two identical pucks collide on an air hockey table. one puck was originally at rest. if the incoming puck has a speed of 6 m/s and scatters to an angle of 30 degree, what is the velocity (magnitude and direction) of the second puck if it scatters at 60 degree?

Answer 1

momentum is conserved
... initial momentum-x = 6 m ... initial momentum-y = 0

let m be the mass of a puck
... a is the moving puck , b is the stationary (2nd) puck

the sum of the final x-momenta is equal to the initial x-momentum
... [va * m * cos(30º)] + [vb * m * cos(60º)] = 6 * m
... [va * cos(30)] + [vb * cos(60º)] = 6

the sum of the final y-momenta is equal to the initial y-momentum
... [va * m * sin(30º)] - [vb * m * sin(60º)] = 0
... [va * sin(30º)] = [vb * sin(60º)]

solve the system of equations for vb (and va)

Answer 2

Please I need final answer with magnitude and direction.

Thank you