Bounce of football (kinetic and potential energy)

[Mcoroklo]

Homework Statement

A football (european) that hits ortogonally a surface will reflect such that the kinetic energy of the ball after the collision will be a fraction of the kinetic energy of the ball before the collision. The football makes a free fall from 50 cm to a floor and jumps to the height of 40 cm. After that the football is kicked into a wall (of the same material as the floor) from 5 meters. The ball hits the wall ortogonally and lands on the floor in the distance a from the wall. Find a.

Homework Equations

We assume that 4/5 of the football's energy is conserved in a collision because

U₁ = 4/5*U₂
where U1 is the potential energy before the collision, U2 after.

The Attempt at a Solution

The ball will have 4/5 of its energy conserved after the collision with the wall. This will make the ball go 4/5 the distance it was shot from, making it reach 4 m.

This might be right, but we suspect it's too easy. The fact the Ekin energi has v² makes me suspect our current suggestion is wrong. a might also depend on how hard the ball is shot, but the excercise doesn't state any force that makes the ball go to and collide with the wall.


Thanks a lot! :-)

 

[anathema]

Thank you for your question. I would like to offer some insights into your problem and provide a potential solution.

First, let's define some variables:
- m: mass of the football
- v1: initial velocity of the football before collision with the wall
- v2: final velocity of the football after collision with the wall
- h1: initial height of the football before collision with the floor
- h2: final height of the football after collision with the floor
- d: distance between the wall and the point where the ball lands on the floor

Now, let's apply the conservation of energy principle to the two collisions:
1. Collision with the floor:
In this case, we can use the conservation of mechanical energy, which states that the sum of kinetic energy (KE) and potential energy (PE) is conserved.
KE1 + PE1 = KE2 + PE2
Since the ball is dropped from rest, KE1 = 0. Also, the potential energy at the initial and final positions can be expressed as:
PE1 = mgh1
PE2 = mgh2
Therefore, we have:
mgh1 = KE2 + mgh2
Substituting the given values (h1 = 50 cm, h2 = 40 cm) and solving for KE2, we get:
KE2 = 0.1*mgh1

2. Collision with the wall:
In this case, we can use the conservation of kinetic energy, which states that the kinetic energy before the collision is equal to the kinetic energy after the collision.
KE1 = KE2
Substituting the given values (KE1 = 0.1*mgh1, KE2 = 4/5*KE1), we get:
0.1*mgh1 = 4/5*0.1*mgh1 + 0.5*m*v2^2
Simplifying and solving for v2, we get:
v2 = √(0.2*g*h1)

Now, we can use the equation of motion for constant acceleration to find the distance d:
d = v2^2*sin(2θ)/g
Since the ball hits the wall orthogonally, we have θ = 90°, and thus:
d = v2^2/g = 0.2*h1

Substituting the given value

 

[kerimek]

I would approach this problem by first understanding the concept of kinetic and potential energy. Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or state. In the case of a football, when it is kicked, it has both kinetic and potential energy. As it travels, its kinetic energy decreases due to friction and air resistance, while its potential energy increases as it reaches a higher height. When it bounces off a surface, its kinetic energy decreases even more due to the transfer of energy to the surface, but its potential energy decreases as it loses height.

In this problem, we are given that the football bounces off the floor and reaches a height of 40 cm. This means that the potential energy at the highest point is equal to the potential energy at the starting point (before the ball was kicked). We are also given that 4/5 of the ball's energy is conserved after the collision with the wall. This means that the remaining 1/5 of the energy is lost due to friction and other factors.

Using the equation U1 = 4/5*U2, we can calculate the potential energy before and after the collision with the wall. Since the potential energy at the highest point is equal to the potential energy at the starting point, we can set U1 = U2 and solve for the distance a.

a = 4 meters

Therefore, the ball will travel a distance of 4 meters from the wall before it lands on the floor. This distance may vary depending on the initial force with which the ball is kicked, but the concept remains the same. The key takeaway is that the ball loses energy as it bounces off the wall, and this loss of energy is reflected in the distance it travels before landing on the floor.