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Chapter 9: Problem 13

(II) A child in a boat throws a 5.70 -kg package out horizontally with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\), Fig. 9-37. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is \(24.0 \mathrm{~kg}\) and that of the boat is \(35.0 \mathrm{~kg}\).

Short Answer

Expert verified
The velocity of the boat immediately after is approximately \(-0.966 \text{ m/s}\) opposite to the package's direction.

Step by step solution

01

Identify the Principle

This problem involves conservation of momentum. The total momentum of an isolated system before an event must equal the total momentum after the event.
02

Calculate Initial Momentum

Initially, the child, package, and boat are all at rest, so the initial total momentum is zero. Thus, the initial momentum: \[ p_{initial} = 0 \]
03

Determine the System Components

The system includes the child, the package, and the boat. So the total mass of the boat with the child is \( M_{boat+child} = 24.0 \text{ kg} + 35.0 \text{ kg} = 59.0 \text{ kg} \). The package has a mass of \( 5.70 \text{ kg} \).
04

Account for Final Momentum

The package is thrown with a velocity of \( 10.0 \text{ m/s} \) in one direction, so the final momentum of the package is \( p_{package} = 5.70 \times 10.0 = 57.0 \text{ kg} \cdot \text{m/s} \). Since the initial momentum is zero, the final momentum of the boat and child system is in the opposite direction.
05

Apply Conservation of Momentum

By conservation of momentum: \[ 0 = 59.0 \times v_{boat+child} + 57.0 \] Solving for \( v_{boat+child} \), we get: \[ v_{boat+child} = -\frac{57.0}{59.0} \].
06

Calculate Velocity

Computing the expression from the previous step gives: \[ v_{boat+child} \approx -0.966 \text{ m/s} \]. The negative sign indicates the direction is opposite to the package's motion.

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Key Concepts

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

Newton's laws of motion
Newton's laws of motion form the foundation of classical mechanics, describing how objects move under the influence of forces. Among these, the third law states that for every action, there is an equal and opposite reaction. This principle is visible in the exercise where a child throws a package from a boat. By pushing the package forward, an equal and opposite force pushes the boat backward.
This concept explains how actions and reactions are equal in magnitude but opposite in direction. Therefore, when the child exerts a force on the package, the package exerts an equal force back on the child and boat system in the opposite direction.
In practical terms, when one body applies a force to another body, the second body applies an equal force in the opposite direction to the first body. This is the reason why the boat, initially at rest, moves backward when the child throws the package.
Isolated System
An isolated system is one where no external forces act, meaning that the total momentum within the system is conserved. This concept is crucial in understanding the exercise, in which the child, boat, and package together form an isolated system just before and after the package is thrown.
In such a system, the internal interactions (like throwing a package) do not alter the total momentum. This is because there are no external influences, such as water resistance or air drag, considered in this scenario.
  • The total momentum before and after throwing the package remains constant.
  • This isolated system assumption allows for the use of conservation of momentum in solving the problem.

In reality, while external forces can complicate real-world scenarios, idealized problems like these assume isolation to simplify calculations and better understand the core physics concepts without unnecessary interference.
Momentum Calculations
Momentum is a measure of an object's motion, which depends on both its mass and velocity. It is a vector quantity, meaning it has both size and direction. In the context of the exercise, calculating initial and final momentum is key to solving the problem.
Initial momentum is calculated by considering the boat, child, and package as one system initially at rest. Thus, their combined initial momentum is zero.
  • The formula for momentum is given by: \[ p = m imes v \]
  • For the package, with mass 5.70 kg and velocity 10 m/s, the momentum is \[ p_{package} = 5.70 imes 10 = 57.0 ext{ kg} \cdot \text{m/s} \]
  • Conservation of momentum allows us to equate the initial and final total momentum, leading to the calculation of the boat and child's velocity.

By setting the total initial momentum (0) equal to the sum of the momenta after the event, we calculate the unknown velocity of the boat and child. The resulting negative velocity indicates that their motion is opposite to that of the thrown package.

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Most popular questions from this chapter

(II) A tennis ball of mass \(m=0.060 \mathrm{~kg}\) and speed \(v=25 \mathrm{~m} / \mathrm{s}\) strikes a wall at a \(45^{\circ}\) angle and rebounds with the same speed at \(45^{\circ}\) (Fig. 9-38). What is the impulse (magnitude and direction) given to the ball?

Suppose two asteroids strike head on. Asteroid A \(\left(m_{\mathrm{A}}=7.5 \times 10^{12} \mathrm{~kg}\right)\) has velocity \(3.3 \mathrm{~km} / \mathrm{s}\) before the collision, and asteroid B \(\left(m_{\mathrm{B}}=1.45 \times 10^{13} \mathrm{~kg}\right)\) has velocity \(1.4 \mathrm{~km} / \mathrm{s}\) before the collision in the opposite direction. If the asteroids stick together, what is the velocity (magnitude and direction) of the new asteroid after the collision?

(II) Two billiard balls of equal mass move at right angles and meet at the origin of an \(x y\) coordinate system. Initially ball \(\mathrm{A}\) is moving upward along the \(y\) axis at \(2.0 \mathrm{~m} / \mathrm{s},\) and ball \(\mathrm{B}\) is moving to the right along the \(x\) axis with speed \(3.7 \mathrm{~m} / \mathrm{s}\). After the collision (assumed elastic), the second ball is moving along the positive \(y\) axis (Fig. \(9-43\) ). What is the final direction of ball \(\mathrm{A},\) and what are the speeds of the two balls?

(II) Mass \(M_{\mathrm{A}}=35 \mathrm{~kg}\) and mass \(M_{\mathrm{B}}=25 \mathrm{~kg} .\) They have velocities \((\) in \(\mathrm{m} / \mathrm{s}) \overrightarrow{\mathbf{v}}_{\mathrm{A}}=12 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{v}}_{\mathrm{B}}=-20 \hat{\mathbf{i}}+14 \hat{\mathbf{j}}\) Determine the velocity of the center of mass of the system.

(II) Find the center of mass of the ammonia molecule. The chemical formula is \(\mathrm{NH}_{3}\). The hydrogens are at the corners of an equilateral triangle (with sides \(0.16 \mathrm{nm}\) ) that forms the base of a pyramid, with nitrogen at the apex \((0.037 \mathrm{nm}\) vertically above the plane of the triangle).
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