In Motion!

Momentum and collisions

Momentum affects collisions

Conservation of momentum

• Unless an external force acts upon an object or system of objects, momentum is conserved (Newton’s first law restated)

Let’s look at the example of two kids on skates facing each other (one boy, and one

girl).

• The girl’s mass is 55 kg, and the boy’s is 60.0 kg. The push causes the boy to slide backwards at 1.0 m/s. How fast does the girl slide?

 

The girl gives the boy a little push, they each move backwards (Newton’s third law).

REMEMBER: Momentum is conserved (there is no external force, like wind causing their momentum to change). Momentum before the push = momentum after the push!

Pboy + Pgirl = 0

(originally they have no motion, therefore no momentum)

After the push, the momentum of the system (the boy and the girl) is conserved (it still adds up to zero)

 Pboy + Pgirl = 0 mboyvboy + mgirlvgirl = 0

 

(60)(-1) + (55)(vgirl) = 0  -60 + 55 vgirl = 0 60= 55vgirl

vgirl = 60/55 = 1.1 m/s

Practice:

• Two ice skaters, a woman (mass=54 kg), and a man (mass= 88 kg) are initially facing each other on smooth ice where friction is negligible. Starting from rest they push off from each other. The woman moves away with a velocity of 2.5 m/s. Find the velocity of the man.

Solution

Given: mm = 88 kg, mw = 54 kg, vw =

2.5 m/s

vm = ?

  Pm + Pw = 0

  Pm = - Pw

mm(vm) = - (mw(vw) )

88(vm) = -(54)(2.5)

vm = -1.5 m/s

• Momentum is conserved in all situation. The total (sum) momentum of the system before the collision is equal to the total (sum) momentum of the system after the collision.

Conservation of Momentum Examples Elastic and Inelastic

CollisionsRemember momentum is conserved!

Psystem before = Psystem after

Inelastic Collisions (objects join together)Example: A sports car (mass=825 kg) travelling at 12 m/s crashes into a slower moving truck (mass = 1225kg) which is moving at 6.9 m/s in the same direction. If the two vehicles remain locked together (inelastic), how fast will they be going?

Given: mc=825 kg, mt= 1225 kg, vcar before=

12m/s, vtruck before = 6.9 m/s vcar&truck after =?

Before Collision After CollisionPc + Pt =

Pcar before = (825)(12)

= 9900 kg∙m/s

Ptruck before = 1225 (6.9)

=8452.5 kg∙m/s

P system before

= 9900 + 8452.5

= 18352.5 kg∙m/s

P (c & t together)

= 18352.5 kg∙m/s

= m(c & t)∙ v(c & t)

18352.5 = (825+1225)v

18352.5 = (2050)v

v = 18352.5/2080

v = 8.9 m/s

+ =

Elastic Example

• Example: A 125 g ball travelling at 55 cm/s to the right collides head–on with another ball at rest (m=215g). If the collision is totally elastic, calculate the velocity of the 125g ball after the collision if the 215g ball travels to the right at 17 m/s.

Given:

Initial Final

m1 = 125 g = .125 kg

m2 = 215 g = .215 kg

m1 = 125 g = .125 kg

m2 = 215 g = .215 kg

v1 = +55 cm/s

= .55 m/s

v1 = ?

v2 = 0 v2 = 17 m/s

Momentum of system before collision = momentum of system

after collision

p1i + p2i = p1f + p2f

p1i = m1(v1i) = (.125)(.55) = 0.06875 kg m/s

p2i = m2 v2i = 0

p1f = (.125) (v1f)

p2f = (.215)(17) = 3.655 kg m/s

p1i + p2i = p1f + p2f

0.06875 + 0 = (.125) (v1f) + 3.655

v1f = -29 m/s

Quick Quiz

A 125 g ball travelling at 5 m/s to the right collides head–on with another ball at rest (m=200g). If the collision is totally elastic, calculate the velocity of the 125g ball after the collision if the 200g ball travels to the right at 17 m/s.

Breaking Distance and Friction

• Braking Distance

• Braking distance is proportional to the square of the velocityd v2

This means if you are going twice as fast, it takes 4 X the distance to come to a stop!

• Braking distance also depends on the friction between the two surfaces (high friction, short breaking distance and vice-versa)