Question

A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator-

a) in a time $t_1$ if the elevator is stationary and

b) in time $t_2$ if it is moving uniformly, then

• A. $t_1=t_2$
• B. $t_1>t_2$
• C. $t_1<t_2$
• D. $t_1<t_2$ or $t_1>t_2$ depending on whether the lift is going up or down

Answer 1

Answer: (A)

$t_1=t_2$

In the frame of the man, both the situations are identical.
The coin is falling under gravity in both the cases with zero initial velocity and it has to cover a height of $h$ to reach the bottom of the elevator.
Hence time taken in both the situation will be same.

Also:

If the lift is stationary, the time $t_1$ to cover a height $h$ falling freely with acceleration $g$ can be calculated by

$h=\frac{1}{2}g{t}_1^2$
or $t_1=\sqrt{\frac{2h}{g}}$ ......(i)
If the lift is going upwards with speed $u$, the speed of the coin at the moment it leaves the hands is also $u$ in upward direction. Since, the displacement of the coin is in downward direction, the initial velocity of the coin in the upward direction should be taken as negative.
Thus, the equation of displacement of the coin can also be written as, $-u{t}_2+\frac{1}{2}gt{2}^2$
i.e. $h=\frac{1}{2}gt{2}^2$
Comparing two equations, we get, $t_1=t_2$
Actually, it is so because in the second case, the average speed of the coin in upward direction is same as that of the base of the lift, i.e., the average relative velocity between the coin and the base of the lift is throughout zero.