Linear Momentum 2D Collisions Extended or Composite ...nebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture33.pdf · Linear Momentum 2D Collisions Extended or Composite Systems Center

Linear Momentum2D Collisions

Extended or Composite SystemsCenter of Mass

Lana Sheridan

De Anza College

Nov 13, 2017

Last time

• inelastic collisions

• perfectly inelastic collisions

• the ballistic pendulum

Overview

• 2D collisions

• center of mass

• finding the center of mass

Collisions in 2 Dimensions

The conservation of momentum equation is a vector equation.

It will apply for any number of dimensions that are relevant in aquestion.

pi = pf ⇒ m1v1i +m2v2i = m1v1f +m2v2f

In particular, we can write equations for each component of themomentum. In 2-d, with x and y components:

x : m1v1ix +m2v2ix = m1v1fx +m2v2fx

y : m1v1iy +m2v2iy = m1v1fy +m2v2fy

If it is an elastic collision:

Ki = Kf ⇒ 1

2m1(v1i )

2 +1

2m2(v2i )

2 =1

2m1(v1f )

2 +1

2m2(v2f )

2









Ki = Kf ⇒ 1

2m1(v1i )

2 +1

2m2(v2i )

2 =1

2m1(v1f )

2 +1

2m2(v2f )

2









Ki = Kf ⇒ 1

2m1(v1i )

2 +1

2m2(v2i )

2 =1

2m1(v1f )

2 +1

2m2(v2f )

2

Collisions in 2 DimensionsAs an example, consider the case of a glancing collision.

264 Chapter 9 Linear Momentum and Collisions

Finalize The negative value for v2f means that block 2 is still moving to the left at the instant we are considering.

(C) Determine the distance the spring is compressed at that instant.

Conceptualize Once again, focus on the configuration of the system shown in Figure 9.10b.

Categorize For the system of the spring and two blocks, no friction or other nonconservative forces act within the sys-tem. Therefore, we categorize the system as an isolated system in terms of energy with no nonconservative forces acting. The system also remains an isolated system in terms of momentum.

Analyze We choose the initial configuration of the system to be that existing immediately before block 1 strikes the spring and the final configuration to be that when block 1 is moving to the right at 3.00 m/s.

S O L U T I O N

Write the appropriate reduction of Equation 8.2:

DK 1 DU 5 0

Evaluate the energies, recognizing that two objects in the system have kinetic energy and that the potential energy is elastic:

3 112m1v1f

2 1 12m2v2f

2 2 2 112m1v1i

2 1 12m2v2i

2 2 4 1 112kx2 2 0 2 5 0

Solve for x 2: x2 5 1k 3m1 1v1i

2 2 v1f2 2 1 m2 1v2i

2 2 v2f2 2 4

Substitute numerical values:

x2 5 a 1600 N/m

b5 11.60 kg 2 3 14.00 m/s 22 2 13.00 m/s 22 4 1 12.10 kg 2 3 12.50 m/s 22 2 11.74 m/s 22 4 6S x 5 0.173 m

Finalize This answer is not the maximum compression of the spring because the two blocks are still moving toward each other at the instant shown in Figure 9.10b. Can you determine the maximum compression of the spring?

9.5 Collisions in Two DimensionsIn Section 9.2, we showed that the momentum of a system of two particles is con-served when the system is isolated. For any collision of two particles, this result implies that the momentum in each of the directions x, y, and z is conserved. An important subset of collisions takes place in a plane. The game of billiards is a famil-iar example involving multiple collisions of objects moving on a two-dimensional surface. For such two-dimensional collisions, we obtain two component equations for conservation of momentum:

m1v1ix 1 m2v2ix 5 m1v1fx 1 m2v2fx

m1v1iy 1 m2v2iy 5 m1v1fy 1 m2v2fy

where the three subscripts on the velocity components in these equations repre-sent, respectively, the identification of the object (1, 2), initial and final values (i, f ), and the velocity component (x, y). Let us consider a specific two-dimensional problem in which particle 1 of mass m1 collides with particle 2 of mass m2 initially at rest as in Figure 9.11. After the collision (Fig. 9.11b), particle 1 moves at an angle u with respect to the horizontal and particle 2 moves at an angle f with respect to the horizontal. This event is called a glancing colli-sion. Applying the law of conservation of momentum in component form and noting that the initial y component of the momentum of the two-particle system is zero gives

Dpx 5 0 S pix 5 pfx S m1v1i 5 m1v1f cos u 1 m2v2f cos f (9.25)

Dpy 5 0 S piy 5 pfy S 0 5 m1v1f sin u 2 m2v2f sin f (9.26)

▸ 9.7 c o n t i n u e d

m1

m2

Before the collision

After the collision

v2f cos

v1f cos

v1f sin

2f sin

θφ

φ

φ

θ

θ

v

a

b

v1iS

v1fS

v2fS

Figure 9.11 An elastic, glancing collision between two particles.







