A ball bearing is dropped from rest at a point A. At the instant it passes a mark 4.5 m below A, another ball bearing is released from rest from a position 5.5 m below A.
1) At what time after its release will the second ball bearing be overtaken by the first?
2) How far does the second bearing fall in that time?
I will glad to help on any of your questions where work done by you is shown, but there do not appear to be any.
15
To answer these questions, we need to consider the motion of both ball bearings.
Let's start by analyzing the motion of the first ball bearing. As it is dropped from rest at point A, it will accelerate due to gravity. The equation that describes this motion is:
y1 = 0.5 * g * t^2
Where:
- y1 is the distance fallen by the first ball bearing at time t
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time elapsed since the ball was dropped
Next, let's examine the motion of the second ball bearing. It is released from rest at a position 5.5 m below point A. The equation that describes its motion is:
y2 = 5.5 + 0.5 * g * t^2
Where:
- y2 is the distance fallen by the second ball bearing at time t
- 5.5 is the initial distance of the second ball bearing below point A
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time elapsed since the ball was released
To find the time at which the second ball bearing is overtaken by the first, we need to equate the positions of the two ball bearings:
0.5 * g * t^2 = 5.5 + 0.5 * g * t^2 - 4.5
Simplifying the equation, we get:
0.5 * g * t^2 = 1
Dividing both sides by 0.5 * g, we find:
t^2 = 1 / (0.5 * g)
Taking the square root of both sides, we can solve for t:
t = √(2 / g)
Plugging in the value for acceleration due to gravity (g ≈ 9.8 m/s^2), we get:
t ≈ √(2 / 9.8) ≈ √(1 / 4.9) ≈ √0.204 ≈ 0.452 s
So, the first ball bearing will overtake the second ball bearing after approximately 0.452 seconds.
To find out how far the second ball bearing falls in that time, we can substitute this value of t into the equation for y2:
y2 = 5.5 + 0.5 * g * t^2
y2 = 5.5 + 0.5 * 9.8 * (0.452)^2
y2 ≈ 5.5 + 0.5 * 9.8 * 0.204
y2 ≈ 5.5 + 1. up to 4 decimal places.
Therefore, the second ball bearing falls approximately 6.5 meters in the time it takes for the first ball bearing to overtake it.