Conservation of Linear Momentum
DATA ANALYSIS
Mass Measurements:
M (kg) uM (kg) uM/M
Mass 1
Mass 2
Speed Before collision:
v (m/s) uv (m/s) uv/v
Mass 1
Mass 2
Angle Before Collision:
θ (deg) uθ (rad)
Mass 1
Mass 2
Speed After Collision:
v (m/s) uv (m/s)
Mass 1
Mass 2
Angles After Collision:
θ (deg) uθ (rad)
Mass 1
Mass 2
Sample Calculation: (Mass 1 before collision)
π
uθ = uθ (deg) = _______________________________________________
180
65
Department of Physics and Astronomy
Conservation of Linear Momentum
Components of Momentum Before Collision:
Px (kg m/s) uPx (kg m/s) Py (kg m/s) uPy (kg m/s)
Mass 1
Mass 2
System
Components of Momentum After Collision:
Px (kg m/s) uPx (kg m/s) Py (kg m/s) uPy (kg m/s)
Mass 1
Mass 2
System
Sample Calculation: (Px and uPx of Mass 1 before the collision)
Px = mv cos θ = _______________________________________________________
u Px = (v cos θ ⋅ u m )2 + (m cos θ ⋅ u v )2 + (mv sin θ ⋅ uθ )2
=________________________________________________________________
Sample Calculation: (Px and uPx of the System before the collision)
Px = Px , mass1 + Px , mass 2 = _________________________________________________
u Px = u Px, mass1 + u Px, mass2 = _______________________________________________
2 2
66
University of North Carolina
Conservation of Linear Momentum
Total Momentum of the System:
P (kg m/s) uP (kg m/s) θ (deg) uθ (deg)
BEFORE
AFTER
Sample Calculation: (Before the collision. Note: Px and Py below are Px,system and Py,system)
P = Px2 + Py2 = _______________________________________________________
2 2
P 2 Py 2
u P = x u Px + u Py
P
P
= ________________________________________________________________
θ = tan −1 (Py Px ) = _____________________________________________________
2 2
Py u Px u Py
u Py = +
P P P
Px
x x y
= _____________________________________________________________
−1
P
2
u θ = 1 + Y u (where uθ is in radians)
PX Py P
x
=________________________________________________________________
180
convert uθ from rad to deg. : uθ (deg) = uθ (rad) = _________________________
π
67
Department of Physics and Astronomy
Conservation of Linear Momentum
Kinetic Energy Before Collision:
K(J ) uK (J )
Mass 1
Mass 2
System
Kinetic Energy After Collision:
K(J ) uK (J )
Mass 1
Mass 2
System
Sample Calculation: (Mass 1 Before Collision)
1
K= Mv 2 = ______________________________________________________
2
2
v2 2
uK =
2 u m + (Mv ) 2 u v2
= ____________________________________________________________
Sample Calculation: (System Before Collision)
K = K mass1 + K mass 2 = _______________________________________________
u k = u K , mass1 + u K , mass 2 = ___________________________________________
2 2
2 2
u M , mass1 u v , mass1
where u K , mass1 = K mass1
M + 2
and similarly for Kmass2
mass1 v mass1
68
University of North Carolina