Coin Drop in Elevator: A Comprehensive Analysis of Motion in Physics
The concept of motion in an accelerating frame, such as an elevator, is a classic example used in physics to illustrate the principles of mechanics. In this article, we will explore the scenario where a man drops a coin from a certain height in a moving elevator. We'll use the equations of motion to calculate the time it takes for the coin to reach the floor.
Setup and Initial Conditions
Consider a man moving up in an elevator with a speed of 10 m/s. At a certain moment, he drops a coin from a height of 2.45 meters above the floor of the elevator. We are interested in determining the time it takes for the coin to reach the floor of the elevator.
To solve this problem, we need to use the equations of motion, considering the initial conditions and the acceleration due to gravity.
Equations of Motion
Given: Initial height, h 2.45 m Initial velocity of the coin when dropped, u -10 m/s (negative because the coin is moving downward relative to the upward motion of the elevator) Acceleration due to gravity, a 9.81 m/s2
Step 1: Use the Second Equation of Motion
The second equation of motion relates distance, initial velocity, time, and acceleration:
s ut 1/2 at2
Where:
s is the distance fallen, which is -2.45 m (since it is downward) u is the initial velocity -10 m/s a is the acceleration due to gravity 9.81 m/s2 t is the time in secondsSubstituting the known values into the equation:
-2.45 -10t 0.5 × 9.81t2
This simplifies to:
-2.45 -10t 4.905t2
Rearranging gives us:
4.905t2 - 10t - 2.45 0
Step 2: Solve the Quadratic Equation
Using the quadratic formula:
t [-b ± √(b2 - 4ac)] / 2a
Where:
a 4.905 b -10 c -2.45Calculating the discriminant:
b2 - 4ac (-10)2 - 4 × 4.905 × -2.45 100 48.078 148.078
Substituting into the quadratic formula:
t [10 ± √148.078] / 2 × 4.905
Calculating the two possible values for t:
t [10 12.17] / 9.81 ≈ 2.25 seconds
t [10 - 12.17] / 9.81 ≈ -0.22 seconds (not physically relevant)
Since the coin is dropped, the relevant time is the positive value:
t ≈ 2.25 seconds
Alternative Scenario: Downward Acceleration and Lift Motion
Another interesting scenario involves applying a downward acceleration of 2 m/s2 on the coin while the lift moves up with a uniform speed of 10 m/s. The initial speed of the coin is 10 m/s upward but accelerated downward with an acceleration of 12 m/s2.
Equations of Motion for this Scenario
In time t, the displacement of the coin is:
s(coin) 10t - 0.5 × 12t2
The lift moves up a distance:
s(lift) 10t
The relative displacement is:
s(coin) - s(lift) -x
Substituting the known values:
10t - 0.5 × 12t2 - 10t -x
This simplifies to:
-6t2 -x
2 - x 6t2
Solving for t:
2 - x 6t2
t2 1/3
t ≈ 0.58 seconds