Download Lecture 17

Physics I
95.141
LECTURE 17
11/8/10
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Outline/Administrative Notes
• Outline
– Ballistic Pendulums
– 2D, 3D Collisions
– Center of Mass and
translational motion
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
• Notes
– HW Review Session
on 11/17 shifted to
11/18.
– Last day to withdraw
with a “W” is 11/12
(Friday)
Ballistic Pendulum
• A device used to measure the speed of a
projectile.
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M
vo
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
M+m
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Ballistic Pendulum
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95.141, F2010, Lecture 17
Department of Physics and Applied Physics
M+m
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Ballistic Pendulum
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95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Exam Prep Problem
• You construct a ballistic “pendulum” out of a rubber block (M=5kg)
attached to a horizontal spring (k=300N/m). You wish to determine
the muzzle velocity of a gun shooting a mass (m=30g). After the
bullet is shot into the block, the spring is observed to have a
maximum compression of 12cm. Assume the spring slides on a
frictionless surface.
• A) (10pts) What is the velocity of the block + bullet immediately
after the bullet is embedded in the block?
• B) (10pts) What is the velocity of the bullet right before it collides
with the block?
• C) (5pts) If you shoot a 15g mass with the same gun (same
velocity), how far do you expect the spring to compress?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Exam Prep Problem
• k=300N/m, m=30g, M=5kg, Δxmax=12cm
• A) (10pts) What is the velocity of the block + bullet immediately
after the bullet is embedded in the block?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Exam Prep Problem
• k=300N/m, m=30g, M=5kg, Δxmax=12cm
• B) (10pts) What is the velocity of the bullet right before it
collides with the block?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Exam Prep Problem
• k=300N/m, m=30g, M=5kg, Δxmax=12cm
• C) (5pts) If you shoot a 15g mass with the same gun (same
velocity), how far do you expect the spring to compress?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Collisions
• In the previous lecture we discussed collisions in
and the role of Energy in collisions.
– Momentum always conserved!


