant figures, the magnitude of \(\mathbf{F}\) when \(t = 2\). [5]

When \(t = 0\), \(P\) is at the point \(A\). The position vector of \(A\) with respect to a fixed origin \(O\) is \((3\mathbf{i} - 4\mathbf{j})\) m. When \(t = 4\), \(P\) is at the point \(B\).

1. Find the position vector of \(B\). [5]

A car of mass 1000 kg is moving along a straight horizontal road with a constant acceleration of \(j\) m s\(^{-2}\). The resistance to motion is modelled as a constant force of magnitude 1200 N. When the car is travelling at 12 m s\(^{-1}\), the power generated by the engine of the car is 24 kW.

1. Calculate the value of \(j\). [4]

When the car is travelling at 14 m s\(^{-1}\), the engine is switched off and the car comes to rest, without braking, in a distance of \(d\) metres. Assuming the same model for resistance,

1. use the work-energy principle to calculate the value of \(d\). [3]

1. Give a reason why the model used for the resistance to motion may not be realistic. [1]

A uniform ladder \(AB\), of mass \(m\) and length \(2a\), has one end \(A\) on rough horizontal ground. The other end \(B\) rests against a smooth vertical wall. The ladder is in a vertical plane perpendicular to the wall. The ladder makes an angle \(α\) with the horizontal, where \(\tan α = \frac{4}{3}\). A child of mass \(2m\) stands on the ladder at \(C\) where \(AC = \frac{1}{4}a\), as shown in Fig. 1. The ladder and the child are in equilibrium. By modelling the ladder as a rod and the child as a particle, calculate the least possible value of the coefficient of friction between the ladder and the ground. [9]

A van of mass 1500 kg is driving up a straight road inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac{1}{12}\). The resistance to motion due to non-gravitational forces is modelled as a constant force of magnitude 1000 N. Given that initially the speed of the van is 30 m s\(^{-1}\) and that the van's engine is working at a rate of 60 kW,

1. calculate the magnitude of the initial deceleration of the van. [4]

When travelling up the same hill, the rate of working of the van's engine is increased to 80 kW. Using the same model for the resistance due to non-gravitational forces,

1. calculate in m s\(^{-1}\) the constant speed which can be sustained by the van at this rate of working. [4]
2. Give one reason why the use of this model for resistance may mean that your answer to part (b) is too high. [1]

A particle \(P\) of mass 0.3 kg is moving under the action of a single force \(\mathbf{F}\) newtons. At time \(t\) seconds the velocity of \(P\), \(\mathbf{v}\) m s\(^{-1}\), is given by $$\mathbf{v} = 3t\mathbf{i} + (6t - 4)\mathbf{j}.$$

1. Calculate, to 3 significant figures, the magnitude of \(\mathbf{F}\) when \(t = 2\). [5]

When \(t = 0\), \(P\) is at the point \(A\). The position vector of \(A\) with respect to a fixed origin \(O\) is \((3\mathbf{i} - 4\mathbf{j})\) m. When \(t = 4\), \(P\) is at the point \(B\).

1. Find the position vector of \(B\). [5]

A car of mass 1000 kg is moving along a straight horizontal road with a constant acceleration of \(f\) m s\(^{-2}\). The resistance to motion is modelled as a constant force of magnitude 1200 N. When the car is travelling at 12 m s\(^{-1}\), the power generated by the engine of the car is 24 kW.

1. Calculate the value of \(f\). [4]

When the car is travelling at 14 m s\(^{-1}\), the engine is switched off and the car comes to rest, without braking, in a distance of \(d\) metres. Assuming the same model for resistance,

1. use the work-energy principle to calculate the value of \(d\). [3]
2. Give a reason why the model used for the resistance to motion may not be realistic. [1]

A car of mass 1000 kg is moving along a straight horizontal road. The resistance to motion is modelled as a constant force of magnitude \(R\) newtons. The engine of the car is working at a rate of 12 kW. When the car is moving with speed 15 m s\(^{-1}\), the acceleration of the car is 0.2 m s\(^{-2}\).

