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Chapter 7: Problem 14

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

Short Answer

Expert verified
The basketball was dropped from a height of approximately 1.37 meters.

Step by step solution

01

Understand the given values

We know the mass of the basketball is \(m = 0.60\, \text{kg}\) and the magnitude of its momentum just before hitting the floor is \(p = 3.1\, \text{kg}\cdot\text{m/s}\). We need to find the height \(h\) from which it was dropped.
02

Relate momentum to velocity

Momentum \(p\) can be expressed as the product of mass and velocity: \(p = m \cdot v\). Thus, the velocity \(v\) of the ball just before impact is given by \(v = \frac{p}{m}\). Substituting the given values, we calculate the velocity: \(v = \frac{3.1}{0.60}\).
03

Calculate the velocity

Calculating the above expression gives us \(v = 5.17\, \text{m/s}\). So, the velocity of the basketball just before hitting the floor is \(5.17\, \text{m/s}\).
04

Apply energy conservation principles

The basketball's initial potential energy when dropped is converted into kinetic energy just before it hits the floor. The initial potential energy is \(mgh\) and the kinetic energy just before impact is \(\frac{1}{2}mv^2\). Equating the potential and kinetic energies, we have \(mgh = \frac{1}{2}mv^2\).
05

Solve for initial height \(h\)

Since mass \(m\) cancels out, the equation simplifies to \(gh = \frac{1}{2}v^2\). Solving for \(h\) gives us \(h = \frac{v^2}{2g}\), where \(g = 9.8\, \text{m/s}^2\). Substituting values, \(h = \frac{(5.17)^2}{2 \times 9.8} = \frac{26.77}{19.6}\).
06

Calculate the height

Dividing the values gives \(h = 1.37\, \text{m}\). Therefore, the basketball was dropped from a height of approximately \(1.37\, \text{meters}\).

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, highlighting the motion of an object as it relates to its mass and velocity. When we say an object has momentum, we are referring to how hard it is to stop that object. The formula for momentum is given by:
  • Momentum (p) = mass (m) 脳 velocity (v)
If you know the momentum and the mass of an object, you can calculate its velocity. In the basketball example, the ball's momentum just before impact is 3.1 kg路m/s. Given its mass of 0.60 kg, we derive its velocity from:
  • v = p / m = 3.1 / 0.60 鈮� 5.17 m/s
Thus, the concept of momentum helps us understand the motion of objects in detail by considering both their mass and speed. It's especially important in collisions and scenarios where objects change speed or direction.
Kinetic Energy
Kinetic energy is energy due to motion. Any moving object has kinetic energy, and it is calculated using the object's mass and the square of its velocity. The formula for kinetic energy is:
  • KE = 1/2 脳 m 脳 v虏
This form of energy comes into play when objects speed up or slow down. In the dropped basketball instance, right before impact, all the potential energy it initially had is converted into kinetic energy. So, its kinetic energy becomes:
  • KE = 1/2 脳 0.60 kg 脳 (5.17 m/s)虏 鈮� 8 J (Joules)
Kinetic energy illustrates how mass and velocity together define an object's energy state while moving. It reflects how much work force it can exert on another object upon collision and how it behaves when forces act upon it.
Potential Energy
Potential energy is the stored energy of position. When an object is at a height, it possesses gravitational potential energy. Potential energy related to height is given by:
  • PE = m 脳 g 脳 h
Here, m is the mass, g is the acceleration due to gravity (9.8 m/s虏 on Earth), and h is the height above the ground. For our falling basketball, the potential energy initially stored gets converted fully into kinetic energy as the ball descends.
To find the initial height h from which the ball was dropped, we equate the potential energy to the kinetic energy at the bottom:
  • m 脳 g 脳 h = 1/2 脳 m 脳 v虏
  • g 脳 h = 1/2 脳 v虏 (mass cancels out)
Solving for h gives:
  • h = v虏 / (2 脳 g) 鈮� (5.17)虏 / (2 脳 9.8) 鈮� 1.37 m
This equation illustrates the principle of conservation of energy, where the initial position-based energy transforms entirely into energy of motion without any loss in a vacuum.

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Most popular questions from this chapter

A 60.0 -kg person, running horizontally with a velocity of \(+3.80 \mathrm{m} / \mathrm{s}\), jumps onto a \(12.0-\mathrm{kg}\) sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast \(30.0 \mathrm{m}\) on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

A 46-kg skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of 21.2 m/s. Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude as, but opposite direction to, the force that the wall applies to her).

A \(40.0-\mathrm{kg}\) boy, riding a \(2.50-\mathrm{kg}\) skateboard at a velocity of \(+5.30 \mathrm{m} / \mathrm{s}\) across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is \(6.00 \mathrm{m} / \mathrm{s}, 9.50^{\circ}\) above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's veloc- ity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

One average force \(\overrightarrow{\mathbf{F}}_{1}\) has a magnitude that is three times as large as that of another average force \(\overrightarrow{\mathbf{F}}_{2} .\) Both forces produce the same impulse. The average force \(\overrightarrow{\mathbf{F}}_{1}\) acts for a time interval of \(3.2 \mathrm{ms}\). For what time interval does the average force \(\overrightarrow{\mathbf{F}}_{2}\) act?

An 85-kg jogger is heading due east at a speed of 2.0 m/s. A 55-kg jogger is heading \(32^{\circ}\) north of east at a speed of 3.0 m / s. Find the magnitude and direction of the sum of the momenta of the two joggers.
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