n apple and feather free fall in a evacuated chamber. The apple and feather are released from the top. Suppose the camera opened flash every 1 s.
(1)Suppose A1 is the distance from t = 0s to t = 1 s, A2 is the distance from t = 1s to t =2s, A3 is the distance from t = 2s to t = 3s …… what is the ratio of A1: A2: A3……?
(2) If camera opened flash every 0.06s, is the ratio of the distance between every 0.06 s the same as above or change? Why?
construct a table,
d=1/2 g t^2
fill in the distance for t=0,1,2,3,...
then add another columm, dt, or the distnace between that time and the past.
For instance, dt at t=3 will be d(3)-d(2).
Note the ratio.
To find the ratio of distance for each time interval, we need to understand the motion of the objects falling in an evacuated chamber.
In the absence of air resistance, objects of different masses will fall with the same acceleration due to gravity. This acceleration is approximately 9.8 m/s^2. Let's assume the distance covered by the objects is directly proportional to the square of the time.
(1) For a time interval of 1 second, we can calculate the distances covered by the apple and the feather. Let's denote the distance covered by the apple in this interval as D1_apple, and the distance covered by the feather as D1_feather.
Since the distance is proportional to the square of time, we have:
D1_apple = (1^2) a = a,
D1_feather = (1^2) a = a.
So, the ratio of A1 (distance covered by the apple) to A1 (distance covered by the feather) is 1:1.
Similarly, for subsequent intervals:
D2_apple = (2^2) a = 4a,
D2_feather = (2^2) a = 4a.
The ratio of A2:A1 for the apple and feather is again 1:1.
Following the same pattern, we can see that the ratio of distances covered in each time interval remains constant. So, the ratio A1:A2:A3... is always 1:1:1...
(2) If the camera opens the flash every 0.06s, the time intervals become much shorter. Let's consider the new time interval, denoted as t1 = 0.06s to t2 = 0.12s. Again, let's calculate the distances covered by the apple and the feather in this interval, denoted as D2_apple (short interval) and D2_feather (short interval).
Using the same logic as before, we find:
D2_apple (short interval) = (0.06^2) a = 0.0036a,
D2_feather (short interval) = (0.06^2) a = 0.0036a.
Comparing the distances covered in the old time interval (1s) to the distances covered in the new time interval (0.06s), we can see that the ratio has changed. The ratio A1:A2:A3... is no longer 1:1:1..., it will be different for the shorter time intervals.
This change occurs because the objects cover less distance in the shorter time intervals, due to the decreased duration of each interval. Consequently, the ratio of distances covered within shorter time intervals will be smaller compared to the ratio in longer time intervals.
In summary, when the camera opens the flash every 1s, the ratio of distances covered by the apple and the feather is always 1:1. However, if the camera opens every 0.06s, the ratio will change because the time intervals have become shorter.