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 | \(\quantity{4}{s}\) |

There are many ways to solve these problems. We’ll approach the first couple of examples using different mechanical principles and different notation. Both approaches could be used for all four problems, but a particular example may tempt you to choose one approach rather than another.

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?

Can we say anything about the approximate values of \(F\) and \(\theta\)? Thinking about this now may help us to check that our final answers are sensible. For example, \(F\sin \theta = 60\), so the magnitude of the force is greater than \(\quantity{60}{N}.\) Is \(\theta\) likely to be greater than or less than \(45^{\circ}\)?

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.

Why do we want to put the particle in equilibrium in this direction?

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.

Does this combination of magnitude and direction seem reasonable?

Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.

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.

Compare these approaches with each other and our approach to the first example.

How did using momentum and impulse compare with using acceleration?

How did using a triangle of forces compare with using components?

- How did using \(\mathbf{i}\) and \(\mathbf{j}\) affect the type of calculations we had to do? Note that we chose to define \(\mathbf{i}\) and \(\mathbf{j}\) in the directions given, but we could have chosen different directions for these. For example, we could have chosen \(\mathbf{i}\) to be in the direction of the initial velocity.

Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.

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.

Before tackling the next two examples, it would be worth considering how they are similar to or different from the first two and each other. How will this affect

the approach you take?

your choice of notation?

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 \(\quantity{3}{s}\)?

Describe how the particle behaves over time if this extra force is applied, assuming all the given forces acting on the particle remain unchanged. Compare this with how the particle would have behaved if the extra force had not been applied.

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 \(\quantity{8}{m\,s^{-1}}.\)