Chapter

Linear Momentum

Blinn College - Physics 1401 - Terry Honan

. - Momentum and Newton’s Second Law

Definition of Momentum

So far we have written the second law in terms of the acceleration of a particle.  It turns out that Newton wrote it differently; his preferred form of the second law was written in terms of the time rate of change of the momentum of the body.

Momentum is a vector quantity describing the dynamics of a moving body.  It is defined simply as the mass times the velocity and we use the symbol (OverVector[p]) to denote it.

(OverVector[p])=m (OverVector[v])

We will often refer to this as linear momentum; this will distinguish it from angular momentum which will be discussed later.

The Second Law

Newton’s second law can be written in terms of the momentum. If the mass of a body is constant then Δ (OverVector[p])=m ((Subscript[(OverVector[v]), f])-(Subscript[(OverVector[v]), i]))=m Δ (OverVector[v]) and we can write

((Δ(OverVector[p]))/(Δ t))=m ((Δ(OverVector[v]))/(Δ t))=m (Subscript[(OverVector[a]), ave])   ⟹  (Subscript[(OverVector[F]), net,ave])=((Δ(OverVector[p]))/(Δ t))    and    (Subscript[(OverVector[F]), net])=(lim_(Δ t->0))((Δ(OverVector[p]))/(Δ t))

This form of the second law was Newton’s preferred way to write it. Although this is equivalent to the (Subscript[(OverVector[F]), net])=m (OverVector[a]) form when the mass is constant, when the mass is changing this new momentum form is the proper one.

. - Impulse and Momentum

Impulse and Average Force

Collisions are usually quick things but they are not instantaneous.  When a bat hits a baseball, the ball rides along the bat for a period of time.  An impulsive force is a large force acting over a short period of time.  We will define the impulse as the area under the force versus time graph.  Suppose there is a collision between (Subscript[t, i]) and (Subscript[t, f]) the force is zero other than in that time interval; the x-component of the impulse (Subscript[I, x]) is the area under the graph of (Subscript[F, x]) vs. time between (Subscript[t, i]) and (Subscript[t, f]).

The average force is defined as the impulse divided by the time Δ t, where the width of the interval is Δ t=(Subscript[t, f])-(Subscript[t, i]). It follows that the area under the force (Subscript[F, x]) versus time curve is also the area of the rectangle with Δ t at its base and with a height of (Subscript[F, ave,x])

(OverVector[I])=(Subscript[(OverVector[F]), ave])Δ t   and   (Subscript[I, x])=(Subscript[F, ave,x])Δ t=(Area under (Subscript[F, x]) vs. t)

The Impulse-Momentum Theorem

The impulse-momentum theorem is an immediate consequence of Newton's second law.

(Subscript[(OverVector[I]), net])=Δ(OverVector[p])

Here the net impulse (Subscript[(OverVector[I]), net]) is the impulse of the net force (Subscript[(OverVector[F]), net]) and the change in the momentum can be written Δ(OverVector[p])=m ((Subscript[(OverVector[v]), f])-(Subscript[(OverVector[v]), i])). Usually, the “net” subscript is omitted when writing the impulse-momentum theorem. The physical significance of this is when there is an impulsive force, it typically is much larger than any other forces acting over the short time of the collision and the other forces can be neglected. For example, when a bat hits a baseball, that force is much larger than gravity or some other force during the collision.

Example . - Hitting a Baseball

Solution

(Subscript[F, ave]) Δ t=I=Δ p=m ((Subscript[v, f])-(Subscript[v, i])).

m=0.145(QuantityUnit),  (Subscript[v, i])=-40(QuantityUnit)/(QuantityUnit), (Subscript[v, f])=50(QuantityUnit)/(QuantityUnit)  and  Δ t=0.035(QuantityUnit) .

(Subscript[F, ave])=(m/(Δ t)) ((Subscript[v, f])-(Subscript[v, i]))=373(QuantityUnit)

. - A System of Particles

Introduction

So far our discussion of dynamics has applied only to point particles.  When we discussed the dynamics of extended objects we treated them as particles.  Why was this proper?  In all the examples considered the body was not rotating, so each point on it had the same acceleration.  This allowed us to treat it as a particle.  If a body rotates then different points have different accelerations and we must be more careful.  We can treat it as a system of particles.

