If a sphere and paper are dropped from 10 meters and the sphere hit the ground in 1.4 seconds and the paper hit in 2 seconds, why did the impact time in case of the piece of paper change in comparison to the sphere?
A. a=(Wt-R)/m;a=g B. a=(Wt-R)/m;a>g
C. a=(Wt-R);a<g D. a=(Wt-R)/m;a<g
To understand why the impact time of the paper changed compared to the sphere, we need to analyze the forces acting on each object during their fall.
First, let's understand the equation used in the answer choices:
a = (Wt - R) / m
Where:
a = acceleration
Wt = weight (mass x gravity)
R = resistive force (friction, air resistance, etc.)
m = mass
g = acceleration due to gravity
Now, let's analyze the situation:
1. The Sphere:
When the sphere is dropped, it experiences two forces: weight (downward force) and air resistance (upwards force). Initially, its acceleration is equal to the acceleration due to gravity (g), as there is no significant air resistance. Therefore, the equation for the sphere becomes:
a = (Wt - R) / m = (mg - R) / m = g
Since the acceleration of the sphere is equal to g, its impact time will be determined solely by the vertical distance it falls.
2. The Paper:
When the paper is dropped, it also experiences weight and air resistance. However, due to its larger surface area, it encounters more air resistance compared to the sphere. As a result, the air resistance affects its motion more significantly and reduces its acceleration.
Therefore, the equation for the paper becomes:
a = (Wt - R) / m = (mg - R) / m < g
Since the acceleration of the paper is less than g due to the stronger influence of air resistance, its impact time is longer compared to the sphere.
Based on the analysis, the correct answer would be:
B. a = (Wt - R) / m; a > g