Classical Mechanics with Maple

Section 3.3: Linear Momentum

Dr. Harald Kammerer
maple@jademountain.de

Initialisation

> restart;

> libname:="C:/mylib/m6dynlib","C:/mylib/m6dynfig",libname:

> with(linalg):with(plots):with(plottools):with(dynamics);with(figures_chapter_3);

Warning, the protected names norm and trace have been redefined and unprotected

Warning, the name changecoords has been redefined

Warning, the name arrow has been redefined

3.3 Linear Momentum

Above we considered the integral of a force along a path between two positions. Now we consider the integral over time. In general the force acting on a mass particle is variable over the time, . The linear momentum of the force is defined as

We apply this definition on the law of motion

This yields

Finally we get

Notice that and consequently also are vectors.

Example: Straight Central Impact

The linear momentum is very useful to describe what happens in the case of the impact of two mass particles. Here we consider the straight central impact. To imagine the situation we consider two balls that are moving translatoric on an straight trajectory. Both centers of gravity shall be before and after the impact on this trajectory (see Fig. 10). We use the coordinate x to describe the motion. Before the impact the velocities of the mass particles m1 and m2 are v1 and v2 .

> display(Fig_3_7(),scaling=constrained,axes=none,title="Figure 10");

We divide the process into three sections:

1. The motion before the impact, described by v1 and v2,

2. the impact and

3. the motion after the impact, described by v1q and v2q.

The impact happens in the time interval t1 to t2 . During this time an inner force acts between both particles. When we cut both particles free the force acts on every particle in the oppostite direction with the same absolute value. For both particles we get the linear momentum

>

>

with the total linear momentum during the time periode . This linear momentum has the same absolute value for both particles but points in the opposite direction, same as the passive force between both particles. For further calculation we need some information about the force F(t). To get this we divide the time periode into two parts and . During the first part both particles are compressed, during the second period both particles are decompressed. Consequently we separate two partial linear momentums L1 and L2 for the process of compression and the decompression. Further we set for the following consideration. Then we can write

At the end of the compression period both particles have the same velocity vm . This yields for the partial linear momentum L1

>

>

Next we need a relation between the partial linear momentum L2 during the decompression and the partial linear momentum L1 . Here we use a hypothesis from Newton

with the parameter e which depends mainly on the characteristics of the material of both particles. This parameter is called the coefficient of elasticity (don't confound with the modulus of elasticity E ). One limit case is given by totatally elastic material. Then there is and so . This is nearly valid whe one drops a rubber ball on the hard ground. There is no enduring deformation at the end of the impact process. The second limit case is given by totally inelastic material. Then there is and so . This is valid when plasticine drops to the ground. It stays there after the impact without further motion. Real materials are always between these two limit cases and so we can say in general and . The total linear momentum during the total impact process is then given by and so

>

For given e we can solve the equations EQ1a, EQ1b, EQ2a, EQ2b and EQ3

> sol:=solve({EQ1a, EQ1b, EQ2a, EQ2b, EQ3},{vm,L,L1,v1q,v2q});

> assign(sol):

At last we get for the velocities after the impact

> v1q:=collect(v1q,{v1,v2});

> v2q:=collect(v2q,{v1,v2});

The relations for this calculation are included in the dynamics package. See impact to learn how to use these functions.

Now we consider a simple way to get the coefficient of elasicity e for a ball falling on the ground. We assume that the ground doesn't move before and after the impact. We drop the ball with the mass m from the height h1=1m . Then we measure the maximum height that it reaches after the impact on the ground. We assume the ball reaches the height h2=0.3m after the impact.

(See Fig. 11. Click the picture and press "play" in the menu bar.)

>

>

> display(Fig_3_8(h1example,h2example),insequence=true,scaling=constrained,axes=framed,title="Figure 11");

To get the coefficient of elasticity e we consider the situation immediately before and immediately after the impact. To calculate the velocities we consider the conservation of energy. We choose the ground as the reference horizontal with . At the initial position and at the end of the considered motion, when the ball reaches the position the velocity of the ball is and .

So we get the kinetic energy at the beginning of the motion

> Tstart:=kinetic_energy(m,0);

and also at the end of the time period

> Tend:=kinetic_energy(m,0);

The kinetic energy imediately before the impact is

> T1:=kinetic_energy(m,v1);

and imediately after the impact

> T2:=kinetic_energy(m,v2);

The gravitational potential is at the begin of the motion

> Vstart:=gravitational_energy(m,g,h1);

At the end we have the gravitational potential

> Vend:=gravitational_energy(m,g,h2);

The potential before impact is zero

> V1:=0;

The conservation of energy before the impact yields

> Energy_conservation_1:=Tstart+Vstart=T1+V1;

and we get for the velocity before the impact

> v1:=solve(Energy_conservation_1,v1);

We need the negative solution, because the coordinate z and also the velocity was choosen upwards positive but the velocity points downwards in reality. That means

> v1:=-sqrt(v1[1]**2);

After impact, the potential is also 0.

> V2:=0;

After the impact the conservation of energy yields

> Energy_conservation_2:=T2+V2=Tend+Vend;

and we get for the velocity before the impact

> v2:=solve(Energy_conservation_2,v2);

Now we need the positive solution, that means

> v2:=sqrt(v2[1]**2);

The procedure impact yields for the velocities after the impact

> v[ball_after_impact]:=expand(impact(m[ball],m[ground],v[ball],v[ground],e)[1]);

> v[ground_after_impact]:=expand(impact(m[ball],m[ground],v[ball],v[ground],e)[2]);

with the mass of the ground and the velocitiy of the ground before the impact. The mass of the ball is and its velocity before the impact is . It is evident that we can assume and without making significant errors. So we get

> v[ball_after_impact]:=limit(v[ball_after_impact],{v[ground]=0,m[ground]=infinity});

> v[ground_after_impact]:=limit(v[ground_after_impact],{v[ground]=0,m[ground]=infinity});

Applying this result we get with and

>

At last we get for the cooefficient of elasticity

> e:=simplify(solve(eq,e),symbolic);

For the concrete values from the example above we get

> subs({h1=h1example,h2=h2example},e);

>