A system is a collection of point particles.  This could represent a huge number of particles, like every atom in a solid or fluid or it could be a small number like the earth, moon and sun.  We arbitrarily divide the world into a system and everything else.  We then break up the forces into internal forces, which are between particles of our system and external force between particles of our system and outside.

Center of Mass

We will now define the center of mass of a system of particles as a weighted average of the positions.  Consider the system to be a discrete distribution, that is a collection of point masses, which consist of masses (Subscript[m, 1]) at postion vector (Subscript[(OverVector[r]), 1]),  (Subscript[m, 2]) at (Subscript[(OverVector[r]), 2]), etc.  The total mass of our system is M:

M=(Subscript[m, 1])+(Subscript[m, 2])+...

We define the position vector of the center of mass by

(Subscript[(OverVector[r]), cm])=(1/M)((Subscript[m, 1]) (Subscript[(OverVector[r]), 1])+(Subscript[m, 2]) (Subscript[(OverVector[r]), 2])+...)

Note that the position vectors (Subscript[(OverVector[r]), i]) point from the origin of our coordinate system to the mass (Subscript[m, i]).  Since the x component of the position vector is just x, the x component of the center of mass is just

(Subscript[x, cm])=(1/M)((Subscript[m, 1]) (Subscript[x, 1])+(Subscript[m, 2]) (Subscript[x, 2])+...)

where the y and z components satisfy similar expressions.

Because of how the instantaneous velocity is related to the position vector and how the instantaneous acceleration is related to the velocity, we can write similar expressions for the velocity and acceleration of the center of mass.

(Subscript[(OverVector[v]), cm])=(1/M)((Subscript[m, 1]) (Subscript[(OverVector[v]), 1])+(Subscript[m, 2]) (Subscript[(OverVector[v]), 2])+...)   and   (Subscript[(OverVector[a]), cm])=(1/M)((Subscript[m, 1]) (Subscript[(OverVector[a]), 1])+(Subscript[m, 2]) (Subscript[(OverVector[a]), 2])+...)

Example . - The Earth-Moon System

(Subscript[M, E])=5.97×(10^24)(QuantityUnit),  (Subscript[M, M])=7.35×(10^22)(QuantityUnit)   and  (Subscript[R, E,M])=3.85×(10^8)(QuantityUnit) .

Solution

({{(Subscript[x, cm])=(((Subscript[M, E]) (Subscript[x, E])+(Subscript[M, M]) (Subscript[x, M]))/((Subscript[M, E])+(Subscript[M, M])))=(((Subscript[M, E]) 0+(Subscript[M, M]) (Subscript[R, E,M]))/((Subscript[M, E])+(Subscript[M, M])))=((Subscript[M, M])/((Subscript[M, E])+(Subscript[M, M])))(Subscript[R, E,M])}, {=0.0122 (Subscript[R, E,M])=4.68×(10^6)(QuantityUnit)}})

A Three Particle System

Consider a three particle system with masses (Subscript[m, 1]),  (Subscript[m, 2]) and (Subscript[m, 3]).  For the forces on (Subscript[m, 1])can be written as a sum of internal forces (Subscript[(OverVector[F]), 1,2]) and (Subscript[(OverVector[F]), 1,3]) and external forces (Subscript[(OverVector[F]), 1]^ext), representing everything outside our system acting on (Subscript[m, 1]).  The forces for (Subscript[m, 2]) and (Subscript[m, 3]) break up similarly giving

(Subscript[(OverVector[F]), net,1])=(Subscript[(OverVector[F]), 1]^ext)+ (Subscript[(OverVector[F]), 1,2]) + (Subscript[(OverVector[F]), 1,3])=(lim_(Δ t->0))((Δ(Subscript[(OverVector[p]), 1]))/(Δ t))=(Subscript[m, 1]) (Subscript[(OverVector[a]), 1])

