/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A basketball (m 0.60 kg) is drop... [FREE SOLUTION] | 91Ó°ÊÓ

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A basketball (m 0.60 kg) is dropped from rest. Just before striking the floor, the ball has a momentum whose magnitude is 3.1 kg m/s. At what height was the basketball dropped?

Short Answer

Expert verified
The basketball was dropped from a height of approximately 1.362 meters.

Step by step solution

01

Understanding Conservation of Energy

The ball is dropped from rest, so its initial kinetic energy is zero. As it falls, its potential energy converts into kinetic energy. Just before striking the floor, all the potential energy has been converted into kinetic energy. Using the conservation of energy principle, we can equate the initial potential energy to the final kinetic energy.
02

Relate Momentum to Kinetic Energy

The momentum of the ball just before striking the floor is given as 3.1 kg m/s. Kinetic energy (K.E) can be calculated using:\[ K.E = \frac{p^2}{2m} \]where \( p \) is the momentum and \( m \) is the mass of the ball. Let's calculate the kinetic energy using this formula.
03

Calculate Kinetic Energy

Substitute the values into the kinetic energy formula:\[ K.E = \frac{(3.1)^2}{2 \times 0.60} \]\[ K.E = \frac{9.61}{1.2} = 8.0083 \text{ J} \]The kinetic energy just before the collision is approximately 8.0083 Joules.
04

Calculate Initial Potential Energy

The initial potential energy (P.E) when the ball is at height \( h \) is given by the formula:\[ P.E = mgh \]where \( g = 9.8 \text{ m/s}^{2} \) (acceleration due to gravity). Since the initial potential energy equals the final kinetic energy: \[ mgh = 8.0083 \]
05

Solve for Height

Rearrange the potential energy equation to solve for \( h \):\[ h = \frac{8.0083}{m \times g} \]Substitute the known values (\( m = 0.60 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \)):\[ h = \frac{8.0083}{0.60 \times 9.8} \]\[ h = \frac{8.0083}{5.88} \]\[ h = 1.362 \text{ m} \]The basketball was dropped from a height of approximately 1.362 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum
Momentum is a measure of the motion of an object and is the product of its mass and velocity. Momentum, denoted as \( p \), is given by the formula:
  • \( p = mv \)
Here, \( m \) is the mass, and \( v \) is the velocity of the object.
In our exercise, the basketball has a momentum of 3.1 kg m/s just before it hits the ground. This value indicates the amount of motion the ball possesses, taking into account both its mass and speed at that moment. The concept of momentum is crucial in understanding how moving objects behave, particularly during collisions or impacts.
Since momentum is conserved in isolated systems (no net external force acting), it plays a significant role in analyzing motion. In this case, while falling, the basketball gains momentum, which peaks just before impact.
Potential Energy
Potential energy is the energy stored in an object due to its position or configuration. For objects in a gravitational field, such as a basketball above the ground, potential energy is given by the formula:
  • \( P.E = mgh \)
where \( m \) is the mass, \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height above the ground.
In our exercise, the basketball's potential energy at the initial height is converted entirely into kinetic energy as it falls under the influence of gravity. This transformation is explained by the conservation of energy principle, where the potential energy at the start is equal to the kinetic energy just before hitting the ground.
To find the initial height from which the basketball was dropped, we equate the potential energy to the known kinetic energy at the end of the drop.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is mathematically described by the formula:
  • \( K.E = \frac{1}{2}mv^2 \)
Kinetic energy can also be related to momentum using the formula:
  • \( K.E = \frac{p^2}{2m} \)
where \( p \) is the momentum and \( m \) is the mass.
In our situation, the basketball has a kinetic energy of approximately 8.0083 Joules just before it strikes the floor. This value is derived from the given momentum value and confirms that all potential energy at the start has been converted into kinetic.
The beauty of kinetic energy is that it reflects the work an object can do during its motion. As the basketball speeds towards the floor, it transitions from stationary potential to fully active kinetic energy, showcasing the transformations predicted by physics principles.

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