psystem  psystem
1D,
– If Kinetic Energy is conserved in a collision, then we call this an
elastic collision, and we can write:
mAv A  mBvB  mAvA  mBvB
– Which simplifies to:
1
1
1
1
m Av 2A  m B v B2  m Av A2  m B v B2
2
2
2
2
v A  vB  vB  vA
– If Kinetic Energy is not conserved, the collision is referred to as
an inelastic collision.
– If the two objects travel together after a collision, this is known as
a perfectly inelastic collision.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Collision Review
• Imagine I shoot a 10g projectile at 450m/s towards a
10kg target at rest.
– If the target is stainless steel, and the collision is elastic, what
are the final speeds of the projectile and target?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Collision Review
• Imagine I shoot a 10g projectile at 450m/s towards a
10kg target at rest.
– If the target is wood, and projectile embeds itself in the target,
what are the final speeds of the projectile and target?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Additional Dimensions
• Up until this point, we have only considered collisions in
one dimension.
• In the real world, objects tend to exist (and move) in
more than one dimension!
• Conservation of momentum holds for collisions in 1, 2
and 3 dimensions!
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
2D Momentum Conservation
• Imagine a projectile (mA) incident, along the x-axis, upon
a target (mB) at rest. After the collision, the two objects
go off at different angles  A , B
• Momentum is a vector, in order for momentum to be
conserved, all components (x,y,z) must be conserved.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
2D Momentum Conservation
• Imagine a projectile (mA) incident, along the x-axis, upon
a target (mB) at rest. After the collision, the two objects
go off at different angles  A , B
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Conservation of Momentum (2D)
• Solving for conservation of momentum gives us
2 equations (one for x-momentum, one for ymomentum).
• We can solve these if we have two unknowns
• If the collision is elastic, then we can add a third
equation (conservation of kinetic energy), and
solve for 3 unknowns.
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Department of Physics and Applied Physics
Example problem
• A cue ball travelling at 4m/s strikes a billiard ball at rest (of equal
mass). After the collision the cue ball travels forward at an angle of
+45º, and the billiard ball forward at -45º. What are the final speeds of
the two balls?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Example Problem II
• Now imagine a collision between two masses (mA=1kg and mB=2kg)
travelling at vA=2m/s and vB= -2m/s along the x-axis. If mA bounces
back at an angle of -30º, what are the final velocities of each ball?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Example Problem II
•
Now imagine a collision between two masses (mA=1kg and mB=2kg)
travelling at vA=2m/s and vB= -2m/s on the x-axis. If mA bounces back at an
angle of -30º, what are the final velocities of each ball, assuming the
collision is elastic?
mAv A  mBvB  mAvA cos A  mBvB cos B
vA  3.26 m s
vB  0.82 m s
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
0  mAvA sin  A  mBvB sin  B
 B  63.8
Simplification of Elastic Collisions
• In 1D, we showed that the conservation of
Kinetic Energy can be written as:
v A  vB  vB  vA
• This does not hold for more than one
dimension!!
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Problem Solving: Collisions
1. Choose your system. If complicated (ballistic
pendulum, for example), divide into parts
2. Consider external forces. Choose a time interval where
they are minimal!
3. Draw a diagram of pre- and post- collision situations
4. Choose a coordinate system
5. Apply momentum conservation (divide into component
form).
6. Consider energy. If elastic, write conservation of
energy equations.
7. Solve for unknowns.
8. Check solutions.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Center of Mass
• Conservation of momentum is powerful for collisions and
analyzing translational motion of an object.
• Up until this point in the course, we have chosen objects
which can be approximated as a point particle of a
certain mass undergoing translational motion.
• But we know that real objects don’t just move
translationally, they can rotate or vibrate (general
motion)  not all points on the object follow the same
path.
• Point masses don’t rotate or vibrate!
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Center of Mass
• We need to find an addition way to describe
motion of non-point mass objects.
• It turns out that on every object, there is one
point which moves in the same path a particle
would move if subjected to the same net Force.
• This point is known as the center of mass (CM).
• The net motion of an object can then be
described by the translational motion of the CM,
plus the rotational, vibrational, and other types of
motion around the CM.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Example
F
F
F
F
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Center of Mass
• If you apply a force to an non-point object, its
center of mass will move as if the Force was
applied to a point mass at the center of mass!!
• This doesn’t tell us about the vibrational or
rotation motion of the rest of the object.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Center of Mass (2 particles, 1D)
• How do we find the center of mass?
• First consider a system made up of two point masses,
both on the x-axis.
x=0
mA
xB
xA
xB
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
x-axis
Center of Mass (n particles, 1D)
• If, instead of two, we have n particles on the x-axis, then
we can apply a similar formula to find the xCM.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Center of Mass (2D, 2 particles)
• For two particles lying in the x-y plane, we can find the
center of mass (now a point in the xy plane) by
individually solving for the xCM and yCM.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Center of Mass (3D, n particles)
• We can extend the previous CM calculations to nparticles lying anywhere in 3 dimensions.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Example
• Suppose we have 3 point masses (mA=1kg, mB=3kg and
mC=2kg), at three different points: A=(0,0,0), B=(2,4,-6)
and C=(3,-3,6).
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Department of Physics and Applied Physics
Solid Objects
• We can easily find the CM for a collection of point
masses, but most everyday items aren’t made up of 2 or
3 point masses. What about solid objects?
• Imagine a solid object made out of an infinite number of
point masses. The easiest trick we can use is that of
symmetry!
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Department of Physics and Applied Physics
CM and Translational Motion
• The translational motion of the CM of an object is directly
related to the net Force acting on the object.


MaCM   Fext
• The sum of all the Forces acting on the system is equal
to the total mass of the system times the acceleration of
its center of mass.
• The center of mass of a system of particles (or objects)
with total mass M moves like a single particle of mass M
acted upon by the same net external force.
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Example
• A 60kg person stands on the right most edge of a uniform board of
mass 30kg and length 6m, lying on a frictionless surface. She then
walks to the other end of the board. How far does the board move?
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Solid Objects (General)
• If symmetry doesn’t work, we can solve for CM
mathematically.
– Divide mass into smaller sections dm.
dm

r
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Solid Objects (General)
• If symmetry doesn’t work, we can solve for CM
mathematically.
– Divide mass into smaller sections dm.
xCM
xCM 
1
xdm

M
1

M
 x dm
i
i
i
yCM 
1
M
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
 ydm
zCM 
1
zdm

M
Example: Rod of varying density
• Imagine we have a circular rod (r=0.1m) with a mass
density given by ρ=2x kg/m3.
x
L=2m
95.141, F2010, Lecture 17
Department of Physics and Applied Physics
Example: Rod of varying density
• Imagine we have a circular rod (r=0.1m) with a mass
density given by ρ=2x kg/m3.
x
L=2m
95.141, F2010, Lecture 17
Department of Physics and Applied Physics