Transcript ch6

Physics 231
Topic 6: Momentum and Collisions
Alex Brown
October 7 2015
MSU Physics 231 Fall 2015
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Momentum
F =ma
Newton’s 2nd law
F = m v/t
a=v/t
F = m (vfinal - vinital)/t
Define p = mv
p: momentum (kg m/s)
F= (pfinal - pinitial)/t
F= p/t = m v/t
p = m v
impulse = mass times change of v
The net force acting on an object equals
the change in momentum (p) in time period (t).
Since velocity is a vector, momentum is also
a vector, pointing in the same direction as v.
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Impulse
F=p/t Force=change in (mv) per time
period (t).
p=Ft The change in momentum equals
the force acting on the object times
how long you apply the force.
Definition: p = Impulse
F2
F1
What if the force is not constant within the time
period t?
p = (F1 t + F2 t + F3 t)
F3
= area under F vs t plot
s s s
t
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car hitting haystack and stoping
car hitting wall
The change in momentum (impulse) is the same, but the force
reaches a much higher value when the car hits a wall!
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Some examples
A tennis player receives a shot
approaching him (horizontally) with
50m/s and returns the ball in the
opposite direction with 40m/s. The mass
of the ball is 0.060 kg.
A) What is the impulse delivered
by the ball to the racket?
B) What is the change in kinetic energy
of the ball?
A) Impulse=change in momentum (p).
p = m (vfinal - vinitial) = 0.060(-40-50) = -5.4 kg m/s
B) KEi = 75 J KEf = 48 J (no PE!)
Enc = KEi - KEf = 27 J (energy lost to heat)
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Child safety
A friend claims that it is safe to go on a car trip with your
child without a child seat since he can hold onto your 12kg
child even if the car makes a frontal collision (lasting 0.05s
and causing the vehicle to stop completely) at v=50 km/h
(about 30 miles/h). Is he to be trusted?
F = p/t
= m (vf-vi) /t
m = 12kg
force = impulse per time period
vf = 0 and vi = 50 km/h = 13.9 m/s
t=0.05s
F = 12(-13.9)/0.05 = -3336 N
This force corresponds to lifting a mass of 340 kg!!!
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Question (assume t is the same for both cases)
The velocity change is largest in case:
A B
The acceleration is largest in case:
A B
The momentum change is largest in case:
A B
The impulse is largest in case:
A B
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Conservation of Momentum
F21 t = p1 = p1f - p1i
p1i
p2i
Newton’s 3rd law: Force of 2 on 1 = F21
m1
F12 = - F21
m2
m1
F12 t = p2 = p2f – p2i
F12 t = - F21 t
p1f - p1i = - p2f + p2i
p1f
m1
m2
m2
F12
p2f
Rewrite: p1i + p2i = p1f + p2f
CONSERVATION OF MOMENTUM
(only for a closed system)
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p1 = m1 v1
p2 = m2 v2
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question
An object is initially at rest in outer space (no gravity).
It explodes and breaks into two pieces, not of equal size.
The magnitude of the momentum of the larger piece is …
a) Greater than that of the smaller piece
b) Smaller than that of the smaller piece
c) The same as that of the smaller piece
Initial momentum is zero
0 = plarge,final + psmall,final
So plarge,final = - psmall,final
Magnitude is the same, direction opposite.
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Quiz
An object is initially at rest in outer space (no gravity).
It explodes and breaks into three pieces, not of equal size.
The momentum of the largest piece:
a) Must be larger than that of each of the smaller pieces
b) Must be greater than that of each of the smaller pieces
c) Must be equal in magnitude but opposite in direction to the sum
of the smaller pieces.
d) Cannot be determined from the momenta of the smaller pieces
The conservation of momentum must hold, even with
more than 2 objects. Although the momenta of each
piece could be pointing in a different direction, the
sum must be equal to zero after the explosion. So, the
momentum of one of the pieces must always be
opposite in direction, but equal in magnitude to the sum
of the momenta of the other two pieces.
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Moving in space
An astronaut (100 kg) is drifting away
from the spaceship with v=0.2 m/s. To get
back he throws a wrench (2 kg) in the
direction away from the ship. With what
velocity does he need to throw the wrench
to move with v=0.1 m/s towards the ship?
