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A second ball is dropped down an elevator shaft 1 s after the first ball is dropped. (a) What happens to the distance between the balls as time goes on? (b) How does the ratio 1'[/1'2 of the speed of the first ball to the speed of the second ball change as time goes on? Neglect air resistance, and give qualitative answers.

A second ball is dropped down an elevator shaft 1 s after the first ball is dropped. (a) What happens to the distance between the balls as time goes on? (b) How does the ratio 1'[/1'2 of the speed of the first ball to the speed of the second ball
change as time goes on? Neglect air resistance, and give qualitative answers.

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
6 years ago
Assumption:
We assume that the distance travelled by the first ball is represented by s1 whereas the distance travelled by the second ball is s2.
We also assume that the time taken by ball one to travel distance s1 is given by t1 (in seconds) whereas the time taken by the second ball to travel distance s2 is given by t2(in seconds).
We also assume that both the balls have not hit the ground, because the time taken by the balls to reach the ground will be same as they fall from the same height. We therefore highlight the relative analysis of the motion of the second ball with respect to the first.
(a) The distance travelled by the first ball in time t1 can be calculated from the equation of kinematics,as:
s_{1} = s_{01} + v_{0y1}t_{1} + \frac{1}{2}gt^{2}_{1}
Where s01 represents the initial distance of the ball with respect to observer and v0y1 represents the initial speed of the ball.
As given in the question, the ball was dropped down the elevator shaft, therefore the initial speed of the ball is zero. Also the value of s01 is zero, since the observer is at the same initial position of the ball.
Substituting the initial conditions in the equation above, we have

s_{1} = \frac{1}{2}gt^{2}_{1} …… (1)
Now we calculate the distance travelled by the second ball from the same equation of kinematics as:
s_{2} = s_{02} + v_{0y2}t_{2} + \frac{1}{2}\ gt_{2}^{2}
Where s02 represents the initial distance of the ball with respect to observer and v0y2 represents the initial speed of the ball.
As given in the question, the ball was dropped down the elevator shaft, therefore the initial speed v0y2of the ball is zero. Also the value of s02 is zero, since the observer is at the same initial position of the ball.
Substituting the initial conditions in the equation above, we have
s_{2} = \frac{1}{2} gts_{2}^{2} …… (2)
However the second ball was released one second later than the first ball, therefore the second ball has travelled distance s2 in time t1 – 1.
Therefore we can relate the time t1 and t2 as:
t2 = t1 – 1
It is important to note that the condition is valid as long as there is no air resistance.
Equating the above relation in equation (2), we have
s_{2} = \frac{1}{2}g (t_{1} - 1)^{2} (3)
The difference of equation (1) and equation (3) represents the distance (say s) between the two balls, and is given as:

235-1534_17.PNG

What we calculated here is as spectacular as it appears, the distance between the two balls does not remain constant under the action of g . Rather the distance increases over time t1 and the factor by which it increases is always an odd number.
For various values of t1 say 2,3,4.... , the distances between the balls increases by a series of odd numbers as 3,5,7.
For t1 = 1 s , the second ball was not released, therefore the distance of separation between the two is equal to the distance travelled by the first ball during that time which is g/2.
Let us assume that the final speed of the first ball after time t1 is given by v1 whereas the final speed of the second ball after time t1 is given by v2.
Therefore, from the equation of kinematics, we have
v1 = v0y1 + gt1
v2 = v0y2 + gt2
Where v0y1 and v0y2 represents the initial speed of the ball, which in this case is zero.
Therefore we have
v1 = gt1 …… (4)
v2 = gt2 …… (5)
Taking the ratio of equation (4) and (5), we have
\frac{v_{1}}{v_{2}} = \frac{gt_{1}}{gt_{2}}
\frac{v_{1}}{v_{2}} = \frac{t_{1}}{t_{2}} …… (6)

However the second ball was released one second later than the first ball, therefore relative to first ball, the second ball is one second behind.
Therefore we can relate the time t1 and t2 as:
t2 = t1 – 1
Equating the above relation in equation (6), we have
\frac{v_{1}}{v_{2}} = \frac{t_{1}}{t_{1}-1}
It is clear from the equation above that the final speed of first ball will always be greater than the final speed of the second ball.
One should realize that the final speed of the balls depends on the time for which acted on them. The second ball, released after 1 sec, has experienced g for a shorter time (considering both the balls have not hit the ground) and therefore has less final speed relatively.
It should be noted that if the value of t1 is very large, then the final speed of the second ball is approximately equal to that with of first ball. This is because the ration of time for the balls have experienced g is getting closer to 1.

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