Question

A ball is dropped from a height of \(2.2 \mathrm{~m}\) and rebounds to a height of \(1.9 \mathrm{~m}\) above the floor. Assume the ball was in contact with the floor for \(96 \mathrm{~ms}\) and determine the average acceleration (magnitude and direction) of the ball during contact with the floor.

Short answer

The average acceleration of the ball during contact with the floor is -65 m/s^2 downwards.

Step-by-step solution

Convert time

Convert the time of contact from milliseconds to seconds for consistent units. \(1 ms = 10^{-3} s\), therefore \(96 ms = 0.096 s.\)

Calculate total distance

Calculate the total distance traveled by the ball during contact with the floor. This is the sum of the initial height (2.2 m) and the rebound height (1.9 m), equals to 2.2 m + 1.9 m = 4.1 m. In considering direction, the distance fallen is made negative because it is downwards, therefore, -2.2 m (falling) + 1.9 m (rebound) = -0.3 m.

Apply equation of motion

Apply the equation relating distance, time and acceleration: \(d = V_{i}t + 0.5at^{2}\). Here, the ball was dropped, so initial velocity (\(V_{i}\)) is 0. Insert known values into the equation: -0.3 = 0*(0.096) + 0.5*a*(0.096)^2. Solving for a, we get \(a = -0.3 / (0.5*(0.096)^2) = -65 m/s^2\).

Key concepts

Motion in Two Dimensions

Motion in two dimensions includes any scenario where an object moves in more than one direction, such as horizontally and vertically at the same time. A classic example is a ball being thrown in a projectile motion, which has both a horizontal component (due to the initial throw) and a vertical component (due to gravity).
Understanding two-dimensional motion involves breaking down the movement into separate components. Typically, this involves analyzing the motion in the horizontal direction independently of the motion in the vertical direction. In our exercise scenario, the ball moves only in one dimension (up and down), but the principles of separating components can still be applied when analyzing more complex two-dimensional situations.
In the case of the ball, when it impacts the floor, it temporarily has a horizontal component of zero, and the motion we consider is vertical. The vertical component is influenced by gravity, which pulls the ball down, and by the rebound force of the floor, sending it upwards.

Kinematic Equations

Kinematic equations describe the relationships between displacement, velocity, acceleration, and time without taking forces into account. These equations are essential tools for solving motion problems in physics, especially when dealing with constant acceleration. There are four key kinematic equations, each serving a different purpose based on the known and unknown variables of the problem.
In our textbook exercise, the equation used relates acceleration (\(a\)), time (\(t\) ), and displacement (\(d\) ). The equation is expressed as: \[ d = V_{i}t + 0.5at^{2} \] where \(V_{i}\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. By rearranging the formula to solve for acceleration, we can find the average acceleration of the ball during the brief moment it contacts the floor. Using this equation provides a clear-cut solution to the exercise, demonstrating how kinematic equations are powerful tools for analyzing motion.

Free Fall

Free fall refers to the motion of an object falling solely under the influence of gravity, with no air resistance affecting its motion. The acceleration due to gravity near the Earth's surface is approximately \(9.8 \text{m/s}^{2}\), and it acts downward, towards the center of the Earth. In free fall, an object's velocity increases by this acceleration with each second of fall.
In our exercise, we see an example of free fall when the ball is dropped from a height of \(2.2 \text{m}\) and subsequently rebounds to a height of \(1.9 \text{m}\), disregarding any air resistance. Although this scenario includes a rebound and is not a classic free fall, the initial drop can be considered as such until the moment it hits the floor. Then, as it makes contact with the floor, it experiences a force that results in an average acceleration far greater than \(9.8 \text{m/s}^{2}\) in the opposite direction during the very short time interval of contact, demonstrating the powerful effect of accelerative forces acting on objects in motion.