1. Show that \(R = 600\). [4]

The car now moves with constant speed \(U\) m s\(^{-1}\) downhill on a straight road inclined at \(\theta\) to the horizontal, where \(\sin \theta = \frac{1}{30}\). The engine of the car is now working at a rate of 7 kW. The resistance to motion from non-gravitational forces remains of magnitude \(R\) newtons.

1. Calculate the value of \(U\). [5]

A car of mass 800 kg is moving at a constant speed of 15 m s\(^{-1}\) down a straight road inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac{3}{4}\). The resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 900 N.

1. Find, in kW, the rate of working of the engine of the car. [4]

When the car is travelling down the road at 15 m s\(^{-1}\), the engine is switched off. The car comes to rest in time \(T\) seconds after the engine is switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 900 N.

1. Find the value of \(T\). [4]

A particle \(P\) of mass 0.5 kg is moving under the action of a single force \(\mathbf{F}\) newtons. At time \(t\) seconds, \(\mathbf{F} = (1.5t^2 - 3)\mathbf{i} + 2t\mathbf{j}\). When \(t = 2\), the velocity of \(P\) is \((-4\mathbf{i} + 5\mathbf{j})\) m s\(^{-1}\).

1. Find the acceleration of \(P\) at time \(t\) seconds. [2]
2. Show that, when \(t = 3\), the velocity of \(P\) is \((9\mathbf{i} + 15\mathbf{j})\) m s\(^{-1}\). [5]

When \(t = 3\), the particle \(P\) receives an impulse \(\mathbf{Q}\) N s. Immediately after the impulse the velocity of \(P\) is \((-3\mathbf{i} + 20\mathbf{j})\) m s\(^{-1}\). Find

1. the magnitude of \(\mathbf{Q}\), [3]
2. the angle between \(\mathbf{Q}\) and \(\mathbf{i}\). [3]

A car of mass 1000 kg is moving at a constant speed of 16 m s\(^{-1}\) up a straight road inclined at an angle \(\theta\) to the horizontal. The rate of working of the engine of the car is 20 kW and the resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 550 N.

1. Show that \(\sin \theta = \frac{1}{14}\). [5]

When the car is travelling up the road at 16 m s\(^{-1}\), the engine is switched off. The car comes to rest, without braking, having moved a distance \(y\) metres from the point where the engine was switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 550 N.

1. Find the value of \(y\). [4]

A cyclist and her cycle have a combined mass of \(75\) kg. The cyclist is cycling up a straight road inclined at \(5°\) to the horizontal. The resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude \(20\) N. At the instant when the cyclist has a speed of \(12\) m s\(^{-1}\), she is decelerating at \(0.2\) m s\(^{-2}\).

1. Find the rate at which the cyclist is working at this instant. [5]

When the cyclist passes the point \(A\) her speed is \(8\) m s\(^{-1}\). At \(A\) she stops working but does not apply the brakes. She comes to rest at the point \(B\). The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude \(20\) N.

1. Use the work-energy principle to find the distance \(AB\). [5]

A car has mass 1200 kg. The maximum power of the car's engine is 32 kW. The resistance to motion due to non-gravitational forces is modelled as a force of constant magnitude 800 N. When the car is travelling on a horizontal road at constant speed \(V\) m s\(^{-1}\), the engine of the car is working at maximum power.

