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Chapter 2: Problem 21

A coin is dropped in a lift. It takes time \(t_{1}\) to reach the floor when lift is stationary. It takes time \(t_{2}\) when lift is moving up with constant acceleration, then (A) \(t_{1}>t_{2}\) (B) \(t_{2}>t_{1}\) (C) \(t_{1}=t_{2}\) (D) \(t_{1} \gg t_{2}\)

Short Answer

Expert verified
The short version of the answer is: By comparing the expressions for time taken in both cases when the lift is stationary and when it is moving upwards with constant acceleration, we find that \(t_1 > t_2\). Thus, the correct answer is (A).

Step by step solution

01

Identify the variables

First, let's identify the variables associated with the problem: \(t_1\) is the time taken by the coin to reach the floor when the lift is stationary, \(t_2\) is the time taken by the coin to reach the floor when the lift is moving upwards with a constant acceleration, and \(g\) is the acceleration due to gravity.
02

Analyze the case when the lift is stationary

When the lift is stationary, the coin simply falls under the effect of gravity. We can use the second equation of motion to find the time taken to reach the floor: \(s = ut + \frac{1}{2}at^2\) Here, \(s\) is the distance fallen, \(u\) is the initial velocity, \(t\) is the time taken, and \(a\) is the acceleration. Since the coin is initially at rest, \(u = 0\). The acceleration is the acceleration due to gravity, which is \(g\). So, the equation becomes: \(s = \frac{1}{2}gt_1^2\)
03

Analyze the case when the elevator is moving upwards with constant acceleration

Now, let's consider the case when the lift is moving upwards with a constant acceleration. The coin's motion will be influenced by both the gravitational force acting downwards and the lift's acceleration acting upwards. The net acceleration acting on the coin will be given by: \(a_{net} = g - a\) Where \(a\) is the acceleration of the lift. Now, let's use the second equation of motion again to find the time taken to reach the floor: \(s = ut + \frac{1}{2}a_{net}t^2\) Here, \(u\) is still zero since, initially, the coin is at rest. The acceleration is the net acceleration acting on the coin, which is \(a_{net}\). So, the equation becomes: \(s = \frac{1}{2}(g - a)t_2^2\)
04

Compare the times in both cases

Now, we need to compare the expressions we derived in Step 2 and Step 3 to see which one is greater: \(\frac{1}{2}gt_1^2 = s\) and \(\frac{1}{2}(g - a)t_2^2 = s\) Divide both equations to get the ratio between \(t_1\) and \(t_2\): \(\frac{t_1^2}{t_2^2} = \frac{g}{g - a}\) Since \(g, a > 0\), we can deduce that: \(g > g - a\) And so, \(t_1^2 > t_2^2\) Taking the square root of both sides: \(t_1 > t_2\) Hence, the correct answer is: (A) \(t_1 > t_2\)

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