V. Analysis
A.Vertical displacemen t= (1/2) x (Vertical Acceleration) x (Time)2
0.92m = (1/2) x (9.8m/s2) x (Time)2
Time = ((2 x 0.92m)/(9.8m/s2))1/2 = 0.43s
Horizontal displacement = (Initial horizontal velocity) x (Time)
0.43m = (Initial horizontal velocity) x (0.43s)
Initial horizontal velocity = Initial velocity = (0.43m/0.43s) = 1.0m/s
Initial Momentum = (Mass) x (Initial Velocity)
P0 = (0.008kg) x (1.0m/s) = 0.008kgm/s
Time =((2 x Displacement)/(Acceleration))1/2
Using vertical displacement and acceleration:
Time = ((2 x 0.92m)/(9.8m/s2))1/2 = 0.43s
Final velocities
Stationary Ball (Ball 1): (0.32m/0.43s) = 0.73m/s = Final Velocity1
Rolling Ball (Ball 2): (0.072m/0.43s) = 0.17m/s = Final Velocity2
Final momentum = ((Mass1) x (VF1)) + ((Mass2) x (VF2))
Mass1=Mass2
PF = (Mass) x (VF1 + VF2)
Vector Addition of Velocities
0.73sin(13.1)= 0.17m/s0.73cos(13.1)= 0.71m/s
0.17sin(48.7)= -0.13m/s0.17cos(48.7)= 0.11m/s
Resultant xcomp = (0.17m/s + (-0.13m/s)) = 0.04m/s
Resultant ycomp = (0.71 m/s + 0.11 m/s) = 0.82m/s
Resultant Final Velocity = ((0.04m/s)2 + (0.82m/s)2)1/2 = 0.68m/s
PF = (0.008kg) x (0.68m/s) = 0.0054kgm/s
B.To calculate the initial momentum, the mass of the object and the velocity of the object must be calculated. To find the velocity, kinematics was used. Because the object has no initial vertical velocity and only horizontal velocity, the initial velocity can be found by dividing the horizontal displacement by the amount of time the object was in the air. However, there are two unknown variables in that equation, so time must be calculated using vertical kinematics. Time can be isolated and solved for by taking the square root of two times the vertical displacement divided by the vertical acceleration. Then, by plugging this value into the original horizontal kinematic equation, the initial velocity can be calculated. The initial momentum is the product of the initial...

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