A person standing on the floor of an elevator drops a coin. The coin touches the floor of the elevator in time $t_{1}$ when the elevator is stationary and time $t_{2}$ when the elevator is moving uniformly then:
A. $t_{1} = t_{2}$
B. $t_{1} > t_{2}$
C. $t_{1} < t_{2}$
D. Either B or C depending upon the direction of motion of the elevator.

Answer

Verified

498.3k+ views

Hint:This problem will become very easy if we solve it from the frame of reference of the elevator and not the inertial frame. Thus, we assume a frame of reference that is moving along with the elevator and one of the cases when the elevator is stationary.

Complete step by step answer:
Both the situations are identical in the frame of the man. In both cases, the coin is falling under gravity. With zero initial velocity, and it has to cover a height of h to reach the bottom of the elevator. Therefore, time taken in both situations will be the same.

Therefore the time coin touches the floor of the elevator in time

t_{1}

when the elevator is stationary and time

t_{2}

when the elevator is moving uniformly then

t_{1} = t_{2}

.

Therefore, the correct option is (A).

Note:Many problems such as these can be easily solved by assuming a proper frame of reference. Always solving in the inertial frame of reference can bring in unnecessary variables and complications in the calculations of some problems. However, while assuming a non-inertial frame of reference, one must always first determine the acceleration of the frame, if any and assign proper pseudo forces opposite to the acceleration of the frame on all bodies in the frame. However, not all moving non-inertial frames have to be assigned pseudo forces. Only the frames which are accelerating have to be assigned pseudo forces, those with constant velocity do not.