Galfer in motion Liceo Scientifico “Galileo Ferraris” gets in motion.
In Motion!
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Transcript of In Motion!
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In Motion!
Momentum and collisions
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Momentum affects collisions
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Conservation of momentum
• Unless an external force acts upon an object or system of objects, momentum is conserved (Newton’s first law restated)
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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).
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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
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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.
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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
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• 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.
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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 =?
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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
+ =
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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.
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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
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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
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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.
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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)