S O L U T I O N


DK 1 DU 5 0


3 112m1v1f

2 1 12m2v2f

2 2 2 112m1v1i

2 1 12m2v2i

2 2 4 1 112kx2 2 0 2 5 0


2 2 v1f2 2 1 m2 1v2i

2 2 v2f2 2 4


x2 5 a 1600 N/m









▸ 9.7 c o n t i n u e d

m1

m2


After the collision

v2f cos

v1f cos

v1f sin

2f sin

θφ

φ

φ

θ

θ

v

a

b

v1iS

v1fS

v2fS


The velocity of particle 1 is in the x-direction.

x-components:

m1v1i = m1v1f cos θ+m2v2f cosφ

y -components:

0 = m1v1f sin θ−m2v2f sinφ

Collisions in 2 DimensionsAs an example, consider the case of a glancing collision.







S O L U T I O N


DK 1 DU 5 0


3 112m1v1f

2 1 12m2v2f

2 2 2 112m1v1i

2 1 12m2v2i

2 2 4 1 112kx2 2 0 2 5 0


2 2 v1f2 2 1 m2 1v2i

2 2 v2f2 2 4


x2 5 a 1600 N/m









▸ 9.7 c o n t i n u e d

m1

m2


After the collision

v2f cos

v1f cos

v1f sin

2f sin

θφ

φ

φ

θ

θ

v

a

b

v1iS

v1fS

v2fS








S O L U T I O N


DK 1 DU 5 0


3 112m1v1f

2 1 12m2v2f

2 2 2 112m1v1i

2 1 12m2v2i

2 2 4 1 112kx2 2 0 2 5 0


2 2 v1f2 2 1 m2 1v2i

2 2 v2f2 2 4


x2 5 a 1600 N/m









▸ 9.7 c o n t i n u e d

m1

m2


After the collision

v2f cos

v1f cos

v1f sin

2f sin

θφ

φ

φ

θ

θ

v

a

b

v1iS

v1fS

v2fS


The velocity of particle 1 is in the x-direction.

x-components:

m1v1i = m1v1f cos θ+m2v2f cosφ

y -components:

0 = m1v1f sin θ−m2v2f sinφ

Example 9.8 - Car collision

A 1500 kg car traveling east with a speed of 25.0 m/s collides atan intersection with a 2500 kg truck traveling north at a speed of20.0 m/s. Find the direction and magnitude of the velocity of thewreckage after the collision, assuming the vehicles stick togetherafter the collision.


Conceptualize Figure 9.12 should help you conceptualize the situation before and after the collision. Let us choose east to be along the positive x direction and north to be along the positive y direction.

Categorize Because we consider moments immediately before and immediately after the collision as defining our time interval, we ignore the small effect that friction would have on the wheels of the vehicles and model the two vehicles as an isolated system in terms of momentum. We also ignore the vehicles’ sizes and model them as particles. The collision is perfectly inelastic because the car and the truck stick together after the collision.

Analyze Before the collision, the only object having momentum in the x direction is the car. Therefore, the magnitude of the total initial momentum of the system (car plus truck) in the x direction is that of only the car. Similarly, the total initial momentum of the system in the y direction is that of the truck. After the collision, let us assume the wreckage moves at an angle u with respect to the x axis with speed vf .

S O L U T I O N

25.0i m/sˆ

20.0j m/sˆ

y

xu

vfS

Figure 9.12 (Example 9.8) An eastbound car colliding with a north-bound truck.

Apply the isolated system model for momen-tum in the x direction:

Dpx 5 0 S o pxi 5 o pxf S (1) m1v1i 5 (m1 1 m2)vf cos u

Apply the isolated system model for momen-tum in the y direction:

Dpy 5 0 S o pyi 5 o pyf S (2) m2v2i 5 (m1 1 m2)vf sin u

Divide Equation (2) by Equation (1):m2v2i

m1v1i5

sin ucos u

5 tan u

Solve for u and substitute numerical values: u 5 tan21am2v2i

m1v1ib 5 tan21 c 12 500 kg 2 120.0 m/s 211 500 kg 2 125.0 m/s 2 d 5 53.18

Use Equation (2) to find the value of vf and substitute numerical values:

vf 5m2v2i1m1 1 m2 2 sin u

512 500 kg 2 120.0 m/s 211 500 kg 1 2 500 kg 2 sin 53.18

5 15.6 m/s

Finalize Notice that the angle u is qualitatively in agreement with Figure 9.12. Also notice that the final speed of the combination is less than the initial speeds of the two cars. This result is consistent with the kinetic energy of the system being reduced in an inelastic collision. It might help if you draw the momentum vectors of each vehicle before the col-lision and the two vehicles together after the collision.