(Subscript[(OverVector[F]), net,2])=(Subscript[(OverVector[F]), 2]^ext)+ (Subscript[(OverVector[F]), 2,1]) + (Subscript[(OverVector[F]), 2,3])=(lim_(Δ t->0))((Δ(Subscript[(OverVector[p]), 2]))/(Δ t))=(Subscript[m, 2]) (Subscript[(OverVector[a]), 2])

(Subscript[(OverVector[F]), net,3])=(Subscript[(OverVector[F]), 3]^ext)+ (Subscript[(OverVector[F]), 3,1]) + (Subscript[(OverVector[F]), 3,2])=(lim_(Δ t->0))((Δ(Subscript[(OverVector[p]), 3]))/(Δ t))=(Subscript[m, 3]) (Subscript[(OverVector[a]), 3]).

To concentrate on the bulk motion of our system we sum over these expressions.  The crucial point is that the internal forces cancel by Newton's third law.   (Subscript[(OverVector[F]), 1,2])+ (Subscript[(OverVector[F]), 2,1])= (OverVector[0]),   (Subscript[(OverVector[F]), 1,3])+ (Subscript[(OverVector[F]), 3,1])= (OverVector[0])  and   (Subscript[(OverVector[F]), 2,3])+ (Subscript[(OverVector[F]), 3,2])= (OverVector[0]).

(Subscript[(OverVector[F]), 1]^ext)+ (Subscript[(OverVector[F]), 2]^ext) + (Subscript[(OverVector[F]), 3]^ext)=(lim_(Δ t->0))((Δ((Subscript[(OverVector[p]), 1])+(Subscript[(OverVector[p]), 2])+(Subscript[(OverVector[p]), 3])))/(Δ t))=(Subscript[m, 1]) (Subscript[(OverVector[a]), 1])+(Subscript[m, 2]) (Subscript[(OverVector[a]), 2])+(Subscript[m, 3]) (Subscript[(OverVector[a]), 3])

The General System

For a general system we define (Subscript[(OverVector[F]), net]^ext) as the net force

(Subscript[(OverVector[F]), net]^ext)=(Subscript[(OverVector[F]), 1]^ext)+(Subscript[(OverVector[F]), 2]^ext)+...

(Subscript[(OverVector[p]), tot]) as the total momentum and M as the total mass

(Subscript[(OverVector[p]), tot])=(Subscript[(OverVector[p]), 1])+(Subscript[(OverVector[p]), 2])+...

M=(Subscript[m, 1])+(Subscript[m, 2])+...

Taking two derivatives of our definition of the center of mass,  M (Subscript[(OverVector[r]), cm])=(Subscript[m, 1]) (Subscript[(OverVector[r]), 1])+(Subscript[m, 2]) (Subscript[(OverVector[r]), 2])+...,  gives:

M (Subscript[(OverVector[a]), cm])=(Subscript[m, 1]) (Subscript[(OverVector[a]), 1])+(Subscript[m, 2]) (Subscript[(OverVector[a]), 2])+...,

since the second derivative of the position vector (OverVector[r]) is the acceleration (OverVector[a]).  With this we get the two expressions for the second law for a system of particles.

(Subscript[(OverVector[F]), net]^ext)=(lim_(Δ t->0))((Δ(Subscript[(OverVector[p]), tot]))/(Δ t))    or    (Subscript[(OverVector[F]), net,ave]^ext)=((Δ(Subscript[(OverVector[p]), tot]))/(Δ t))

(Subscript[(OverVector[F]), net]^ext)=M (Subscript[(OverVector[a]), cm])

Suppose a uniform stick is thrown so that it has both translational and rotational motion. The net external force acting on it is (Subscript[(OverVector[F]), net]^ext)=M (OverVector[g]), The second expression above implies that  (Subscript[(OverVector[a]), cm])=(OverVector[g]), so the stick’s center of mass will follow the parabolic path of a projectile while the stick rotates about that path.

