Chapter 7
The Giancoli onLine Tutor

Linear Momentum

I. Center-of-Mass

Consider the masses arranged in the following pattern:

To find the Center-of-Mass you would first find its location on the x-axis by using

X_cm = (2*x₂+1*x₁+4*x₄+5*x₅)/(2+1+4+5)

where x₂ refers to the location of the 2 kg mass on the x-axis, etc. The denominator is the sum of the masses. This is a normal weighted average. If the mass in the lower left hand corner is at x=0,y=0, the 1 kg mass at x=0,y=1, the 4 kg mass at x=2,y=1, and the 5 kg mass at x=2,y=0, then

X_cm=(2*0+1*0+4*2+5*2)/12= 1.5

The result for Y_cm is .417, check it!

Interactive Example

Complete the following table and click outside to get your results.

Mass = 2 kg 6 kg 3 kg 7 kg

Location x= y= x= y= x= y= x= y=

Xcm =

Ycm =

Mass =	2 kg	6 kg	3 kg	7 kg
Location	x=	y=	x=	y=	x=	y=	x=	y=
Xcm =
Ycm =

II. Momentum

The momentum of a mass is a vector with magnitude equal to the product its mass and speed (m|v|), and direction in the same direction as the velocity (v). The momentum of a group of particles is the vector sum of the momenta of each of the particles. You add them just as you added the vectors in Chapter 3.

Interactive Example

Particle 1 has a mass of 5 kg and a velocity of 2 m/s, North (as on a map this is taken as up).
Particle 2 has a mass of 2 kg and a velocity of 5 m/s, East (as on a map this is taken as right).
Particle 3 has a mass of 10 kg and a velocity of 1 m/s. North East.

Enter your results below and click outside each time to get your results.

X Y

Particle 1

Particle 2

Particle 3

Total P

Now find the magnitude and direction of the Total Momentum and enter the result below.

Magnitude:

Direction:

Magnitude:
Direction:

The total momentum of an isolated system (group) of particles is a constant; it does not change in either its magnitude (size) or direction. The individual momenta might radically change, but the total remains the same.

A system of particles is isolated if there are no external forces, or if the external forces cancel out, or if they are negligible.

III. Impulse

The momentum p (=mv) of a particle changes when it is acted on by a force over a period of time.

Dp = F D t

F D t is called the Impulse.

When a ball bounces off a wall it experiences an impulse. If the ball is a soft rubber ball the impact with the wall takes longer than if it where a hard ball. The change in momentum is the same in both cases so the impulse is the same in both cases.

(Remember that momentum is a vector so if the ball is going left to right before it hits the wall (positive momentum), its momentum will be negative after it bounces back. The change in momentum will be
p_after-p_before= -p-p=-2p)

Since the impulse is the same for both balls mentioned above and the first one occurs over a longer time (D t is longer) the force (F) on it must be the smaller. This situation is illustrated in the following two movies. The solid ball is the hard one. Both balls have the same mass and speeds before and after collision. In the Graph the force is plotted on the vertical axis and the time on the horizontal axis. The area under the graph in both cases is the impulse and is the same in both cases.

If you not have the quicktime plugin you can view a GIF animation version of the movie here and here.

IV. Collisions

Momentum in Collisions

In the study of collisions we will limit ourselves to one dimension so that direction will always be given by "+" (left to right) and "-" (right to left). In all collisions momentum is conserved, that is, it is the same before and after. So we always have

m₁v₁+m₂v₂=m₁v'₁+m₂v'₂

where v is the velocity before and v' is the velocity after.

Tips:

Draw a picture!

The first ball is always travelling left to right. v₁ always equals +v₁. The plus sign gives its direction and v₁ is its magnitude.
The second ball can be going in either direction depending on the problem statement. v₂ always equals +or- v₂ or zero
Make sure to use the minus sign above if it moving to the left.
Assume that both balls are travelling to the right after the collision. v'₁ always equals +v'₁ and v'₂ always equals +v'₂
In purely algebraic terms the equation always reduces to m₁v₁+or-m₂v₂=m₁v'₁+m₂v'₂

The only decision you have to make is on the + or -
- If the problem statement contains something like " overtakes the second ball" that means the second ball is going to the right and you need the plus sign.
- If the problem statement contains something like "head-on collision" that means that the second ball is travelling to the left and you need the minus sign.
- If the problem says that the second ball is at rest you, of course, use zero.
This is one equation and therefore you can only find one unknown!