(a) 0.1 m/s (b) 0.2 m/s (c) 5 m/s (d) 16 m/s
(e) this will never work?
Initial momentum: mavai + mwvwi =100*0.2+2*0.2 = 20.4 kg m/s
After throw: mavaf + mwvwf = 100*(-0.1) + 2*vwf kg m/s
Conservation of momentum: mavai + mwvwi = mavaf + mwvwf
20.4 = -10 + 2*vwf
vwf=15.7 m/s
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Perfectly inelastic collisions
Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f
After the collision m1 and m2 form one new object with mass
M = m1 + m2 and one velocity vf =v1f = v2f
m1v1i + m2v2i = vf (m1 + m2)
vf = (m1v1i + m2v2i)/ (m1 + m2)
after
before
Demo: perfect inelastic collision on airtrack
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Perfectly inelastic collisions
Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f
After the collision m1 and m2 form one new object with mass
M = m1 + m2 and one velocity vf =v1f = v2f
m1v1i + m2v2i = vf (m1 + m2)
vf = (m1v1i + m2v2i)/ (m1 + m2)
Let
c = m2/m1
then
vf = (v1i + c v2i)/ (1+c)
If m1 = m2 (c=1) then vf = (v1i + v2i)/ 2
If c=1 and v2i = 0 then
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vf = v1i/2
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Perfect inelastic collision: an example
50 m/s 20 m/s
25 m/s
A car collides into the back of a truck and their bumpers get
stuck. What is the ratio of the mass of the truck and the car?
(mtruck/mcar = m2/m1 =c) What is the fraction of KE lost?
vf = [v1i + cv2i] / (1+c)
solve for c
c = [v1i - vf] / [vf – v2i] = (50-25)/(25-20) = 5
Before collision: KEi = ½m1(50)2 + ½5m1(20)2
After collision:
KEf = ½6m1(25)2
Ratio: KEf/KEi= [6*(25)2]/[(50)2+5*(20)2] = 0.83
17% of the KE is lost (damage to cars!)
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Types of collisions
Inelastic collisions
Elastic collisions
•Momentum is conserved
•Momentum is conserved
•Some energy is lost in the •No energy is lost in the
collision: kinetic energy is collision: kinetic energy is
not conserved
conserved
•Perfectly inelastic: the
objects stick together.
KEi - KEf = Enc
KEi - KEf = 0
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Inelastic Solution
Conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f
After the collision m1 and m2 form one new object with mass
M = m1 + m2 and one velocity vf =v1f = v2f
m1v1i + m2v2i = vf (m1 + m2)
vf = [m1 v1i + m2v2i] / (m1+m2)
With
KEi - KEf = Enc
m2 = c m1
vf = [v1i + cv2i] / (1+c)
Special case
m2 = m1
(c=1)
vf = [v1i + v2i] / 2
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Elastic collisions
Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f
Conservation of KE: ½m1v1i2+½m2v2i2=½m1v1f2+½m2v2f2
Two equations and two unknowns
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Elastic Solution
v1f = [(m1 –m2) v1i + 2m2v2i] / (m1+m2)
v2f = [(m2 –m1) v2i + 2m1v1i] / (m1+m2)
For
KEi - KEf = 0
m2 = c m1
v1f = [(1-c) v1i + 2cv2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)
Special case
m2 = m1
(c=1)
v1f = v2i
v2f = v1i
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Elastic collisions of equal mass
Given m2=m1 and v1i=0
What are the velocities of m1 and m2 after the collision in
terms of the initial velocity of m2 if m1 is originally at rest?
m1
m2
For
m2 = c m1
v1f = [(1-c) v1i + 2cv2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)
m1
m2
c=1
v1f = v2i
v2f = v1i = 0
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inelastic collision
With
m2 = c m1
vf = [v1i + cv2i] / (1+c)
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elastic collisions
For
m2 = c m1
v1f = [(1-c) v1i + 2cv2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)
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elastic collisions
For
m2 = c m1
v1f = [(1-c) v1i + 2cv2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)
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Quiz
In a system with 2 moving objects when a collision occurs
between the objects:
a) the total kinetic energy is always conserved
b) the total momentum is always conserved
c) the total kinetic energy and total momentum are always
conserved
d) neither the kinetic energy nor the momentum is conserved
For both inelastic and elastic collisions, the momentum is
conserved.