Conservation of Linear Momentum

The conservation of momentum is a consequence of the momentum form of the second law for a system.

(Subscript[(OverVector[F]), net]^ext)=(lim_(Δ t->0))((Δ(Subscript[(OverVector[p]), tot]))/(Δ t))

If there are no external forces acting on a system them the total momentum of the system is conserved.

(Subscript[(OverVector[F]), net]^ext)=(OverVector[0])  ⟹   Δ(Subscript[(OverVector[p]), tot])=(OverVector[0])

This is the second fundamentally conserved quantities encountered in our course.  To see how this is fundamental, imagine enlarging the system to  include everything; there is then, by definition, no external force and thus the total momentum is conserved.

The conservation is also true component by component.  If there is no external force in some direction, say the x direction, then the x component of the total momentum is conserved.

(Subscript[F, net,x]^ext)=0  ⟹   Δ (Subscript[p, tot,x])=0

Example . - Throwing and Catching a Ball on Frictionless Ice

Solution

(Subscript[F, net]^ext)=0  ⟹  Δ (Subscript[p, tot])=0  ⟹  (Subscript[p, tot,i])=(Subscript[p, tot,f])  ⟹  0=M (-V)+m v  ⟹  V=(m/M)v

Solution

(Subscript[F, net]^ext)=0  ⟹  Δ (Subscript[p, tot])=0  ⟹  (Subscript[p, tot,i])=(Subscript[p, tot,f])  ⟹  0+m v=(M+m) V  ⟹  V=(m/(M+m))v

Solution

(Subscript[F, net,x]^ext)=0  ⟹  Δ (Subscript[p, tot,x])=0  ⟹  (Subscript[p, tot,x,i])=(Subscript[p, tot,x,f])  ⟹  0=M (-V)+m v cos θ  ⟹  V=(m/M)v cos θ

Symmetries and Conservation Laws - Noether's Theorem

It is a very deep and fundamental matter that symmetries give rise to conserved quantities.  This result is known as Noether's theorem.  The mathematician Amalie Noether demonstrated around 1920 that to every symmetry there is a conservation law.  For example, the invariance or symmetry of the laws of physics under time translations, that the laws are the same now as a few minutes from now, implies that there is a conserved quantity; this is energy!  The symmetry that the laws of physics are invariant under spatial translations implies a conserved quantity, in this case linear momentum.  Rotational symmetry implies conservation of angular momentum.

. - Two-Body Collisions

Introduction

We now consider two-body collisions as a special case of our more general discussion.  If there are no external forces act while two bodies collide, then the total momentum of the two-body system is conserved.  Even if there are external forces, often we can neglect them and consider momentum conserved.  Consider a mid-air collision between two bodies.  Gravity acts as an external force during the collision but usually, to a good approximation, the collision is so fast that the large impulsive internal forces dominate the gravity force and gravity can be neglected.  We can equate the total momentum just before and just after the collision.

Momentum Conservation

Mass (Subscript[m, 1]) moving at (Subscript[(OverVector[v]), 1,i]) collides with mass (Subscript[m, 2]) moving at (Subscript[(OverVector[v]), 2,i]).  After the collision the velocities are (Subscript[(OverVector[v]), 1,f]) and  (Subscript[(OverVector[v]), 2,f]).  The conservation of momentum for a two-body collision has the form

(Subscript[m, 1]) (Subscript[(OverVector[v]), 1,i])+(Subscript[m, 2]) (Subscript[(OverVector[v]), 2,i])=(Subscript[m, 1]) (Subscript[(OverVector[v]), 1,f])+(Subscript[m, 2]) (Subscript[(OverVector[v]), 2,f]).

The left-hand side is the total initial momentum and the right hand side is the total final momentum.

In the case of a one dimensional collision then the above expression applies but we may omit the vector arrows.  In one dimension a vector is a real number and the sign gives the direction.  The vector nature of momentum and velocity is reflected in their signs.

Elastic Collisions - Kinetic Energy Conservation

Typically energy is lost in a collision.  Often to a reasonable approximation we can consider conservation of energy.  The relevant energy is kinetic.

