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Chapter 6: Problem 4

A \(0.10\) -kg ball is thrown straight up into the air with an initial speed of \(15 \mathrm{~m} / \mathrm{s}\). Find the momentum of the ball (a) at its maximum height and (b) halfway to its maximum height.

Short Answer

Expert verified
The momentum of the ball at its maximum height is \(0 kg.m/s\) and its momentum halfway to its maximum height is calculated as the product of its mass and the velocity obtained in step 3.

Step by step solution

01

Determine Velocity at Maximum Height

When an object achieves maximum height, its vertical velocity becomes zero because at this point, it momentarily stops before starting to fall back down. So, velocity (v) = 0 m/s at the maximum height.
02

Calculate Momentum at Maximum Height

The momentum (p) of the ball at maximum height can be calculated using the formula \(p = mv\). Since the velocity at maximum height is zero, multiplying the mass, \(0.10 kg\), by the velocity, \(0 m/s\), the momentum is \(0 kg.m/s\). So momentum of the ball at its maximum height is \(0 kg.m/s\).
03

Determine Velocity Halfway to Maximum Height

In ascending motion the acceleration, \(a\), is equal to \(-g\), where \(g = 9.8 m/s^2\) is the acceleration due to gravity. We can determine the remaining velocity at the halfway point by applying the equation of motion \(v^2 = u^2 + 2as\). Since we are considering halfway point, distance \(s\) will be half of the maximum height which can be obtained from the equation \(H_{max} = (u^2)/(2g)\). Plugging in given values, we simplify to obtain the value of velocity.
04

Calculate Momentum Halfway to Maximum Height

We can then calculate momentum halfway to the maximum height by applying the formula \(p = mv\). The mass \(m\) is \(0.10 kg\) and the velocity \(v\) is the value obtained in step 3.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is fundamental in physics. It states that the total energy in a closed system remains constant, although it can switch forms. For example, when the ball is thrown up, kinetic energy (from motion) is converted into potential energy (due to height).
At the highest point, all the kinetic energy has converted into potential energy.
  • Total Energy = Kinetic Energy + Potential Energy
  • Total Energy stays constant; it only changes forms.
So, understanding conservation of energy helps us know where the energy in the system goes, even if velocity and momentum change.
Kinematics
Kinematics is all about describing motion. It deals with the speeds and distances, but not the forces that cause them. Kinematics provides equations and concepts for understanding how objects move. For instance, when the ball is thrown upwards, it moves in a vertical motion. The initial velocity is important, and we need to consider gravity that acts downward.
  • Initial Velocity: The speed at which the ball is thrown.
  • Acceleration: Due to gravity (-9.8 m/s虏 when going up).
  • Maximum Height: Reached when the velocity of the ball becomes zero.
Kinematics allows us to calculate various parameters such as speed, distance, and time and assists in understanding the journey of the ball through the air.
Equations of Motion
Equations of motion are vital for calculating different properties of moving objects. They allow us to link different variables like velocity, acceleration, and time. In the case of the ball thrown upwards, we use these equations to find out how fast the ball moves at different points.
  • First Equation: Allows finding final velocity. Given by \( v = u + at \) where \( v \) is final velocity, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
  • Second Equation: Describes displacement. \( s = ut + \frac{1}{2} a t^2 \).
  • Third Equation: Used here, relates velocity with displacement: \(v^2 = u^2 + 2as\).
These equations simplify the process of understanding how the ball's speed and position change with time, especially when finding the velocity halfway to maximum height.

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Most popular questions from this chapter

8\. A \(75.0\) -kg ice skater moving at \(10.0 \mathrm{~m} / \mathrm{s}\) crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at \(5.00 \mathrm{~m} / \mathrm{s}\). Suppose the average force a skater can experience without breaking a bone is \(4500 \mathrm{~N}\). If the impact time is \(0.100 \mathrm{~s}\), does a bone break?

A rifle with a weight of \(30 \mathrm{~N}\) fires a \(5.0-\mathrm{g}\) bullet with a speed of \(300 \mathrm{~m} / \mathrm{s} .\) (a) Find the recoil speed of the rifle. (b) If a \(700-\mathrm{N}\) man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.

IA 0.280-kg volleyball approaches a player horizontally with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of \(22.0 \mathrm{~m} / \mathrm{s}\). (a) What impulse is delivered to the ball by the player? (b) If the player's fist is in contact with the ball for \(0.0600 \mathrm{~s}\), find the magnitude of the average force exerted on the player's fist.

The front \(1.20 \mathrm{~m}\) of a \(1400-\mathrm{kg}\) car is designed as a "crumple zone" that collapses to absorb the shock of a collision. If a car traveling \(25.0 \mathrm{~m} / \mathrm{s}\) stops uniformly in \(1.20 \mathrm{~m}\), (a) how long does the collision last, (b) what is the magnitude of the average force on the car, and (c) what is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity.

A A \(45.0\) -kg girl is standing on a 150 -kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of \(1.50 \mathrm{~m} / \mathrm{s}\) to the right relative to the plank. (a) What is her velocity relative to the surface of the ice? (b) What is the velocity of the plank relative to the surface of the ice?
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