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Chapter 9: Problem 5

(II) A 145-g baseball, moving along the \(x\) axis with speed \(30.0 \mathrm{~m} / \mathrm{s}\), strikes a fence at a \(45^{\circ}\) angle and rebounds along the \(y\) axis with unchanged speed. Give its change in momentum using unit vector notation.

Short Answer

Expert verified
Change in momentum: \(-3.07 \hat{\imath} + 1.28 \hat{\jmath} \).

Step by step solution

01

Identify Initial Momentum

The initial momentum of the baseball can be calculated using the formula for momentum, \( p = m imes v \), where \( m \) is the mass and \( v \) is the velocity. The mass of the baseball \( m \) is 145 g, which is 0.145 kg, and the initial velocity along the \( x \) axis is 30.0 m/s. Thus, the initial momentum \( p_i = 0.145 \, \text{kg} \times 30.0 \, \text{m/s} = 4.35 \, \text{kg m/s} \). Since the motion is at a \( 45^{\circ} \) angle in the \( x \) direction, the momentum in unit vector form is \( \mathbf{p}_{i} = 4.35 \, \mathrm{cos}(45^{\circ}) \hat{\imath} + 0 \hat{\jmath} \).
02

Calculate Components of Initial Momentum

The component of initial momentum along the \( x \) axis (\( \hat{\imath} \)) is \( 4.35 \, \mathrm{cos}(45^{\circ}) \), which equals \( 3.07 \, \text{kg m/s} \), and along the \( y \) axis (\( \hat{\jmath} \)) is \( 4.35 \, \mathrm{sin}(45^{\circ}) \), also \( 3.07 \, \text{kg m/s} \). So, the initial momentum vector in unit vector notation is \( \mathbf{p}_{i} = 3.07 \hat{\imath} + 3.07 \hat{\jmath} \).
03

Determine Final Momentum

After hitting the fence, the baseball rebounds along the \( y \) axis with the same speed. The momentum is now zero along the \( x \) axis and remains 4.35 kg m/s along the \( y \) axis (since the speed is unchanged and now only along \( y \)). The final momentum vector is \( \mathbf{p}_{f} = 0 \hat{\imath} + 4.35 \hat{\jmath} \).
04

Calculate Change in Momentum

The change in momentum \( \Delta \mathbf{p} \) is equal to the final momentum \( \mathbf{p}_{f} \) minus the initial momentum \( \mathbf{p}_{i} \). This means \( \Delta \mathbf{p} = (0 \hat{\imath} + 4.35 \hat{\jmath}) - (3.07 \hat{\imath} + 3.07 \hat{\jmath}) = -3.07 \hat{\imath} + 1.28 \hat{\jmath} \). Thus, the change in momentum is \(-3.07 \hat{\imath} + 1.28 \hat{\jmath} \).

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Key Concepts

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

Collisions
Collisions are scenarios where two objects come into contact and exert forces on each other, affecting their motion. In physics, understanding collisions helps us analyze how momentum and energy transfer between objects. There are two main types of collisions: elastic and inelastic.
  • Elastic Collisions: Both momentum and kinetic energy are conserved. After an elastic collision, objects rebound without lasting deformation or generation of heat.
  • Inelastic Collisions: Momentum is conserved, but kinetic energy is not. The objects may stick together or change shape, resulting in some mechanical energy being converted into heat or sound.
In our baseball example, when it hits the fence, this is a type of collision. The baseball rebounds without losing speed, implying that most of the kinetic energy is retained, akin to an elastic collision. However, as it changes direction upon rebounding, understanding momentum change becomes crucial to fully grasp the dynamics.
Vector Notation
Vectors are mathematical entities used to represent quantities that have both magnitude and direction. In physics, vectors are essential for describing motion and other forces acting in multiple dimensions.
Vector notation simplifies the representation and calculation of physical quantities such as velocity and momentum.
  • A vector in two dimensions is expressed as a combination of components along the horizontal (\( \hat{\imath} \)) and vertical (\( \hat{\jmath} \)) axes.
  • For example, the vector \( \mathbf{v} = v_{x} \hat{\imath} + v_{y} \hat{\jmath} \) shows velocity components along these axes.
In the exercise, initial and final velocities were represented using unit vector notation. This allows us to break down the momentum into components along the \( x \) and \( y \) axes, simplifying the calculation of changes during the collision.
Conservation of Momentum
The conservation of momentum is a fundamental principle in physics stating that the total momentum of a closed system remains constant if no external forces act on it. This concept is crucial in analyzing collisions.
Here's how it works:
  • Before Collision: The system's total momentum is the sum of momentums of individual objects involved.
  • After Collision: Total momentum remains the same as before, though individual momenta of objects may change.
In our baseball scenario, even though the ball changes direction upon collision, the overall system's momentum along certain directions remains conserved. By knowing this, we calculate the change in the baseball's momentum after it rebounds, emphasizing momentum conservation's role in predicting the behavior of objects.

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Most popular questions from this chapter

(I) An atomic nucleus at rest decays radioactively into an alpha particle and a smaller nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is \(2.8 \times 10^{5} \mathrm{~m} / \mathrm{s}\) ? Assume the recoiling nucleus has a mass 57 times greater than that of the alpha particle.

(II) \(\mathrm{A} 3500\) -kg rocket is to be accelerated at 3.0 \(\mathrm{g}\) at take-off from the Earth. If the gases can be ejected at a rate of \(27 \mathrm{kg} / \mathrm{s},\) what must be their exhaust speed?

(III) A 280 -kg flatcar \(25 \mathrm{~m}\) long is moving with a speed of \(6.0 \mathrm{~m} / \mathrm{s}\) along horizontal frictionless rails. A \(95-\mathrm{kg}\) worker starts walking from one end of the car to the other in the direction of motion, with speed \(2.0 \mathrm{~m} / \mathrm{s}\) with respect to the car. In the time it takes for him to reach the other end, how far has the flatcar moved?

A 22 -g bullet strikes and becomes embedded in a \(1.35-\mathrm{kg}\) block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.28 , and the impact drives the block a distance of \(8.5 \mathrm{~m}\) before it comes to rest, what was the muzzle speed of the bullet?

(II) A square uniform raft, \(18 \mathrm{~m}\) by \(18 \mathrm{~m}\), of mass \(6200 \mathrm{~kg}\), is used as a ferryboat. If three cars, each of mass \(1350 \mathrm{~kg}\), occupy the NE, SE, and SW corners, determine the CM of the loaded ferryboat relative to the center of the raft.
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