(1/2)(Subscript[m, 1]) (Subscript[v, 1,i]^2)+(1/2)(Subscript[m, 2]) (Subscript[v, 2,i]^2)=(1/2)(Subscript[m, 1]) (Subscript[v, 1,f]^2)+(1/2)(Subscript[m, 2]) (Subscript[v, 2,f]^2)   (elastic)

In other words

(Subscript[K, tot,i])=(Subscript[K, tot,f]).   (elastic)

Inelastic and Totally Inelastic Collisions

When we say a collision is inelastic we mean merely that it is not elastic.

(Subscript[K, tot,i])≠(Subscript[K, tot,f])  (inelastic)

The extreme case of an inelastic collision is called totally inelastic.  The most energy that can be lost in a collision is when the two colliding objects stick together. This means that the two objects have the same final velocity.

(Subscript[(OverVector[v]), 1,f])=(Subscript[(OverVector[v]), 2,f])=(Subscript[(OverVector[v]), f])  (totally inelastic)

The conservation of momentum formula then has the simple form:

(Subscript[m, 1]) (Subscript[(OverVector[v]), 1,i])+(Subscript[m, 2]) (Subscript[(OverVector[v]), 2,i])=((Subscript[m, 1])+(Subscript[m, 2]))  (Subscript[(OverVector[v]), f]).  (totally inelastic)

Example . - A Projectile embeds in a Block

Solution

(Subscript[m, 1])=0.30(QuantityUnit), (Subscript[m, 2])=5.0(QuantityUnit), (Subscript[(OverVector[v]), 1,i])=⟨0,40⟩(QuantityUnit)/(QuantityUnit), (Subscript[(OverVector[v]), 2,i])=⟨-2.0,0⟩(QuantityUnit)/(QuantityUnit)

(Subscript[m, 1]) (Subscript[(OverVector[v]), 1,i])+(Subscript[m, 2]) (Subscript[(OverVector[v]), 2,i])=((Subscript[m, 1])+(Subscript[m, 2]))  (Subscript[(OverVector[v]), f])   ⟹   (Subscript[(OverVector[v]), f])=(((Subscript[m, 1]) (Subscript[(OverVector[v]), 1,i])+(Subscript[m, 2]) (Subscript[(OverVector[v]), 2,i]))/((Subscript[m, 1])+(Subscript[m, 2])))=⟨-1.89,2.26⟩(QuantityUnit)/(QuantityUnit)

Solution

(Subscript[m, 1])=0.30(QuantityUnit), (Subscript[m, 2])=5.0(QuantityUnit), (Subscript[v, 1,i])=40(QuantityUnit)/(QuantityUnit), (Subscript[v, 2,i])=-2.0(QuantityUnit)/(QuantityUnit)

(Subscript[m, 1]) (Subscript[v, 1,i])+(Subscript[m, 2]) (Subscript[v, 2,i])=((Subscript[m, 1])+(Subscript[m, 2]))  (Subscript[v, f])   ⟹   (Subscript[v, f])=(((Subscript[m, 1]) (Subscript[v, 1,i])+(Subscript[m, 2]) (Subscript[v, 2,i]))/((Subscript[m, 1])+(Subscript[m, 2])))=+0.377(QuantityUnit)/(QuantityUnit)

Example . - A Bullet Passes through a Block

Solution

(Subscript[m, 1])=0.0175(QuantityUnit) ,  (Subscript[v, 1,,i])=355(QuantityUnit)/(QuantityUnit) ,  (Subscript[v, 1,f])=125(QuantityUnit)/(QuantityUnit) ,  (Subscript[m, 2])=1.55(QuantityUnit) ,  (Subscript[v, 2,,i])=0 ,  (Subscript[v, 2,f])=?

