1. A Collison

2. A hockey puck with mass 4 kg makes a head on collision with a smaller puck with mass 2 kg. The larger puck was originally travelling east at 3 m/sec. The smaller puck was originally travelling west at 4 m/sec. How fast are they moving after the collision. (Assume a perfectly elastic collision all in a straight line, that is one dimensional.)

3. Fundamentals, Using Math Before collision: m A =4 v A =3 m B =2 v B =−4 After collision: m A =4 v A =? m B =2 v B = ? Momentum In (mv) = Momentum Out (mv) 4∙3 + 2∙(−4) = 4 v A + 2v B KE In ( ½ mv 2 ) = KE Out ( ½ mv 2 ) ½ ∙4∙3 2 + ½ ∙2∙(−4) 2 = ½ ∙4v A 2 + ½ ∙2v B 2 18 + 16 = 2 v A 2 + v B 2 4v A + 2v B = 4 (momentum) 2v A 2 + v B 2 = 34 (energy)

4. Solving Simultaneously 4 v A + 2v B = 4 (momentum) v B = 2 − 2v A 2v A 2 + v B 2 = 34 (energy) 2v A 2 + (2 − 2v A ) 2 = 34 2v A 2 + 4 − 2∙4v A + 4v A 2 = 34 6v A 2 − 8v A − 30 = 0 2∙(3v A 2 − 4v A − 15) = 0 2∙(3v A + 5)∙(v A − 3) = 0

5. Solutions 2∙(3v A + 5)∙(v A − 3) = 0 One of the factors must be zero 3v A + 5 = 0  v A = − 5/3  v B = 2−2v A = 16/3 Or v A − 3 = 0  v A = 3  v B = 2−2v A = − 4

6. Cheating Reading the Book Galileo was challenged. “If the earth is moving so fast around the sun, why don’t we feel it.” He responded by describing a goldfish in its bowl in the captain’s cabin on a ship. Does the goldfish swim faster in the direction of the ship’s motion? Galilean Relativity The Laws of Physics are the same if we add the same amount to all velocities.

7. If v B1 is Zero m A v A1 = m A v A2 + m B v B2 (momentum) ½m A v A1 ² = ½m A v A2 ² + ½ m B v B2 ² (energy) m A v A1 ² − m A v A2 ² = m B v B2 ² m A v A1 − m A v A2 = m B v B2 (v A1 ² − v A2 ²)/(v A1 − v A2 ) = v B2 v A2 + v A1 = v B2 *** (Elastic Collisions) According to Galileo, we can add the same quantity to all velocities, and get a valid equation. If we start with v B1 not equal to zero, we can add −v B1 to all quantities. Then equation *** is true v A2 −v B1 + v A1 −v B1 = v B2 −v B1  v A1 − v B1 = v B2 − v A2

8. The Short Way (Cheating) v A1 − v B1 = v B2 − v A2 3 − (−4) = v B2 − v A2 v B2 − v A2 = 7 4v A2 + 2v B2 = 4 (momentum) v B2 = v A2 + 7 4v A2 + 2(v A2 + 7) = 4 6v A2 = −10 v A2 = −5/3 v B2 = 16/3