Form 4 physics (mechanices)

用戶144644

2011-02-03 4:34 am

A man standing in a stationary lift drops a stone on the floor. The stone takes t seconds to reach the floor. If the lift is moving downwards with constant velocity initially,

A. the stone takes a time less than t seconds to reach the floor.
B. the stone takes also a time of t seconds to reach the floor.
C. the stone takes a time greater than t seconds to reach the floor.
D. the stone may or may not reach the ground depending on the speed of the lift.
E. the stone will drop with constant velocity.

Explain your answer

回答 (1)

天同

2011-02-03 7:58 am

✔ 最佳答案

Let d be the height above the floor of the lift where the stone is dropped.When the lift is at rest, using equation of motion: s = ut + (1/2)at^2, we haved = (1/2)gt^2 -------------------- (1)
where g is the acceleration due to gravity
Suppose the lift is now descending with a constant speed of v, and it takes a time of T for the stone, dropped from the same height d, to reach the floor. During this time interval T, the lift floor has descended a distance of vT downward. Hence, the stone has dropped a total distance of (d + vT).

Using the equation of motion: s = ut + (1/2)at^2
with s = (d + vT), t = T, a = g, u = v
hence, (d+vT) = vT + (1/2)gT^2
i.e. d = (1/2)gT^2 ------------------------ (2)

Equating (1) and (2),
(1/2)gt^2 = (1/2)gT^2
thus, T = t
The answer is option B
In fact, the answer is apparent, as the Laws of Physics should be the same in all inertial reference system. Hence, the time of fall of the stone to the lift floor should not be dependent on the speed of the lift.

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