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Repeat Question 30 taking air resistance into account. Again, give qualitative answers.

Repeat Question 30 taking air resistance into account. Again, give qualitative answers.

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
9 years ago
Assumption:
Both the ball experiences the same acceleration in the upward direction because of the presence of air resistance.
The change in velocity of the first ball is given as:
\frac{dv}{dt} = (g-a_{R})
Here, g is the free fall acceleration and aR is the dacceleration ue to air resistance.
Integrate above expression under the limit that the initial velocity of the first ball at t = 0 is 0 m / s whereas the final velocity of the ball at time t is v1 :
\int_{0}^{v_{1}} dv = \int_{0}^{t} (g - a_{R}) dt
It is important to note that the acceleration due to air resistance is dependent on time, therefore the above integral can be written as:
v_{1} = gt - \int_{0}^{t} a_{R}\ dt | …… (1)

Similarly, the change in velocity of the second ball, released one second later can be written as:
\frac{dv}{dt} = (g - a_{R})
Integrate above expression under the limit that the initial velocity of the second ball at t = 0 is 0 m / s whereas the final velocity of the ball at time is t – 1 is v2 :
\int_{0}^{v_{2}} dv = \int_{0}^{t-1} (g - a_{R}) dt
v_{2} = g (t - 1) - \int_{0}^{t-1} a_{R}dt …… (2)

It is important to note that when the first ball has travelled for time t , the second ball has travelled for time t – 1.
Subtract equations (1) and (2),
v_{1} - v_{2} = gt - \int_{0}^{t} a_{R}dt - g (t - 1) + \int_{0}^{t-1} a_{R}dt
= gt - \int_{0}^{t} a_{R}dt - gt + g (1 s) + \int_{0}^{t-1} a_{R}dt
= g (1\ s ) - \int_{0}^{t} a_{R}dt + \int_{0}^{t-1} a_{R}dt
It is obvious that the value of integral \int_{t}^{0} a_{R}dtwould be greater than the value of integral\int_{0}^{t - 1} a_{R}dt , as a result, the difference v1 – v2 would be less than g (1 sec). Therefore, the objects will come closer to each other over time.

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