Download CHAPTER 9- CONSERVATION of MOMENTUM DEFINITION ... and more Exams Physics in PDF only on Docsity!
CHAPTER 9- CONSERVATION of MOMENTUM
DEFINITION of MOMENTUM In physics we define force as the time rate of change of momentum. Momentum as force is a vector quantity.
F = d P /dt where P = M V
COLLISONS In a collision between objects the internal forces are equal and opposite thus balance. The net work performed is zero! Since F = M dV/dt = d/dt (MV) = 0 we say that the net Momentum = MV in the collision is conserved.
Pi = M VI for the ith particle
d/dt ( P1 + P2 + P3 + ..) = 0 then P1 + P2 + P3 = Constant
Momentum Conservation always hold whether the collision is elastic or inelastic! In elastic collision energy is conserved, in inelastic collisions energy is not conserved.
IMPULSE I = ∆ P We define the impulse on an object felt in a collision as the integral of the force exerted upon it over the time of impact. Below we see that I = ∆ P.
I = ∫ F dt = ∫ (d P /dt) dt = ∆ P = Pf - Pi
In the elastic collision of a ball with a solid wall where V = V’ the Impulse is calculated
I = (Pf – Pi) i = (- m V’ ) - (m V) = -2mV i
Note that the signs (directions) of the velocity vectors are important!
P
P
P
Net forces balance.
m
V
V’
m
before
after
ELASTIC COLLISIONS ∑ P = Constant & ∑ E =Constant
Elastic collisions are impacts where energy as well as momentum is conserved. Note that the signs (directions) of the velocity vectors are important.
m V1 - MV2 = -m V1’ + M V2’ 1/2 m V1^2 + 1/2 MV2^2 = 1/2 m V1’^2 + 1/2 M V2’^2
Solving this set of two equations for V1’ and V2’ the final velocities
V1’ = [(m-M)/(m+M)] V1 + (2M)/(m+M) V2 eq 9- V2’ = (2m)/(m+M)] V1 + [(M-m)/(m+M)] V2 eq 9-
Case I : m=M V2 =0 For equal masses they exchange velocities. V1’ = 0 This we observe when the cue ball hits another V2’ = V1 straight on.
Case II M>>m V2 =0 We expect M to barely move and m to recoil V1’ = -V1 letting M>>m in eq 9- V2’ ≅ 0
Case III m>>M V2=0 We expect both to move to he right V1’ = V V2’ = 2V
m M
V1 V
V1’
m M V2’
before
after
Before
After
Before
After
Before
After
SUM FORMULA
The center of mass of this cluster of blocks, all of mass M=1kg is:
Xcm = (1/8) [ 1x5 + 3x15 + 2x25 + 2x35] m = 19.5 m
Ycm = (1/8) [ 4x5 + 2x15 + 2x25 ] m = 12.5 m
R cm =( 19.5 i + 12.5 j ) m
For solid objects we can approximate the center of mass by using the location of
their centers and using the ∑ formulas.
INTEGRAL FORMULA
In some cases we must use the integral formulas. Consider a solid wedge of uniform density.
ρ = Mass / Area. = 1kg /m^2
dm = ρ dA where dA = y(x) dx
y(x) = (a/b) x the equation of the line
dm = ρ (a/b) x dx
First find the total mass M
M = 0
b
∫ dm =
0
b
∫ ρ^ (a/b) x dx
= ρ (a/b) 0
b
∫ x dx =^ ρ^ (a/b) b
(^2) /2 = 1/2 ρ a b (the area of a triangle times the density!)
Next the Xcm and Ycm
Xcm = 0
b
∫ ρ^ (a/b) x dx = (1/M)
0
b
∫ x dm =
0
b
∫ x^ ρ^ (a/b) x dx = (1/M)^ ρ^ (a /b )
0
b
∫ x
(^2) dx
= (2 / ρ a b ) ρ (a /b ) b^3 /3 = (2/3) b
Ycm = (1/3) a
0 10 20 30 40 m
a
b
y(x)
dx
dA= y(x) dx
