Make it stop!

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Each of the following situations, A-D, shows a system of forces acting on a particle of given mass and the initial velocity of the particle. Work out the single extra constant force that can be added to this system, so that the particle would be brought instantaneously to rest at the specified time.

	Mass and forces	Initial velocity	Time when at rest
A			\(\quantity{4}{s}\)

Without any extra force, this \(\quantity{6}{kg}\) particle is moving with velocity \(\quantity{8}{m\,s^{-1}}\) and accelerating perpendicular to this, due to the \(\quantity{60}{N}\) force.

How will the force of magnitude \(\quantity{F}{N}\) bring the particle to rest?

To work out the actual value of \(F\) and the angle \(\theta\), let’s consider perpendicular components along the dotted lines in the diagram.

Firstly, we know that \(F\sin \theta = 60\) because we want to balance out the \(\quantity{60}{N}\) force.

Secondly, \(F\cos \theta\) needs to provide the acceleration to bring the \(\quantity{6}{kg}\) particle to rest in \(4\) seconds. Using \(v=u+at\) and taking left to be the positive direction, we have \(0=-8+4a.\) So the acceleration we need is \(\quantity{2}{m\,s^{-2}}\) in the opposite direction to the initial velocity. It follows from Newton’s Second Law that \(F\cos \theta=12.\)

We now know the perpendicular components of the force, \(F\cos \theta\) and \(F\sin \theta\), so we can find \(F\) and \(\theta\). \[F=\sqrt{60^2+12^2}=\sqrt{3744}=61.18...\] so the force has magnitude \(\quantity{61.2}{N}\,\,\,(3\,s.f.)\) and we now need to work out its direction.

As \(\tan \theta =\dfrac{60}{12}=5\), we have \(\theta=\arctan{5}=78.69...\) so \(\theta=79^{\circ}\) to the nearest degree.

With the extra force added, the resultant force on the particle is acting in the opposite direction to the initial velocity. Therefore the particle will be restricted to moving in a straight line, reversing direction after \(4\) seconds. Without this force, the particle would have maintained the \(\quantity{8}{m\,s^{-1}}\) component of its initial velocity, but also accelerated in the direction of the \(\quantity{60}{N}\) force at a rate of \(\quantity{10}{m\,s^{-2}}\), so the component of velocity in this direction after time \(t\) seconds would have been \(\quantity{10t}{m\,s^{-1}}.\) Try to sketch this path of the particle.

	Mass and forces	Initial velocity	Time when at rest
B			\(\quantity{2}{s}\)

For this example we’ll use two approaches.

Momentum and impulse, using unit vectors

We’ll use unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) where the directions of \(\mathbf{i}\) and \(\mathbf{j}\) are shown in this diagram.

As we know the mass and initial velocity of the particle, we know its initial momentum. We want to find the force that over \(\quantity{2}{s}\) will provide an impulse which brings the particle to rest.

In terms of \(\mathbf{i}\) and \(\mathbf{j}\), the initial velocity is \(\quantity{(6\cos 30^{\circ}\mathbf{i}+6\sin 30^{\circ}\mathbf{j})}{m\,s^{-1}}.\) This is \(\quantity{\left(6\times\tfrac{\sqrt{3}}{2}\mathbf{i}+6\times\tfrac{1}{2}\mathbf{j}\right)}{m\,s^{-1}}\), which simplifies to \(\quantity{3(\sqrt{3}\mathbf{i}+\mathbf{j})}{m\,s^{-1}}.\)

Therefore the initial momentum is \(\quantity{15\left(\sqrt{3}\mathbf{i}+\mathbf{j}\right)}{Ns}.\)

The total impulse, \(\mathbf{I}\), (due to forces acting for \(\quantity{2}{s}\)) needs to satisfy \(15\left(\sqrt{3}\mathbf{i}+\mathbf{j}\right)+\mathbf{I}=\mathbf{0}.\)

How do the forces acting on the particle provide this impulse?

The \(\quantity{-30\mathbf{j}}{N}\) force contributes an impulse in the direction we want, but this is likely to be too large for the stopping time required. We also need impulse in the direction of \(-\mathbf{i}\) so we need a further force to act on the particle. We’ll write this extra force as \(\quantity{(b\mathbf{i}+c\mathbf{j})}{N}\) where we expect \(b\) to be negative and \(c\) to be positive.

Therefore \(15\left(\sqrt{3}\mathbf{i}+\mathbf{j}\right) + 2(-30\mathbf{j} + b\mathbf{i}+c\mathbf{j})=\mathbf{0}\),

which gives us two equations to solve: \(2b=-15\sqrt{3}\) and \(2(c-30)=-15.\)

Solving these gives us the required force of \(\quantity{(-13.0\mathbf{i}+22.5\mathbf{j})}{N} \,\,\,\text{(3 s.f.)}.\) As expected, \(b\) is negative and \(c\) is positive.

Force triangle

The acceleration needed to bring this particle to instantaneous rest is \(\quantity{3}{m\,s^{-2}}\) in the opposite direction to the initial velocity. This acceleration is provided by a force of magnitude \(\quantity{15}{N}.\)

If there were no other forces acting on the particle, the \(\quantity{15}{N}\) force would be the extra force we’d need. But we need to work out what extra force we can apply so that the \(\quantity{15}{N}\) force is the resultant of the \(\quantity{30}{N}\) force and this extra force.

Using a vector triangle, as shown, we can work out the magnitude \(F\) and direction \(\theta\) of the extra force.

The angle between the \(\quantity{15}{N}\) force and the \(\quantity{30}{N}\) force is \(60^{\circ}\) and \(\cos 60^{\circ}=\tfrac{15}{30}\), so the vector triangle shown is right-angled, with \(\theta=30^{\circ}\) and \(F^2=675.\) It follows that the extra force we need to apply has magnitude \(\quantity{15\sqrt{3}}{N}\) and is directed at \(30^{\circ}\) to the upward dotted line, which means it is perpendicular to the direction of the initial velocity.

If the triangle had not been right-angled, how could we have found \(F\) and \(\theta\)?

With the extra force in place the impulse is directed in the opposite direction to the initial momentum, so the particle will move in a straight line, coming to rest after \(2\) seconds as it reverses direction along this line. Without this extra force, the component of momentum in the direction of \(\mathbf{i}\) would have been maintained, but the particle would have received an impulse in the \(-\mathbf{j}\) direction, first bringing the momentum in this direction to zero, but after this the velocity and momentum would have had components in the positive \(\mathbf{i}\) and negative \(\mathbf{j}\) directions.

	Mass and forces	Initial velocity	Time when at rest
C			\(\quantity{3}{s}\)

Why will the extra force shown bring this particle to rest in \(\quantity{3}{s}\)?

If the extra force is applied, the particle will continue to move in the direction of its initial velocity, travelling in a straight line but reversing direction after \(5\) seconds. Without this force, the component of the particle’s velocity parallel to its initial velocity would have decreased to zero and then reversed direction, whereas the component of velocity perpendicular to this would have continued to increase from zero, so at time \(\quantity{3}{s}\) the particle would have been moving perpendicular to its initial velocity. Can you sketch the two paths of the particle?

	Mass and forces	Initial velocity	Time when at rest
D			\(\quantity{2}{s}\)

Why will the extra force shown bring this particle to rest in \(\quantity{2}{s}\)?

If the extra force is applied, the particle will move in a straight line, reversing direction after \(2\) seconds. If the extra force had not been applied, the particle would have continued to move along the same straight line, but as it would be in equilibrium, it would have maintained its speed of \(\quantity{8}{m\,s^{-1}}.\)