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Momentum p = mv
Impulse
p = pf - pi = Ft
Conservation of momentum (closed system) p1i + p2i = p1f + p2f
Collisions in one dimension given
given m1 and m2 or c = m2 /m1
v1i and v2i
Inelastic (stick together, KE lost)
vf = [v1i + cv2i] / (1+c)
Elastic (KE conserved)
v1f = [(1-c) v1i + 2cv2i] / (1+c)
v2f = [(c-1) v2i + 2v1i] / (1+c)
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Quiz
B
A
Consider a closed system of objects A and B.
Object A is thrown at an object B that was originally at rest, as shown in
the figure. It is not known if A is heavier or lighter than B. After A
collides with B:
a) Object A must go left (back in the direction it came from)
b) Object A must go right
c) Object A could go left or right
d) Object B could go left
e) The total momentum will be different than before the collision
If the collision is perfectly inelastic. A & B stick together and must go
right (conservation of momentum).
If the collision is elastic and mA<<mB A will go left. If mA>>mB both will
go right.
So A could go left or right. B can not go left (A would have to go left
to, which violates momentum conservation.
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Ballistic Pendulum
How high will the block go?
Mblock=1 kg
Mbullet=0.1 kg
Vbullet=20 m/s
h
There are 2 stages:
• The collision
• The Swing of the block
The collision The bullet gets stuck in the block (perfect
inelastic collision). Use conservation of momentum.
m1 v1i + m2 v2i=vf (m1+m2) so: vf=1.8 m/s
The swing of the block Use conservation of Mechanical energy.
(mgh + ½mv2)start of swing = (mgh + ½mv2)at highest point
0 + ½1.1(1.8)2
= 1.1*9.81*h + 0
so h=0.165 m
Why can’t we use Conservation of ME right from the start??
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Transporting momentum
For elastic collision of equal masses
2
1
0
2
1
v2f = 0
v1f = v2i
0
2
v2f= 0
v1f= v2i
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0
v1f = 0
v0f = v1i
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Quiz
The kinetic energy of an object is quadrupled.
Its momentum will change by a factor of…
a) 0
b) 2
c) 4
d) 8
e) 16
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Impact of a meteorite
Estimate what happens if a 2.8x1013 kg meteorite collides with earth:
a) Is the orbit of earth around the sun changed?
b) how much energy is released?
Assume: meteorite has same density as earth, the collision is
inelastic and the meteorite’s v is 10 km/s (relative to earth)
A) Earth’s mass: me = 6x1024kg
mm = 2.8x1013 kg
ve=0
vm=1x104 m/s
Conservation of momentum: meve + mmvm= (me+mm) vme =mme vme
0 + (2.8x1013)(1x104) = (6x1024)vme so
vme=4.7x10-8 m/s (very small)
B) Energy=Kinetic energy loss: (½meve2+½mmvm2) - (½mmevme2)
0 + 0.5 (2.8x1013) (1x104)2 - 0.5 (6x1024) (4.7x10-8)2 = 1.4x1021 J
Largest nuclear bomb existing: 100 megaton TNT = 4.2x1017 J
Energy release: 3300 nuclear bombs!!!!!
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m1
h=5m
problem
m1=5 kg m2=10 kg. M1 slides down and makes
an elastic collision with m2. How high does
m1 go back up?
m2
ME = mgh + (1/2)mv2
Step 1. What is the velocity of m1 just before it hits m2?
Conservation of ME: MEtop= MEbottom = m1gh + 0 = 0 + 0.5m1v2
So v1 = 9.9 m/s
Step 2. Elastic collision with c = 2, v1i = 9.9 and v2i = 0
so v1f = -3.3
Step 3. m1 moves back up; use conservation of ME again.
