A billiard ball rolling across a table at 1.55 m/s makes a head-on elastic colli

Question

A billiard ball rolling across a table at 1.55 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs. (a) The second ball is initially at rest. first ball Correct: Your answer is correct. m/s second ball Correct: Your answer is correct. m/s (b) The second ball is moving toward the first at a speed of 1.10 m/s. first ball Incorrect: Your answer is incorrect. The correct answer is not zero. m/s second ball Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (c) The second ball is moving away from the first at a speed of 0.80 m/s. first ball m/s second ball m/s

Explanation / Answer

Since the initial and final velocity must be equal the velocities/ speeds switch in a way. Use (m1v1 + m2v2)initial = (m1v1 + m2v2)final as your guideline. But in this problem since the masses of the balls are equal you can cancel out the masses. For the first one: the first ball hits a ball that is sitting still (v = 0 m/s). AFTER the hit the second ball should be moving, so the speed was transferred to the second ball, making the first ball's velocity 0 m/s and the second ball's 1.60 m/s. For b: The second ball is moving towards the first, so in a negative direction. So: 1.6 + (-1.15). When they hit, the balls will travel in opposite directions than they initially were, so ball 1 will have a -1.15 m/s speed and ball 2 will have a +1.6 m/s speed. Since the faster one hits a slower one and vice versa, their speeds should speed up or slow down accordingly. For c: The second ball is traveling in the same direction but slower, so it will still get hit by the first ball. So both answers will be positive because basically all that happens is the second ball speeds up after the hit while the first will slow down. So the second ball will be 1.6 m/s and the first will be 0.80 m/s.

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