Elastic Collisions

Elastic collisions (billiard ball collisions, for example) also have another equation that can be used. In an elastic collision Kinetic Energy is also conserved.

m₁v₁²/2+m₂v₂²/2= m₁v'₁²/2+m₂v'₂²/2

Unfortunately using these two equation together is an algebraic nightmare. Fortunately the first person to study elastic collisions, Christiaan Huygens (1629-1695), showed that such collisions always had the equivalent property that the relative velocity after collision is the negative of the relative velocity before collision, which translates to

v₁-v₂ = -(v'₁-v'₂)
or
v₁-v₂ = -v'₁+v'₂

Use this equation and the momentum equations to solve the problem.
Do not use the Kinetic Energy equation directly.
Usually you will be looking for the velocities after collision (v'₁ and v'₂)
There are ways to solve the problem once you have the equations.
1. But if you do not know what else to do: multiply both sides of the relative motion equation by m₁.
2. Add it to the momentum equation.
3. This will eliminate v'₁ and leave v'₂ as the only unknown.
4. Solve the resulting equation for v'₂.
5. Substitute your result for v'₂ into the relative motion equation.
6. Solve for v'₁!

Interactive Example
A 2 kg ball traveling at 4 m/s overtakes a 4 kg ball traveling at 2 m/s, what are the velocities after the collision?

1) Fill in each space below to construct your momentum equation, clicking outside to get your result. (Enter the correct number in the blanks and replace the '?' with the appropriate numbers.)

m₁v₁	m₂v₂	BOOM	m₁v'₁	m₂v'₂
		->

2) Collect the terms on the left and enter in the table below.

3) Next construct the second equation (relative motion equation).

v₁	-v₂	BOOM	-v'₁	v'₂
		->

4) Now collect the terms on the left into a single number

=

5) Carry out step 4.a above and complete the form below.

= v'₂

6) Carry out steps 4.b and c above (add step 2 to step 5) and complete the form below.

= v'₁ v'₂

7) Carry out step 4.d (find v'₂), enter below:

=

8) Carry out step 4.e (put your result from above into step 4), enter below:

=

9) Solve the above for v'1, enter below and you are done!

=

In the previous example the two balls were traveling in the same direction before the colission. What follows is the other major case.

Interactive Example

Just like the one above with one small change.
A 2 kg ball traveling at 4 m/s collides head-on with a 4 kg ball traveling at 2 m/s, what are the velocities after the collision?

1) Fill in each space below to construct your momentum equation, clicking outside to get your result. (Enter the correct number in the blanks and replace the '?' with the appropriate numbers.)

m₁v₁	m₂v₂	BOOM	m₁v'₁	m₂v'₂
		->

2) Collect the terms on the left and enter in the table below.

3) Next construct the second equation (relative motion equation).

v₁	-v₂	BOOM	-v'₁	v'₂
		->

4) Now collect the terms on the left into a single number

=

5) Carry out step 4.a above and complete the form below.

= v'₂

6) Carry out steps 4.b and c above (add step 2 to step 5) and complete the form below.

= v'₁ v'₂

7) Carry out step 4.d (find v'₂), enter below:

=

8) Carry out step 4.e (put your result from above into step 4), enter below:

=

9) Solve the above for v'1, enter below and you are done!

=

Inelastic Collisions

Collisions in which Kinetic Energy is not conserved (Car Crashes, for example) are called inelastic collisions. The only ones we will consider are called totally inelastic collisions. In these collisions the two bodies stick together after collision. So the momentum equation simplifies to

m₁v₁+m₂v₂= (m₁+m₂)v'

If v' is what you are looking for, the above equation becomes:

v'=(m₁v₁+m₂v₂)/(m₁+m₂)

Do not forget to include a minus sign if the second body is moving to the left.

Interactive Example
A 2 kg ball traveling at 4 m/s collides head-on with and sticks to a 4 kg ball traveling at 2 m/s, what is the combined velocity after the collision? Enter your answer below and click outside

Linear Momentum

I. Center-of-Mass

Interactive Example

II. Momentum

Interactive Example

The total momentum of an isolated system (group) of particles is a constant; it does not change in either its magnitude (size) or direction. The individual momenta might radically change, but the total remains the same.

III. Impulse

(Remember that momentum is a vector so if the ball is going left to right before it hits the wall (positive momentum), its momentum will be negative after it bounces back. The change in momentum will be pafter-pbefore= -p-p=-2p)

If you not have the quicktime plugin you can view a GIF animation version of the movie here and here.

IV. Collisions

Momentum in Collisions

Elastic Collisions

Interactive Example

Interactive Example

Just like the one above with one small change.

Inelastic Collisions

Interactive Example

(Remember that momentum is a vector so if the ball is going left to right before it hits the wall (positive momentum), its momentum will be negative after it bounces back. The change in momentum will be
p_after-p_before= -p-p=-2p)