A ball is released at the left end of these different tracks. The tracks are bent from equal-length pieces of channel iron. a. From fastest to slowest, rank the speed of the ball at the right end of the track. b. From longest to shortest, rank the tracks in terms of the ?time for the ball to reach the end. c. From greatest to least, rank the tracks in terms of the ?average speed of the ball. Or do all balls have the same average speed on all three tracks?

Solution Step 1 of 3 a. From fastest to slowest, rank the speed of the ball at the right end of the track. First, assume that there is no friction. On track A the ball accelerates only on the one downward curve, and then travels at a constant velocity. On track B the ball accelerates twice, once during each downward curve, so it ends up with a higher velocity. On track C the ball accelerates twice, but it decelerates once (during the upswing segment), so it’s final velocity is the same after the dip, as it was before the dip. That is the same velocity it had in track A. So to rank the speed from fastest to slowest, Speed on track B >A=C Step 2 of 3 b. From longest to shortest, rank the tracks in terms of the time for the ball to reach the end. On track A the ball accelerates only on the one downward curve, and then travels at a constant velocity. This will be this slowest speed of the three – so it will take the longest time for the ball to reach the end. On track B the ball accelerates twice, once during each downward curve, so it ends up with a higher velocity. Thus it will go at a higher speed than track A, so it takes less time for the ball to reach the end. On track C the ball accelerates twice, but it decelerates once (during the upswing segment), so it’s velocity on the final segment is the same after the dip, as it was before. So it takes same time as that of A. So to rank the time taken by ball to reach end from longest to shortest, Time taken on track A=C > B