Show that the total kinetic energy before is the equal to the total kinetic energy after the collisic IN BOTH MAGNITUDE AND DIRECTION. Show all solutions completely. 0.5m 0.5m Ipl -0.51 kg mis M-1.02 m/s Kinetic Energy = 0.77 J 0.00 s More Data Mass (kg) 0.50 1.50 More Data Mass (kg) Position (m) y 0.00 -0.20 0.50 1.50 -1.00 0.60 Kinetic Energy = 0.77 J 1.00 s M-1.66 mis Ipl-0.83 kg mis Position (m) y -0.80 -0.30 X 0.07 0.03 M-0.82 m/s Ipl-1.23 kg mis V₂ 1.00 -0.80 Velocity (m/s) V₂ -0.18 0.19 Normal Slow M-0.32 m/s Ipl -0.48 kg mis AFTER COLLISION Momentum (kgm/s) P Py 0.50 -0.09 -1.20 0.29 Normal Slow Velocity (m/s) V₂ V₂ -1.56 -0.56 0.05 0.32 Momentum (kgm/s) P Py -0.78 -0.28 0.08 0.48 Balls 2 # Balls 2 0 ◄► # DOO 20 Velocity Momentum Center of Mass Kinetic Energy Values Reflecting Border Path Elasticity Inelastic Constant Size Momenta Diagram total ✔Velocity Momentum Center of Mass ✔ Kinetic Energy ✔Values ✔Reflecting Border Path Elasticity 100% Constant Size Elastic Momenta Diagram 4x 100% Elec Q Q
Show that the total kinetic energy before is the equal to the total kinetic energy after the collisic IN BOTH MAGNITUDE AND DIRECTION. Show all solutions completely. 0.5m 0.5m Ipl -0.51 kg mis M-1.02 m/s Kinetic Energy = 0.77 J 0.00 s More Data Mass (kg) 0.50 1.50 More Data Mass (kg) Position (m) y 0.00 -0.20 0.50 1.50 -1.00 0.60 Kinetic Energy = 0.77 J 1.00 s M-1.66 mis Ipl-0.83 kg mis Position (m) y -0.80 -0.30 X 0.07 0.03 M-0.82 m/s Ipl-1.23 kg mis V₂ 1.00 -0.80 Velocity (m/s) V₂ -0.18 0.19 Normal Slow M-0.32 m/s Ipl -0.48 kg mis AFTER COLLISION Momentum (kgm/s) P Py 0.50 -0.09 -1.20 0.29 Normal Slow Velocity (m/s) V₂ V₂ -1.56 -0.56 0.05 0.32 Momentum (kgm/s) P Py -0.78 -0.28 0.08 0.48 Balls 2 # Balls 2 0 ◄► # DOO 20 Velocity Momentum Center of Mass Kinetic Energy Values Reflecting Border Path Elasticity Inelastic Constant Size Momenta Diagram total ✔Velocity Momentum Center of Mass ✔ Kinetic Energy ✔Values ✔Reflecting Border Path Elasticity 100% Constant Size Elastic Momenta Diagram 4x 100% Elec Q Q
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Transcribed Image Text:INSTRUCTONS:
Show that the total kinetic energy before is the equal to the total kinetic energy after the collision
IN BOTH MAGNITUDE AND DIRECTION.
Show all solutions completely.
0.5 m
0.5 m
Kinetic Energy = 0.77 J
0.00 s
More Data
(1)
v=1.02 m/s
Ipl -0.51 kgm/s
0.50
1.50
Mass (kg)
✔More Data
Mass (kg)
Position (m)
X
y
0.00
-0.20
0.50
1.50
-1.00
0.60
Kinetic Energy = 0.77 J
1.00 s
M=1.66 m/s
Ipl - 0.83 kg mis
Position (m)
y
X
-0.80
0.07
-0.30
0.03
v=0.82 m/s
Ipl = 1.23 kg m/s
1.00
-0.80
Velocity (m/s)
Vy
-0.18
0.19
Normal
Slow
M=0,32 m/s
Ipl -0.48 kg mis
AFTER COLLISION
Momentum (kg m/s)
Px
0.50
-1.20
Normal
Slow
Velocity (m/s)
Vx
-1.56 -0.56
0.05
0.32
Py
-0.09
0.29
Q
Momentum (kg m/s)
Px
-0.78
0.08
Py
-0.28
0.48
Balls
2
Balls
2
#0
OOR
OQE
Velocity
Momentum
Center of Mass
Kinetic Energy
Values
Reflecting Border
Path
Elasticity
Inelastic
Constant Size
Momenta Diagram
OQQQOOK
total
Velocity
Momentum
Center of Mass
Kinetic Energy
Values
Reflecting Border
Path
Elasticity
Inelastic
Constant Size
100%
↑↑*
Elastic
Momenta Diagram
↑ ↑ *
100%
-0
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