Collisions[1]

8/13/2019 Collisions[1]

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Collisions

The previous example is an example of a collision. Collisions play acentral role in many parts of modern physics and are the basis of our currentunderstanding of particle physics. The essential effect of collisions is to

redistribute the total momentum of the colliding objects. We can classify allcollisions into one of three categories:

1. Elasticcollisions, which conserve kinetic energy,2. Inelasticcollisions, which do not conserve kinetic energy, and3. Completely inelastic collisions, in which the objects stick together

afterwards.

No matter what type of collision occurs, we can study them all in the same way.The guiding principle is that of conservation of linear momentum.

A simple way of dealing with collisions is to use the concept of the center

of mass. Suppose we have a collection of n particles, with a mass of m1,m2,...,mnrespectively. Also, let the position of the particles be given by (x1,y1),(x2,y2),...,(xn,yn). Then recall that the center of mass for this system is at

x

m x

m

CM

i i

i

n

i

i

n= =

=

1

1

We can now replace all of our equations of motion with equations involving the

center of mass; thus

A 3-kg object experiences an accel'n of 4 m/s/s. (54)

and


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F ma

m v

t

m

m

m

m a

ext CM

CM

i

v

t

i

n

i

i

n

i i

i

n

i

=

=

=

=

=

=

=

1

1

1

(55)

where aCMfollows from the above definitions. Notice that the force in (55) is theexternal force only. All of the internal forces acting between the particles willcancel out, leaving only the forces acting on the system as a whole. From thiswe see that when a system of particles is acted on by external forces, the center

of mass moves as if all of the mass was centered at the center of mass and all ofthe forces acted as a single resultant force at that point.

Elastic CollisionsNow let us look at collisions, starting with elastic collisions. Elastic

collisions are those that also conserve kinetic energy. It can be shown that dueto the conservation of kinetic energy, the relative velocities before the collisionare equal and opposite to the relative velocities after the collision. Thus, we canwrite for two particles

( )v v v v2 1 2 1 = ' ' (56)

In the center of mass frame, this implies that the two particles will approach eachother at some particular speed and after the collision they will recede from eachother with their velocity vector having the same magnitude but opposite sign.

Example:A ball of mass m1= 0.1 kg traveling with a velocity v1= 0.5 m/sec collides

head on with a ball of mass m2= 0.2 kg which is initially at rest. Calculate thefinal velocities, v1' and v2', in the event that the collision is elastic, in thelaboratory frame of reference.


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m1

m2

m1

m2

v1 v'

1v'

2

Using momentum conservation we have

m v m v m v m v

m v m v m v

1 1 2 2 1 1 2 2

1 1 1 1 2 2

+ = +

= +

' '

' '

solving for v2', get

( )v m

mv v2

1

2

1 1

' '=

We still need v1'. So using (56) we get

( )

( )

( )

( ) ( )

=

=

= +

=

+

=

=

v v v

v m

mv v v

m

mv

m

mv

v m m

m mv

1 2 1

11

2

1 1 1

1

2

1

1

2

1

11 2

1 2

1

1 1

017

' '

' '

'

'

.

-0.1 kg

0.3 kg0.5 ms

ms

and


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( )

( ) ( )

v m

mv

m m

m mv

m

m

m

m mv

m

m mv

2

1

2

1

1 2

1 2

1

1

2

2

1 2

1

1

1 2

1

2

2

2

0 33

'

.

=

+

=+

=+

=

=

0.1 kg

0.3 kg0.5 m

s

ms

Example:Solve the same problem as above, but in the center of mass frame.

In this case, the center of mass is moving with a velocity

( )( )

( ) ( )

v m v m v

m m

m v

m m

i

i

CM =

++

=+

=+

=

1 1 2 2

1 2

1 1

1 2

017

0.1 kg 0.5

0.1 kg 0.2 kg

ms

ms

.

so the velocities of the balls relative to the center of mass is

( ) ( )

u v v

i i

i

CM1 1

0 33

=

=

=

0.5 0.17ms

ms

ms

.

( )

u v v

i

i

CM2 2

0017

=

= =

0.17

m

s

ms

.

Since kinetic energy must be conserved, we get u1' = -u1, u2' = -u2and thus,converting back into the lab frame,


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( ) ( )

v u v

i i

i

CM1 1

017

' '

.

= +

= +

=

0.33 0.17ms

ms

ms

( ) ( )

v u v

i i

i

CM

m

s

2 2

0 33

' '

.

= += +

=

-0.17 0.17ms

ms

Completely Inelastic Collisions

Now let us turn to completely inelastic collisions.

Example:A 15 g bullet is fired into a 10 g wooden block that is mounted on wheels.

If the block moves 180 m in one second, what was the muzzle velocity of thebullet?

mb

vb

mt

V

Using momentum conservation we have

( )( )

m v m v m m V

m v m m V

b b t t b t

b b b t

+ = += +

or

( )

( )( )

( )

vm m V

mb

b t

b

=

+

=

+

=

0.010 kg kg 180

0.015 kg

ms

ms

0015

300

.

The change in kinetic energy is


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( )

( )( ) ( )( )

K m m V m vb t b v

= +

=

=

12

2 12

2

12

212

2

270

0.025 kg 180 0.015 kg 300

J

ms

ms

Example:If the rifle had a mass of 2.5 kg, what was the recoil velocity of the rifle in

the previous example?

The total momentum of the rifle + bullet was zero before the rifle was fired,so by conservation of momentum,

( )

( ) ( )

0

18

= += +

=

=

=

m v m v

m v m v

v m

mv

r r b b

r r b b

r

b

r

b

0.015 kg

2.5 kg300 ms

ms

.

Notice that, for recoils, if we take the ratio of the velocities of the objects we get(since the initial momentum was zero)

v

v

m

m

2

1

1

2

= (57)

similarly, the ratio of the kinetic energies is

12 1 1

2

12 2 2

2

1

2

1

2

2

1

2

2

1

2

2

1

m v

m v

m

m

v

v

m

m

m

m

m

m

=

=

=

(58)

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Collisions[1]