▸ 9.8 c o n t i n u e d

Example 9.9 Proton–Proton Collision

A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.50 3 105 m/s and makes a glancing collision with the second proton as in Figure 9.11. (At close separations, the pro-tons exert a repulsive electrostatic force on each other.) After the collision, one proton moves off at an angle of 37.08 to the original direction of motion and the second deflects at an angle of f to the same axis. Find the final speeds of the two protons and the angle f.

Conceptualize This collision is like that shown in Figure 9.11, which will help you conceptualize the behavior of the system. We define the x axis to be along the direction of the velocity vector of the initially moving proton.

Categorize The pair of protons form an isolated system. Both momentum and kinetic energy of the system are con-served in this glancing elastic collision.

AM

S O L U T I O N

1Serway & Jewett, page 265.

Example 9.8 - Car collisionThis is an inelastic collision.

x-components:

m1v1i = (m1 +m2)vf cos θ (1)

y -components:

m2v2i = (m1 +m2)vf sin θ (2)

Dividing (2) by (1):m2v2im1v1i

= tan θ

θ = tan−1

(m2v2im1v1i

)= 53.1◦

andvf =

m2v2i(m1 +m2) sin(53.1)

= 15.6 m/s

Example 9.8 - Car collisionThis is an inelastic collision.

x-components:

m1v1i = (m1 +m2)vf cos θ (1)

y -components:

m2v2i = (m1 +m2)vf sin θ (2)

Dividing (2) by (1):m2v2im1v1i

= tan θ

θ = tan−1

(m2v2im1v1i

)= 53.1◦

andvf =

m2v2i(m1 +m2) sin(53.1)

= 15.6 m/s

Example 9.14 - Exploding Rocket

A rocket is fired vertically upward. At the instant it reaches analtitude of 1000 m and a speed of vi = 300 m/s, it explodes intothree fragments having equal mass.

One fragment moves upward with a speed of v1 = 450 m/sfollowing the explosion. The second fragment has a speed ofv2 = 240 m/s and is moving east right after the explosion.

What is the velocity of the third fragment immediately after theexplosion?

(What is the change in kinetic energy of the system of the rocketparts?)


pi = pf ⇒ M vi =M

3(v1 + v2 + v3)

Let j point in the the upward vertical direction, and i point east.

v3 = 3vi − v1 − v2

= 3× 300 j− 450 j− 240 i

= (−240 i+ 450 j) m/s

Or, 510 m/s at an angle of 62◦ above the horizontal to the west.

(∆K = Kf − Ki =12M(4502 + 2402 + 5102) − 1

2(3M)(3002) =+1.25× 105M, a positive number)


pi = pf ⇒ M vi =M

3(v1 + v2 + v3)

Let j point in the the upward vertical direction, and i point east.

v3 = 3vi − v1 − v2

= 3× 300 j− 450 j− 240 i

= (−240 i+ 450 j) m/s

Or, 510 m/s at an angle of 62◦ above the horizontal to the west.

(∆K = Kf − Ki =12M(4502 + 2402 + 5102) − 1

2(3M)(3002) =+1.25× 105M, a positive number)

Center of Mass

The center of mass of a system is the average position of thesystem’s mass.

If the system is a point-like particle, the center of mass is just thepoint-particle’s location.

We modeled many systems that were not point-like as if they were.We pretended they were points located at their center of mass.

For translational motion, it is as if all of the system’s mass isconcentrated at that one point and all external forces are appliedat that point.

Center of Mass





Center of Mass





Center of Mass





Center of Mass


That model worked up until now, but we could not modelvibrations, rotations, or deformations.

The translational motion is independent of all these othermotions.

Center of Mass




Center of Mass




Center of Mass

For a solid, rigid object:

center of mass

the point on an object where we can model all the mass as being,in order to find the object’s trajectory; a freely moving objectrotates about this point

The center of mass of the wrench follows a straight line as thewrench rotates about that point.

