Chapter 4, Problem 23A

A track is made of a piece of channel metal bent as shown. A ball is released from rest at the left end of the track and continues past the various points. Rank the ball at points A, B, C, and D, from fastest to slowest. (Again, watch for tie scores.)

To determine

To Rank: The ball at points A, B, C, and D from fastest to slowest.

Answer to Problem 23A

  $C>D=B>A$

Explanation of Solution

Given:

Initial velocity of the ball $=0$

Formula used:

Kinematics equation:

  $\begin{array}{l}v=u+gt\\ d=ut+\frac{1}{2}g{t}^2\end{array}$

Here, u is the initial velocity, g is the acceleration due to gravity, t is the time, v is the final velocity and d is the displacement.

Calculation:

Plugging the values in the formula

  $\begin{array}{l}v=0+gt\\ v=gt\mathrm{......}\left(1\right)\\ d=ut+\frac{1}{2}g{t}^2\\ d=\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2d}{g}}\mathrm{......}\left(2\right)\end{array}$

From equation (1) and (2)

  $\begin{array}{l}v=g\sqrt{\frac{2d}{g}}\\ v=\sqrt{2dg}\end{array}$

Since the acceleration due to gravity is constant, therefore the velocity is directly proportional to the square root of the distance travelled.

Since the depth of the point C is greatest, thus the speed at this point is fastest. After that B and D are at the same height which is less than C and greater than A. Therefore, speed at these points are equal but its magnitude is less than the speed at point C. Depth of the point A is minimum hence, velocity at that is point is minimum.

Conclusion:

Thus the velocity at point C is maximum and at point A is minimum.

  $C>D=B>A$