Further Mechanics | Chapter 1: Momentum and Impulse

Curriculum Objective

• recall and use the definition of linear momentum, and show understanding of its vector nature (in one dimension only);
• recall Newton’s experimental law and the definition of the coefficient of restitution, the property 0 ø e ø 1, and the meaning of the terms ‘perfectly elastic’ (e = 1) and ‘inelastic’ (e = 0);
• use conservation of linear momentum and/or Newton’s experimental law to solve problems that may be modelled as the direct impact of two smooth spheres or the direct or oblique impact of a smooth sphere with a fixed surface;
• recall and use the definition of the impulse of a constant force, and relate the impulse acting on a particle to the change of momentum of the particle (in one dimension only)

Introduction

In Classical Mechanics there exists a quantity called Linear Momentum associated with an object that is defined as the product of the mass and the velocity of the body. Which is to say:

\[\bar p = m\bar v\]

Note that momentum, like velocity, is a vector quantity, hence has a magnitude and a direction associated with it. In a more formal treatment the momentum in the above equation (and the velocity) would be a three dimensional vector, however in this course, you will not be required to work in more than two dimensions.

The physical interpretation of this quantity is the ability of body to retain its velocity. A massive truck moving at the same speed as a sedan, would required a greater force, or more time to be brought to a halt, while the sedan would be comparatively easier to be brought to rest.

Ex.

Given that two bodies, of masses 10 and 6 kg, are moving at the velocities of $6\;m{s^{ - 1}}$ and $12\;m{s^{ - 1}}$ respectively. Calculate their momentum and hence use it to predict which body will come to rest quicker if the same force is applied as resistance to the bodies. 

Sol:

The momentum of the bodies respectively are:

$\begin{array}{l} {p_1} = {m_1}{v_1} = 10 \cdot 6 = 60\;kgm{s^{ - 1}}\\ {p_2} = {m_2}{v_2} = 6 \cdot 12 = 72\;kgm{s^{ - 1}} \end{array}$


By the reasoning already stated above, we conclude the body with greater momentum will take more time to be brought to rest.

Working from Newton's Second Law of Motion, we know that:

\[F = ma = m\frac{{dv}}{{dt}}\]

As we will only be working with discrete changes in momentum, we may do away with the infinitesimal changes and the use of calculus and we may simply denote it as:

\[F = m\frac{{\Delta v}}{{\Delta t}} = \frac{{\Delta \left( {mv} \right)}}{{\Delta t}} = \frac{{\Delta p}}{{\Delta t}}\] where $\Delta$ denotes the change in the quantity. Even though we have forgone calculus, you can see how the treatment wouldn't be much different even if we were working with the derivative.

We conclude then that the rate of change in the momentum is equal to the resultant force acting on the particle.

Further note that this gives us a new unit for the momentum, the Newton-second. We see that the change in time multiplied by force is equal to the change in momentum. As the units of change in a quantity are same as those of the quantity itself, we can conclude safely that momentum can indeed be measured in Newton-Seconds.

As we study discrete changes in momentum, consider the following:

\[F \cdot \Delta t = \Delta \left( {mv} \right) = m{v_f} - m{v_i}\]

The change in momentum is ofttimes an important quantity in our study, hence we give it a special name. The Impulse produced by a force is equal to the change in momentum it causes. We note that its units are Ns, same as momentum, and I denote it with $\varphi $, but I am not sure what your examiner prefers. It would be advisable to define every variable you use in the paper for that very reason. 


Ex. 

A truck of mass 1000 kg travelling at 3 metres per second is brought to rest when it hits a buffer in 2 seconds. What force (assumed to be constant) is exerted by the buffer?

Sol. 

$\begin{array}{l} \varphi = m{v_f} - m{v_i} = 0 - 1000\left( { - 3} \right) = 3000\\ \varphi = Ft = 3000\\ t = 2 \Rightarrow F = 1500\,N \end{array}$

Conservation of Momentum

We know already:

$F = \frac{{dp}}{{dt}}$

Hence if there is no external resultant force, F, the change in momentum for a single body will be zero. Naturally after a single body, we talk about a collection of bodies. 

Of a collection of bodies, or a closed system, we note that by the equation given above without an external force acting on the system there will be no change in the momentum. Of course this doesn't cover the interaction between any bodies in the closed system. We also have to consider the effect of the interaction of the bodies in the closed system with each other.

