Transcript Powerpoint Slide

Lab 6: 1-D inelastic collisions
Only 6 more to go!!
Mo’mentum:
An object of mass m, and velocity v, has momentum.


p  mv
Remember that Newton’s 2nd Law is:


F  ma
  
  v v f  vi

that acceleration is equal to: a 
t
t
Newton’s 2nd Law as:
Also, remember
So we can rewrite
 


 m v m v f  vi  mv f  mvi
F


t
t
t
So now we have Newton’s 2nd Law in terms of momentum:

F


p f  pi
t

p

t
It turns out that the change in momentum is pretty important, in fact
we call the change in momentum, p the “IMPULSE”, and from
Newton’s 2nd law the IMPULSE is:
 
p  F t
Conservation of Mo’mentum:


pi  p f
Consider this situation:
BEFORE COLLISION:
v1i
v2i
m1
m2
AFTER COLLISION:
v1f
m1
m2
v2f
During the collision m1 exerts a force (F1) on m2, likewise m2 exerts a
Force (F2) on m1. According to Newton’s 3rd law, these forces are equal
But opposite, so:
F1 = -F2
Remember that
 
p  F t
So we can write:




F1t  p1  m1v1 f  m1v1i




F2 t  p2  m2 v2 f  m2 v2i
And since F1 = -F2 we get:




m1v1 f  m1v1i  (m2v2 f  m2v2i )




m1v1 f  m1v1i  m2v2 f  m2v2i
If a combine the terms for the initial momentum on the left and the
terms for final momentum on the right we get:




m1v1i  m2 v2i  m1v1 f  m2 v2 f
And this just proves that:


pi  p f
Let’s look at an example to illustrate how this works:
BEFORE COLLISION
+x
Cart #1
Cart #2
mass = m1
velocity = v1i
mass = m2
AFTER COLLISION
+x
Cart #1
velocity = v2i=0
Cart #2
mass = m1+m2
velocity = v(1+2)f
So how would we write the cons. of mo’mentum?
m1v1 + m2v2 = (m1 + m2) v(1+2)f
m1v1 = (m1 + m2) v(1+2)f
; but v2 = 0 so this becomes:
Let’s look at this example:
+x
mG = 5-kg
vB= 300 m/s
mB = 15-grams
What is the recoil velocity of the rifle? We can use the con. of mo’
mentum.:
pi =pf
(mB + mG) V(B+G) = mBVB – mGVG ; V(B+G) = 0 so we have,
0 = mBVB – mGVG ; solving for VG we get,
mB
0.015kg
VG 
VB 
300m / s  0.9m / s
mG
5kg
One more example:
mc = 1500 kg
vc = 10 m/s
When the car collides
with the tree it comes
to a stop in 0.2 s
What is the change in mo’mentum? p = pf – pi = 0 –(1500 kg)(10 m/s)
= -15000 kgm/s
What average force acted on the car during impact?
 p  15000kg m / s
F

 75,000 N
t
0.2s
Kinetic Energy: Kinetic Energy is the energy an object has do to its
motion. Mathematically kinetic energy is expressed as:
KE = ½ mv2
where m is the mass of an object, and v is its velocity
What happens if I triple the speed v?
KEOLD = ½ mv2
KENEW = ½ m(3v)2 = ½ m(9v2) = 9 KEOLD
You can see be tripling the speed I am able to ninedruple the KE
COLLISIONS: There are three types of collisions
1. Elastic:
KE and mo’mentum are both conserved, objects rebound
2. Inelastic: mo’mentum is conserved, some KE is lost but objects still
rebound
3. Perfectly inelastic: maximum amount of KE is lost, mo’mentum is
conserved. Objects stick together
Today we are going to measure the momentum and KE of objects
that experience a perfectly inelastic collision.
Velocity (m/s)
We will need to know the objects’ mass, and their speed both before
and after the collision.
Velocity before collision, VBC
Velocity after collision, VAC
Time (s)