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# A 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball. As...

A 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball. As a result, the incident ball is deflected 30.0 degrees from its initial path (remember the 90 degree rule for same masses). 1. Draw a vector diagram showing the momentum of each ball after the collision 2. What is the velocity of the incident ball after the collision? 3. What is the velocity of the struck ball after the collision?

H: m1v0 = m1v1cos theta1 + m2v2 cos theta2

0.4=0.86v1+v2 cos theta2

V: 0 = m1v1 sin theta1 - m2v2 sin theta 2

0 = 0.5v1+v2 sin theta2

Kinetic: 0.5 m1 v0^2 = 0.5 m1 v1^2 + 0.5 m2 v2^2

0.6 = v1^2 + v2^2

How do I solve the unknowns? theta 2, v1, v2?

Answer #1

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