1. Find the value of \(V\). [3]

The car now travels downhill on a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac{1}{40}\). The resistance to motion due to non-gravitational forces is still modelled as a force of constant magnitude 800 N. Given that the engine of the car is again working at maximum power,

1. find the acceleration of the car when its speed is 20 m s\(^{-1}\). [4]

A particle \(P\) of mass 0.25 kg moves under the action of a single force \(\mathbf{F}\) newtons. At time \(t\) seconds, the velocity of \(P\) is \(\mathbf{v}\) m s\(^{-1}\), where $$\mathbf{v} = (2 - 4t)\mathbf{i} + (t^2 + 2t)\mathbf{j}$$ When \(t = 0\), \(P\) is at the point with position vector \((2\mathbf{i} - 4\mathbf{j})\) m with respect to a fixed origin \(O\). When \(t = 3\), \(P\) is at the point \(A\). Find

1. the momentum of \(P\) when \(t = 3\), [2]
2. the magnitude of \(\mathbf{F}\) when \(t = 3\), [6]
3. the position vector of \(A\). [5]

\includegraphics{figure_1} A metal ball \(B\) of mass \(m\) is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(A\). The ball \(B\) moves in a horizontal circle with centre \(O\) vertically below \(A\), as shown in Fig. 1. The string makes a constant angle \(\alpha°\) with the downward vertical and \(B\) moves with constant angular speed \(\sqrt{(2gk)}\), where \(k\) is a constant. The tension in the string is \(3mg\). By modelling \(B\) as a particle, find

1. the value of \(\alpha\), [4]
2. the length of the string. [5]

A particle \(P\) of mass 2.5 kg moves along the positive \(x\)-axis. It moves away from a fixed origin \(O\), under the action of a force directed away from \(O\). When \(OP = x\) metres the magnitude of the force is \(2e^{-0.1x}\) newtons and the speed of \(P\) is \(v\) m s\(^{-1}\). When \(x = 0\), \(v = 2\). Find

1. \(v^2\) in terms of \(x\), [6]
2. the value of \(x\) when \(v = 4\). [3]
3. Give a reason why the speed of \(P\) does not exceed \(\sqrt{20}\) m s\(^{-1}\). [1]

\includegraphics{figure_3} A particle of mass \(m\) is attached to one end of a light inextensible string of length \(l\). The other end of the string is attached to a fixed point \(O\). The particle is hanging at the point \(A\), which is vertically below \(O\). It is projected horizontally with speed \(u\). When the particle is at the point \(P\), \(\angle AOP = \theta\), as shown in Fig. 3. The string oscillates through an angle \(\alpha\) on either side of \(OA\) where \(\cos \alpha = \frac{2}{3}\).

1. Find \(u\) in terms of \(g\) and \(l\). [4]

When \(\angle AOP = \theta\), the tension in the string is \(T\).

1. Show that \(T = \frac{mg}{3}(9\cos\theta - 4)\). [6]
2. Find the range of values of \(T\). [4]

A toy car of mass \(0.2\) kg is travelling in a straight line on a horizontal floor. The car is modelled as a particle. At time \(t = 0\) the car passes through a fixed point \(O\). After \(t\) seconds the speed of the car is \(v \text{ m s}^{-1}\) and the car is at a point \(P\) with \(OP = x\) metres. The resultant force on the car is modelled as \(\frac{1}{5}x(4 - 3x)\) N in the direction \(OP\). The car comes to instantaneous rest when \(x = 6\). Find

1. an expression for \(v^2\) in terms of \(x\), [7]
2. the initial speed of the car. [2]

\includegraphics{figure_1} A particle \(P\) of mass \(m\) is attached to the ends of two light inextensible strings \(AP\) and \(BP\) each of length \(l\). The ends \(A\) and \(B\) are attached to fixed points, with \(A\) vertically above \(B\) and \(AB = \frac{3}{4}l\), as shown in Fig. 1. The particle \(P\) moves in a horizontal circle with constant angular speed \(\omega\). The centre of the circle is the mid-point of \(AB\) and both strings remain taut.

1. Show that the tension \(AP\) is \(\frac{1}{6}m(3l\omega^2 + 4g)\). [7]
2. Find, in terms of \(m\), \(l\), \(\omega\) and \(g\), an expression for the tension in \(BP\). [2]
3. Deduce that \(\omega^2 \geq \frac{4g}{3l}\). [2]