meh, tried to make a physics/unofficial joke. i got nothing. if a physics 211 student collides with...
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Meh, tried to make a physics/unofficial joke. I got nothing.If a Physics 211 student collides with another Physics 211 student on Unofficial, will the collision be elastic or inelastic? Can we please go over the kinetic energy of the system part?THE THIRD PART IS CONFUSING.. K-REFRENCE FRAME AND BLABLABLA. ARE THERE ANY EXAMPLES?Everything was fine until the energy of a system of particles and thats where I was lost. Could you go over the Pushing a block problem in class? can you go over the bowling ball and ping pong ball example? i didnt quite grasp the concept please go over bouncing ball Everything. This unit makes me want to cry.I get really overwhelmed with all the symbols and subscripts and asterisks* in the prelecture
SERIOUSLY...HOW DO NEUTRINOS TRAVEL FASTER THAN THE SPEED OF LIGHT? I love Impulse! I like today's prelecture. WHY DIDN'T WE LEARN ABOUT SAME RELATIVE SPEEDS BEFORE!!! THIS MAKES ELASTIC PROBLEMS SO MUCH EASIER. I think you should make an announcement about how you can rate problems..
Mechanics Lecture 13, Slide 1
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Physics 211Lecture 13
Today’s Concepts:a) More on Elastic Collisionsb) Average Force during Collisions
Mechanics Lecture 13, Slide 2
In the last slides of the prelecture, it talked about KE in terms of different reference frames and in one of the reference frames the KE = 0 why?
Throw Something
Mechanics Lecture 13, Slide 3
In CM frame, the speed of an object before an elastic collision is the same as the speed of the object after.
|v*1,i - v*
2,i|
But the difference of two vectors is the same in any reference frame.
Rate of approach before an elastic collision is the same as the rate of separation afterward, in any reference frame!
= |-v*1, f + v*
2, f |
= |-(v*1, f - v*
2, f)| = |v*1, f - v*
2, f |
So the magnitude of the difference of the two velocities is the same before and after the collision.
More on Elastic Collisions
m1m2
V*1,i V*
2,i
m2V*1, f V*
2, fm1
Just Remember This
Mechanics Lecture 13, Slide 4
Clicker Question
Consider the two elastic collisions shown below. In 1, a golf ball moving with speed V hits a stationary bowling ball head on. In 2, a bowling ball moving with the same speed V hits a stationary golf ball.
In which case does the golf ball have the greater speed after the collision?
A) 1 B) 2 C) same
V
1 V 2
Mechanics Lecture 13, Slide 5
Clicker Question
A small ball is placed above a much bigger ball, and both are dropped together from a height H above the floor. Assume all collisions are elastic.
What height do the balls bounce back to?
A
H
Before B CAfterMechanics Lecture 13, Slide 6
Explanation
For an elastic collision, the rate of approach before is the same as the rate of separation afterward:
v
v
m
M
Rate of approach= 2V
Rate of separation= 2V
v m
M
3v
v
v
v
Mechanics Lecture 13, Slide 7
A block slides to the right with speed V on a frictionless floor and collides with a bigger block which is initially at rest. After the collision the speed of both blocks is V/3 in opposite directions. Is the collision elastic?A) Yes B) No
CheckPoint
Mechanics Lecture 13, Slide 8
V
V/3V/3
Before Collision
After Collision
m M
Mm
F
t
ti tf
t
totF ma
dt
PdFtot
PddtFtot
12
2
1
tPtPdtFt
t
tot
tFave
P
Impulse
tFdtF ave
t
t
tot 2
1
Forces during Collisions
Mechanics Lecture 13, Slide 9
Fave
tFP ave
Clicker Question
The change in momentum of block B is:
A) Twice the change in momentum of block AB) The same as the change in momentum of block AC) Half the change in momentum of block A
F BF A
tFP ave
Two blocks, B having twice the mass of A, are initially at rest on frictionless air tracks. You now apply the same constant force to both blocks for exactly one second.
air trackair track
Mechanics Lecture 13, Slide 10
Clicker Question
Two boxes, one having twice the mass of the other, are initially at rest on a horizontal frictionless surface. A force F acts on the lighter box and a force 2F acts on the heavier box. Both forces act for exactly one second.
Which box ends up with the biggest momentum?
A) Bigger box B) Smaller box C) same
tFP ave
F 2F M 2M
Mechanics Lecture 13, Slide 11
CheckPoint
A) Four times as long.B) Twice as long.C) The same length.D) Half as long.E) A quarter as long.
Lets try it again !
A constant force acts for a time Dt on a block that is initially at rest on a frictionless surface, resulting in a final velocity V. Suppose the experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity?
Mechanics Lecture 13, Slide 12
F
The experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity?
A) Four times as long.B) Twice as long.C) The same length.
A) Ft=mv, if v is the same, m is doubled, F is cut in half, so t has no choice but to be four times as big
B) The change in momentum is simply the force times the time of collision. Since the force is halved, the time it would take to reach the same momentum must be doubled.
C) Ft=mv, the 2 and the one half cancel
Mechanics Lecture 13, Slide 13
F
Case 1 Case 2
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1 the ball bounces off a cement floor and in Case 2 the ball bounces off a piece of stretchy rubber. In which case is the average force acting on the ball during the collision the biggest?A) Case 1 B) Case 2
Lets try it again !
CheckPoint
Mechanics Lecture 13, Slide 14
Case 1 Case 2
In which case is the average force acting on the ball during the collision the biggest?A) Case 1 B) Case 2
A) The time of contact is less so the force must be greater.
B) Ball is on the ground for a greater amount of time so more force acts on it.
ave
PF
t
Mechanics Lecture 13, Slide 15
CheckPoint Responses
HW Problem with demo
Mechanics Lecture 13, Slide 16
pf
pi
pf-pi
DP = pf -pi
|DPX | = 2mv cos(q)cosq|DPY | = 0
mv
Another way to look at it:
Mechanics Lecture 13, Slide 17
|FAVE | = |DP | /Dt = 2mv cos(q) /Dt
Mechanics Lecture 13, Slide 18
pf
pi
|DP| = |pf –pi| = m|vf –vi|
Dt = DP/FAVE
2 21 1
2 2f iK mv mv
pf -pi
DP = pf -pi
FAVE = DP/Dt
Another way to look at it:
Mechanics Lecture 13, Slide 19
H
h
vi vf
= reading on scale
Scale
f ip m v v
time between bouncest
ave
pF
t
2iv gH 2fv gh
Demo – similar concept
Mechanics Lecture 13, Slide 20
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