5 A ball is dropped from rest from a height of 1 m in a frictionless tube as shown in the figure. If the tube profile is approximated by two straight lines (ignoring the curved portion), the total distance travelled (in m) by the ball is __________ (correct to two decimal places).

The correct answer is 2.41.

Let's analyze the problem step-by-step:

1. Understanding the motion of the ball:
The ball is released from rest at a height of 1.0 m in a vertical, frictionless tube. Under the action of gravity, the ball descends vertically to the bottom of the tube. As indicated in the diagram, the vertical drop height is:
h 1 = 1.0 m
Therefore, the distance travelled by the ball in the vertical section is 1.0 m.

2. Motion along the inclined section:
Since the tube is frictionless, mechanical energy is conserved throughout the motion. The potential energy lost during the vertical drop is completely converted into kinetic energy at the bottom, which is then fully converted back into potential energy as the ball rises along the inclined section.
Consequently, the ball must reach the same vertical height from which it was dropped:
h 2 = 1.0 m

3. Calculating the distance travelled along the incline:
The inclined section of the tube makes an angle of 45° with the horizontal, as shown in the diagram. Let the distance travelled along the inclined section be d. Using basic trigonometry:
sin ( 45 ° ) = h 2 d
Substituting the values:
d = 1.0 sin ( 45 ° ) = 1.0 1 / 2 = 2 m
Using the approximation 2≈1.4142 m:
d ≈ 1.4142 m

4. Calculating the total distance travelled:
The total distance travelled by the ball is the sum of the distance travelled in the vertical section and the distance travelled along the inclined section:
d total = h 1 + d
d total = 1.0 + 1.4142 = 2.4142 m
Rounding to two decimal places, the total distance travelled is 2.41 m.