MEtop= MEbottom = m1gh’ + 0 = 0 + 0.5 m1 v1f2
So h’ = 0.55 m
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Playing with blocks
m1=0.5 kg collision is elastic
m2=1.0 kg
h1=2.5 m
h2=2.0 m
A) determine the velocity of the blocks after the collision
B) how far away from the bottom of the table does m2 land
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Determine the velocity of the blocks after the collision
m1=0.5 kg collision is elastic
m2=1.0 kg
h1=2.5 m
h2=2.0 m
Step 1: determine velocity of m1 at the bottom of the slide
Conservation of ME (mgh+½mv2)top =(mgh+½mv2)bottom
0.5*9.81*2.5 + 0
= 0 + 0.5*0.5*v2
so: v1=7.0 m/s
Step 2: Elastic collision equation with c = 2, v1i = 7 and v2i = 0
Gives v1f = -2.33 and v2f = 4.67
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How far away from the table does m2 land?
m1=0.5 kg collision is elastic
m2=1.0 kg
h1=2.5 m
h2=2.0 m
v1f=-2.3 m/s v2f=4.7 m/s
This is a parabolic motion with initial horizontal velocity.
Horizontal
vertical
x = vxot
y = yo + vyot - ½gt2
x = 4.7 t
0 = 2.0 + 0 - 0.5*9.81*t2
= 2.96 m
t=0.63 s
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(A)
v y  v0 y  at
(B)
y  voy t  at
2
(C)
y  v y t  at
2
(D)
y  12 (voy  v y )t
(E)
y 
1
2
1
2
1
2a
with a = -g
(v y  v0 y )
2
2
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at the top all have vy = 0 (why?)
at the top and bottom vx (Q) < vx (N) (why?)
at the top vx (P) = vx (R) = 0 (why?)
For all (PE+KE) (bottom)
= (PE+KE) (top)
mvo2/2 = m(vox2 + voy2)/2 = mgh + [mvx2/2]
For Q and N get t from (C)
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a) Speed at X compared to V ?
b) Velocity at X compared to P ?
c) Size of total force at R compared to T ?
d) Size of total force at Q compared to T ?
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a) Speed at X compared to V ? X larger than V
b) Velocity at X compared to P ? same
c) Size of total force at R compared to T ? T larger than R
d) Size of total force at Q compared to T ? Q larger than T
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Remember at the a given time
Power = P = F v = m a v
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question
A girl (g) is standing on a plank (p) that is lying on ice
(assume frictionless surface between ice and board). She
starts to walk with 1 m/s, relative to the plank. What is
her speed relative to the ice? Use mg=70 kg mp=100 kg
Choose the ice as a frame of reference
Initial: both she and the plank are at rest.
Final: her velocity relative to the ice:
vgp = vg – vp
vg = v (girl relative to ice)
vp = v (plank relative to ice)
vp = vg - vgp
vgp = v (girl relative to plank)
Conservation of momentum: Initial (not moving): =0
Final: mg vg+ mp vp = mg vg + mp(vg-vgp) = 70 vg + 100(vg-1) = 0
Solve for vg = 0.59 m/s
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Center of Mass/Gravity
We often deal with extended objects
and need to identify the central
(average) location of the mass.
This location, also called the center
of mass (or gravity) can be found by
averaging the x and y positions:
𝒙𝒄𝒎 =
𝒎𝒊 𝒙𝒊
𝒎𝒊
𝒚𝒄𝒎 =
𝒎𝒊 𝒚𝒊
𝒎𝒊
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Center of Mass/Gravity
Given three objects of different mass, find
the center of mass of the system.
M1: mass = m x = -L/2 y = 0
M2: mass = m x = +L/2 y = 0
M3: mass = 2m x = 0
y = L cos(30)
𝒙𝒄𝒎 =
𝒙𝒄𝒎 =
𝒙𝒄𝒎
M3
𝒎𝒊 𝒙𝒊
𝒎𝒊
(𝒎𝟏 𝒙𝟏 + 𝒎𝟐 𝒙𝟐 + 𝒎𝟑 𝒙𝟑 )
𝒎𝟏 + 𝒎𝟐 + 𝒎𝟑
M1
𝑳𝒎 𝑳𝒎
(− 𝟐 + 𝟐 + 𝟎)
𝟎
=
=
=𝟎
𝒎 + 𝒎 + 𝟐𝒎
𝟒𝒎
𝒚𝒄𝒎 =
-L/2
+L/2
M2
(𝟎 + 𝟎 + 𝟐𝒎𝑳𝒄𝒐𝒔(𝜽)) 𝟐𝒎𝑳𝒄𝒐𝒔(𝜽) 𝑳𝒄𝒐𝒔 𝜽
=
=
𝒎 + 𝒎 + 𝟐𝒎
𝟒𝒎
𝟐
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