Question

Two objects are dropped from a bridge, an interval of 1.0 s apart. Air resistance is negligible. During the time that both objects continue to fall, their separation A) increases. B) decreases. C) stays constant. D) increases at first, but then stays constant. E) decreases at first, but then stays constant.

Answer 1

Answer:

A

Explanation:

This question can be modeled by using kinematic equation of motion in vertical direction:

y = y (0) + y'(0)*t + 0.5*a*t^2

Given:

y(0) = 0 for both cases

y'(0) = 0  ... both cases objects were dropped

t taken for first ball to travel distance y_1

t_2 taken for the second ball to travel distance y_2

t -1 = t_2   ..... Eq 1

a = 9.81 m/s^2

Plugging in the respective condition is the above equation of motion:

y_1 = 0.5*a*(t)^2

y_2 = 0.5*a*(t_2)^2

Substituting t_2 = t + 1, we get

y_2 = 0.5*a*(t - 1)^2

Find the difference between position of two objects:

y_1 - y_2 = 0.5*a*(t^2 - (t-1)^2)

dy = 0.5*9.81*( 2*t - 1 )

dy = 9.81*t - 4.905

We can see from above derived expression which shows the separation between the two objects which is a function of time t. The function is an increasing function such that as t increases dy also increases, hence their separation increases with time.