Center of Mass

1Figure fromhttp://www4.uwsp.edu/physastr/kmenning/Phys203/Lect18.html

Center of Mass

For that system of two particles:

xCM =m1x1 +m2x2m1 +m2


is applied at the center of mass, the system moves in the direction of the force with-out rotating (see Fig. 9.13c). The center of mass of an object can be located with this procedure. The center of mass of the pair of particles described in Figure 9.14 is located on the x axis and lies somewhere between the particles. Its x coordinate is given by

xCM ;m1x1 1 m2x2

m1 1 m2 (9.28)

For example, if x1 5 0, x2 5 d, and m2 5 2m1, we find that xCM 5 23d. That is, the

center of mass lies closer to the more massive particle. If the two masses are equal, the center of mass lies midway between the particles. We can extend this concept to a system of many particles with masses mi in three dimensions. The x coordinate of the center of mass of n particles is defined to be

xCM ;m1x1 1 m2x2 1 m3x3 1 c1 mnxn

m1 1 m2 1 m3 1 c1 mn5

ai

mixi

ai

mi

5a

imixi

M5

1M a

imixi

(9.29)

where xi is the x coordinate of the ith particle and the total mass is M ; oi mi where the sum runs over all n particles. The y and z coordinates of the center of mass are similarly defined by the equations

yCM ;1M a

imiyi and zCM ;

1M a

imizi (9.30)

The center of mass can be located in three dimensions by its position vector rSCM. The components of this vector are xCM, yCM, and zCM, defined in Equations 9.29 and 9.30. Therefore,

rSCM 5 xCM i 1 yCM j 1 zCM k 51M a

imixi i 1

1M a

imiyi j 1

1M a

imizi k

rSCM ;1M a

imi r

Si (9.31)

where rSi is the position vector of the ith particle, defined by

rSi ; xi i 1 yi j 1 zi k

Although locating the center of mass for an extended, continuous object is some-what more cumbersome than locating the center of mass of a small number of par-ticles, the basic ideas we have discussed still apply. Think of an extended object as a system containing a large number of small mass elements such as the cube in Figure 9.15. Because the separation between elements is very small, the object can be con-sidered to have a continuous mass distribution. By dividing the object into elements of mass Dmi with coordinates xi, yi, zi, we see that the x coordinate of the center of mass is approximately

xCM <1M

ai

xi Dmi

with similar expressions for yCM and zCM. If we let the number of elements n approach infinity, the size of each element approaches zero and xCM is given pre-cisely. In this limit, we replace the sum by an integral and Dmi by the differential element dm:

xCM 5 limDmi S 0

1M

ai

xi Dmi 51M

3 x dm (9.32)

Likewise, for yCM and zCM we obtain

yCM 51M

3 y dm and zCM 51M

3 z dm (9.33)

CM

CM

CM

a

b

c

The system rotates clockwise when a force is applied above the center of mass.

The system rotates counter-clockwise when a force is applied below the center of mass.

The system moves in the direction of the force without rotating when a force is applied at the center of mass.

Figure 9.13 A force is applied to a system of two particles of unequal mass connected by a light, rigid rod.

Figure 9.14 The center of mass of two particles of unequal mass on the x axis is located at xCM, a point between the particles, closer to the one having the larger mass.

y

m1

x1

x 2

CM

m 2

x

x CM

Center of Mass

For that system of two particles:

xCM =m1x1 +m2x2m1 +m2

For more particles in 1 dimension:

xCM =

∑i mixi∑i mi

or

xCM =1

M

∑i

mixi

where M is the total mass of all the particles in the system.

Center of Mass

This expression also gives us the x coordinate of the center ofmass when we have more dimensions.

xCM =1

M

∑i

mixi

Likewise for y :

yCM =1

M

∑i

miyi

and z :

zCM =1

M

∑i

mizi

where M is the total mass of all the particles in the system.

Center of Mass

Therefore, we can condense all three expressions into a singlevector expression.

rCM =1

M

∑i

mi ri

where ri = xi i+ yi j+ zik is the displacement of particle i from theorigin.

Summary

• 2D collisions

• center of mass

Quiz tomorrow, in class.

3rd Collected Homework due Monday, Nov 20.

(Uncollected) Homework Serway & Jewett,

• Look at example 9.15 on page 275.

• PREV: Ch 9, onward from page 275. Probs: 35, 37, 41, 43

• Ch 9, onward from page 288. Probs: 67, 71, 77, 81

• Read Chapter 9, if you haven’t already.

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Linear Momentum 2D Collisions Extended or Composite ...nebula2.deanza.edu/~lanasheridan/4A/Phys4A-Lecture33.pdf · Linear Momentum 2D Collisions Extended or Composite Systems Center