Let us say that two bodies in a closed system interact in some manner, say collide. We call the objects A and B, assume they comprise the entire system for the moment, and further call the force A exerts on B $F_{AB}$ and the force B exerts on A $F_{BA}$. We know from Newton's third law that:

$F_{AB}=-F_{BA}$

$\frac{dp_{B}}{dt}=-\frac{dp_{A}}{dt}$

$\frac{d}{dt}\left (p_{A}+p_{B}  \right )=0$

We see then that despite the fact that the momentum of the bodies has changed, the momentum of the entire closed system remains the same. For more than two bodies you can easily adapt this argument.

We call this the law of conservation of momentum.

Newton's Law of Restitution

We begin with defining an elastic and inelastic collision. An elastic collision is such that when two objects collide, they bounce back. If however they coalesce after the collision, the impact is inelastic. 

For those studying Physics, you must be familiar of the terms, but defined in the context of Kinetic Energy. Different while they may seem, they don't contradict one another, rather suggest much the same thing.

It is generally held that if after a collision the two objects do bounce back, the separation speed (the speed after the collision) will be equal or less than the approach speed (speed ere collision). Experimental evidence suggests that the ratio of these relative speeds is constant. This is known as the Newton's Law of Restitution. It may be written as follows:

\[\frac{{relative\,separation\,speed}}{{relative\,approach\,speed}} = e\]

"e" is then called the coefficient of restitution and it is constant for particular objects. We also note that it must be restricted to $0 \le e \le 1$ ($e > 1$ may also occur but you need not worry about that). This law holds both for two moving objects, and one moving and one fixed object, the law of conservation of momentum however holds for only the first case (why?).

In this range, consider how e can take the values 0 and 1 in particular. 0 may occur when the collision is inelastic, for if it is inelastic then the objects coalesce and move at the same speed together, hence rendering the relative separation speed zero. 1 happens in the special case called a perfectly elastic collision, wherein the particle retain all their Kinetic Energy, and thereby have the same separation speed as approach speed.

This being a mathematical course you are going to need a more mathematically grounded formula rather than the vague concept of relative approach or separation speed. Note then that the coefficient of restitution is given by:

$\frac{{{v_b} - {v_a}}}{{{u_a} - {u_b}}}$ $= e$

Here we call the two objects a and b. Then ${v_b},{v_a}$ are the velocities of b and a after collision and ${u_b},{u_a}$ are the velocities of the objects b and a before the collision

Also note that for an object that collides with a surface the coefficient of restitution becomes:

$ e =$ $\frac{{{v_a}}}{{{u_a}}}$

Where we have called the object a, and the surface b, and plugged in the values zero for the value. A further consideration that eliminates the negative sign is that $v_a$ is in the opposite direction as $u_a$.

Ex.

Two identical smooth spheres A and B are free to move on a horizontal plane. B is at rest and A is projected with velocity $u$ to strike B directly. B then collides with a vertical wall that is perpendicular to the direction of the motion of the spheres. After rebounding from the wall B again collides with A and is brought to rest by this impact. If the coefficient of restitution has the same value for all the impacts, prove that $e=1$.

Sol.

In this question and all the question of this form that the examiner may ask you, there will be a multitude of variables. You must keep strict track of them, perhaps making a proper note in the beginning and any relationship or special conditions as given in the question. For this question I have:

1) $u$ as given in the question is the initial speed of A before collision

2) $u_1$ and $v_1$ are the speeds of A and B respectively after the first collision

3) $v_2$ is the speed of B after it has rebounded from the wall

4) $u_3$ is the speed of A after the final collision between B and A

The manner I suggest to do this in, would be to assign to every different body a different letter, and then the index would tell you what collision the velocity is consequent to. For example $u_3$ would be the velocity of the Sphere A after the third impact.

Also, note that I said speed, because the direction of the motion make a difference here. Consider two balls approaching one another, then the relative approach speed would be the sum of their speeds, if both are moving in the same direction than it would be the difference. Both of which cases are simply the difference of their velocities. While for calculations velocities would have sufficed, in analysis questions such as these we must need to use the difference in the direction of motion.

First collision

By the Law of Restitution:

$eu = {v_1} - {u_1}$

By the Law of Conversation of Momentum

$mu = m{u_1} + m{v_1} \Rightarrow u = {u_1} + {v_1}$

It follows that:

$\left( {1 + e} \right)u = 2{v_1}$

$\left( {1 - e} \right)u = 2{u_1}$

Second Collision

By Law of Restitution:

$e{v_1} = {v_2}$

Note that the momentum will not be conserved here because there is an external force at work here, that is keeping the vertical wall from moving. Recall that the Law of Conservation of Momentum doesn't hold when there is an external impetus.

Third Collision

By Law of Restitution:

$e\left( {{u_1} + {v_1}} \right) = {u_3}$ 

Here the direction of approach matters; as the spheres are approaching one another the relative approach speed is the sum of their speed, which would be of course the difference of their velocities.