(Subscript[m, 1]) (Subscript[v, 1,i])+(Subscript[m, 2]) (Subscript[v, 2,i])=(Subscript[m, 1]) (Subscript[v, 1,f])+(Subscript[m, 2]) (Subscript[v, 2,f])   ⟹  (Subscript[m, 1]) (Subscript[v, 1,i])+0=(Subscript[m, 1]) (Subscript[v, 1,f])+(Subscript[m, 2]) (Subscript[v, 2,f])   ⟹    (Subscript[v, 2,f])=((Subscript[m, 1])/(Subscript[m, 2])) ((Subscript[v, 1,i])- (Subscript[v, 1,f]))=2.60(QuantityUnit)/(QuantityUnit)

One Dimensional Elastic Collisions

For the case of a one dimensional elastic collision we can solve for the final velocities in terms of the initial velocities and the masses.

(Subscript[m, 1]) (Subscript[v, 1,i])+(Subscript[m, 2]) (Subscript[v, 2,i])=(Subscript[m, 1]) (Subscript[v, 1,f])+(Subscript[m, 2]) (Subscript[v, 2,f])     (momentum eq.)

(1/2)(Subscript[m, 1]) (Subscript[v, 1,i]^2)+(1/2)(Subscript[m, 2]) (Subscript[v, 2,i]^2)=(1/2)(Subscript[m, 1]) (Subscript[v, 1,f]^2)+(1/2)(Subscript[m, 2]) (Subscript[v, 2,f]^2)    (energy eq.)

Given the two initial velocities and the masses, we can solve for the final velocities. This algebra is awkward, so the solutions will be given.

(Subscript[v, i,f])=(((Subscript[m, 1])-(Subscript[m, 2]))/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 1,i])+((2(Subscript[m, 2]))/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 2,i])   and   (Subscript[v, 2,f])=((2(Subscript[m, 1]))/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 1,i])+(((Subscript[m, 2])-m)/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 2,i])

Example . - One-Dimensional Elastic Collision

Solution

(Subscript[m, 1])=m, (Subscript[m, 2])=2 m, (Subscript[v, 1,i])=v, (Subscript[v, 2,i])=-v

(Subscript[v, i,f])=(((Subscript[m, 1])-(Subscript[m, 2]))/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 1,i])+((2(Subscript[m, 2]))/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 2,i])=((m-2m)/(3m))v+((4m)/(3m))(-v)=-(1/3)v-(4/3)v=-(5/3)v

(Subscript[v, 2,f])=((2(Subscript[m, 1]))/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 1,i])+(((Subscript[m, 2])-m)/((Subscript[m, 1])+(Subscript[m, 2])))(Subscript[v, 2,i])=((2m)/(3m))(Subscript[v, 1,i])+((2m-m)/(3m))(-v)=(2/3)v-(1/3)v=(1/3)v

. - Rocket Propulsion

When moving with respect to a medium one can propel something forward by pushing backward.  To walk, you push backward in the floor and the floor then pushes forward on you.  A boat pushes backward on the water and the water pushes forward in it. A plane or jet propels itself similarly by pushing backward on the air.  This leads to an obvious question for rocket propulsion. How does a rocket propel itself forward in the vacuum of space? The answer is that the rocket throws part of itself, its spent fuel, backward and thus propels the rest of the rocket forward.  The mass change of a rocket is an essential part of its propulsion.

Consider a rocket of mass m moving with a velocity v. The rocket propels itself forward by shooting spent fuel backward at a speed of (Subscript[v, e]), the exhaust speed, relative to the rocket. In doing this the mass of the rocket changes. The (positive) mass of ejected fuel is |Δ m|. Ejecting the fuel backward makes the rocket recoil forward with a small Δ v.  Looking at conservation of momentum in the frame where the rocket was initially at rest gives

0=m Δ v-(Subscript[v, e])|Δ m|   ⟹ Δ v=(Subscript[v, e])((|Δ m|)/m)

The force propelling a rocket forward due to this expelled fuel is known as the thrust. From the momentum form of the second law the force on the expelled fuel is

F=((Δ p)/(Δ t))=(Subscript[v, e])((|Δ m|)/(Δ t))

By Newton’s third law, the thrust, the forward force on the rocket is

thrust=(Subscript[v, e])((|Δ m|)/(Δ t))