Question

The two pendulums shown have 14 kg balls supported by rigid mass-less rods. The pendulums rotate on frictionless pivots in the same vertical plane In a circular path of radius 2 meters and collide. If each is at rest and oriented at theta = 38 degrees then released simultaneously, determine the Impulse of the collision in kg-m/s if the coefficient of restitution - 0.6. DO NOT ENTER SIGNS OR UNITS!!

Answer 1

given

m = 14 kg , angle $\theta$ = 38^o , R = 2 m

coefficient of restitution is = 0.6

we have initial velocity u and mass

then

m g ( R + R sin $\theta$ ) = 1/2 m u²

u = ( 2 g R ( 1 + sin $\theta$ ) )^1/2

impulse can be defined as the change in momentum

so impulse = change in momentum

impulse = m ( u + u / 1.66 ) ( since used coefficient of restitution = 0.6 )

= 2.66 m u / 1.66

substituting the " u " value

impulse = 2.66 / 1.66 X m X ( 2 g R ( 1 + sin $\theta$ ) )^1/2   

= 2.66 / 1.66 X 14 X ( 2 X 9.8 X 2 X ( 1 + sin38 ) )^1/2  

= 22.43 X 7.958

impulse = 178.527 kg.m/sec