By Conservation of Momentum:

${v_2} - {u_1} = {u_3}$

Eliminating $u_3$ gives:

${v_2} - {u_1} = e\left( {{u_1} + {v_2}} \right) \Rightarrow {v_2}\left( {1 - e} \right) = {u_1}\left( {1 + e} \right)$

Using previous equations we have built, we can create expressions dependent only on u and e for $u_1$ $u_2$, and then put them into the above equation.

${v_2} = e{v_1} = e\left( {{\textstyle{1 \over 2}}u} \right)\left( {1 + e} \right)$

${u_1} = \left( {1 - e} \right)\left( {{\textstyle{1 \over 2}}u} \right)$

$e\left( {1 + e} \right)\left( {{\textstyle{1 \over 2}}u} \right)\left( {1 - e} \right) = \left( {1 - e} \right)\left( {{\textstyle{1 \over 2}}u} \right)\left( {1 + e} \right)$

Make the necessary eliminations and our question is solved.

Consider the following example from your past papers (Oct/Nov 2014 Paper 21 Q1)


a) Two smooth spheres A and B, of equal radii and masses 2m and m respectively, lie at rest on a smooth horizontal table. The spheres A and B are projected directly towards each other with speeds 4u and 3u respectively. The coefficient of restitution between the spheres is e. Find the set of values of e for which the direction of motion of A is reversed in the collision.

Sol:

1) $v_A$ be the velocity of sphere A after the collision an $v_B$ the velocity of B after the collision.

We assume that sphere A is moving in the positive direction and B is moving in the negative direction. They have the velocities: 4u; and -3u. 

By Conservation:

\[2m\left( {4u} \right) + m\left( { - 3u} \right) = 2m{v_A} + m{v_B} \Rightarrow 8u - 3u = 2{v_A} + {v_B} = 5u\]

By restitution:

\[\frac{{{v_A} - {v_B}}}{{{u_B} - {u_A}}} = \frac{{{v_A} - {v_B}}}{{ - 3u - 4u}} = e\]
\[{v_A} - {v_B} =  - 7eu\]


We are concerned with the motion of the Sphere A, hence we use the above system to find a value for it. With the necessary algebra, you will find that:

\[{v_A} = {\textstyle{1 \over 3}}\left( {5 - 7e} \right)u\]

As by assumption $u > 0$, for the sphere to change direction $v_A$ must be less than zero. Now, by equality, ${\textstyle{1 \over 3}}\left( {5 - 7e} \right)u < 0$. As $u > 0$:

${\textstyle{1 \over 3}}\left( {5 - 7e} \right) < 0 \Rightarrow e > {\textstyle{5 \over 7}}$ 

We conclude: $e \in \left( {{\textstyle{5 \over 7}},1 } \right]$

Oblique collision with a surface

We have covered direct impact between moving spheres, and between one moving sphere and a surface. All that is left is oblique impact with a surface. 

Oblique impact is such that the sphere collides with the surface at an angle. In this case, the problem is no longer a one-dimensional one, but rather a two dimensional one. Hence, as with studying projectile motion, we will resolve the velocity in two different, perpendicular components. Our choice of the components is such that the direction of impact (the normal to the surface) is one and the other then has to be parallel to the surface. For example consider:

 

After the collision the component parallel to the surface will remain the same, i.e. $V\sin (x)$, assuming that the ball bounces back without sliding across the surface. In the line of impact ($V\cos (x)$), however the law of restitution will hold and as before separation speed = e * approach speed.

Consider the following past paper question. (Oct/Nov 14 Paper 21 Q2)

Sol:

Let us begin with the easier parallel component. Initially it is at $4\cos (\alpha )$. Let $\beta$ be the angle that P makes with the horizontal when it bounces back. We know the speed of P is halved, hence 2, and that the parallel component remains constant. Hence:

$4\cos (\alpha ) = 2\cos\left( \beta  \right)$

The normal component then gives us:

$0.4 \cdot 4\sin (\alpha ) = 2sin\left( \beta  \right)$

We wish to find $\alpha$

We know:

${V^2} = {2^2} = {V^2}{\sin ^2}\beta  + {V^2}{\cos ^2}\beta  = {\left( {1.6\sin \alpha } \right)^2} + {\left( {4\cos \alpha } \right)^2}$

Beyond this it is only the matter of using ${\sin ^2}\theta  + {\cos ^2}\theta  = 1$ to create a quadratic involving either sine or cosine, and thence finding $\alpha$. You should find that:

$\alpha  = 1.24$