Problem 68 - University Physics with Modern Physics

Question

A soccer ball rolls out of a gym through the center of a doorway into the next room. The adjacent room is \(6.00 \mathrm{~m}\) by \(6.00 \mathrm{~m}\) with the \(2.00-\mathrm{m}\) wide doorway located at the center of the wall. The ball hits the center of a side wall at \(45.0^{\circ} .\) If the coefficient of restitution for the soccer ball is \(0.700,\) does the ball bounce back out of the room? (Note that the ball rolls without slipping, so no energy is lost to the floor.)

Step-by-Step Solution

Answer

Answer: No, the ball does not bounce back out of the room.

Step 1: Find the initial speed of the ball

First, let's find the time it takes for the ball to travel from the doorway to the center of the room, which is half of the room's width, i.e., 3 meters. Then, we will calculate the initial speed. Let's use the formula: d = (1/2)at^2 Considering d = 3m and a = g*cos(45), where g = 9.81m/s^2 (acceleration due to gravity), we can find the time, t. 3 = (1/2)(9.81*cos(45))t^2 By solving this equation, we get: t = 0.7744 s Now, let's find the initial speed (v) of the ball using the formula: v = at v = (9.81*cos(45))*0.7744 v = 5.38 m/s

Step 2: Calculate the final speed of the ball

Now, we need to find the final speed of the ball after bouncing off the wall. The coefficient of restitution (e) is given as 0.700, and the relation for the final speed (vf) is: vf = ev vf = 0.700*(5.38) vf = 3.77 m/s

Step 3: Calculate the angle of the bounce

Since the ball hits the wall at a 45-degree angle and maintains this angle during the bounce, the angle at which the ball bounces off the wall will also be \(45.0^{\circ}\).

Step 4: Determine the distance traveled after the bounce

Now, we need to determine the distance traveled by the ball after bouncing off the wall and the time it takes to do so. Let's use the following expression to find the time (t_bounce) taken after the bounce: t_bounce = 2*(vf*sin(45))/g t_bounce = 2*(3.77*sin(45))/9.81 t_bounce = 0.5366 s Next, we need to find the horizontal distance (d_bounce) traveled by the ball in this time: d_bounce = vf*cos(45)*t_bounce d_bounce = 3.77*cos(45)*0.5366 d_bounce = 1.81 m Since the doorway is 2 meters wide and located in the center of the 6-meter wall, the room's center lies 1 meter away from the edge of the doorway. Therefore, since the ball travels a horizontal distance of 1.81 meters after bouncing, it does not reach the doorway and doesn't bounce back out of the room.

Key Concepts

Coefficient of Restitution

The coefficient of restitution is a crucial concept in physics that measures how much kinetic energy remains after a collision compared to what was initially present. It is denoted by the symbol \( e \) and is a dimensionless value ranging from 0 to 1.

• An \( e \) value of 0 suggests a perfectly inelastic collision, where objects stick together and no kinetic energy is conserved as usable work.
• An \( e \) value of 1 indicates a perfectly elastic collision, where no kinetic energy is lost.

For our soccer ball problem, the coefficient of restitution is given as 0.700. This implies that 70% of the ball's kinetic energy is retained after hitting the wall. This impacts the ball's speed post-collision, allowing us to calculate new speeds or predict future movements. Being aware of \( e \) is crucial in sports physics, mechanical engineering, and material science, where collision impacts need precise assessment.

Projectile Motion

Projectile motion involves the movement of an object thrown or projected into the air, subject to the only force of gravity. It is a key concept to understand for problems like the one involving the soccer ball.

• The motion occurs in a curved path due to the influence of gravity, causing the object to follow a parabolic trajectory.
• Key components of projectile motion include an initial velocity, angle of projection, and the action of gravitational force.

In this exercise, the soccer ball moves with an initial speed and strikes the wall at a given angle of \(45^{\circ}\). Understanding these parameters helps us determine factors like the angle of motion post-collision and the trajectory of the object after rebounding. Gravity affects the ball throughout, ensuring a consistent downward force during motion. Analyzing these components helps in predicting whether the ball would exit the room upon bouncing back.

Energy Conservation

Energy conservation in physics refers to the idea that within an isolated system, energy remains constant; it can neither be created nor destroyed, only transformed. In the context of this problem, the rolling soccer ball transitions between kinetic energy states.

• The kinetic energy before hitting the wall is partially transferred back after collision, as described by the coefficient of restitution \( e \).
• As the coefficient is less than 1, some energy is dispersed in forms like sound or heat upon impact.

Conserving energy principles allow us to evaluate the soccer ball's speed both before and after hitting the wall. They provide assurances by defining how much of the ball’s energy is viable for further motion, crucial for determining if the ball might leave the room after rebounding.

Kinematics

Kinematics, a branch of mechanics, deals with the motion of objects without considering the forces that cause the motion. This makes understanding the motion trajectories in the soccer ball problem crucial.

• Important kinematic equations relate velocity, acceleration, displacement, and time:
• \( d = v_i t + \frac{1}{2} a t^2 \)
• \( v_f = v_i + at \)
• \( v_f^2 = v_i^2 + 2ad \)

Utilizing kinematics, we can establish the initial speed of the soccer ball before impact and determine its subsequent speed, angle, and travel post-bounce. Calculating these values confirms the scenario's continuation and provides insight into whether rebounding actions remain within expected physical bounds. These relations are fundamental in planning the motion paths of diversified moving objects across various physical situations.