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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.

6 years ago

We assume that the distance travelled by the first ball is represented by s

We also assume that the time taken by ball one to travel distance s

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 t

Where s

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 s

Substituting the initial conditions in the equation above, we have

…… (1)

Now we calculate the distance travelled by the second ball from the same equation of kinematics as:

Where s

As given in the question, the ball was dropped down the elevator shaft, therefore the initial speed v

Substituting the initial conditions in the equation above, we have

…… (2)

However the second ball was released one second later than the first ball, therefore the second ball has travelled distance s

Therefore we can relate the time t

t

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

(3)

The difference of equation (1) and equation (3) represents the distance (say s) between the two balls, and is given as:

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 t

For various values of t

For t

Let us assume that the final speed of the first ball after time t

Therefore, from the equation of kinematics, we have

v

v

Where v

Therefore we have

v

v

Taking the ratio of equation (4) and (5), we have

…… (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 t

t

Equating the above relation in equation (6), we have

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 t

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