/*! 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 20 You are standing on a sheet of i... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 20

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction hctwecn your feet and the icc. A fricnd throws you a \(0.600 \mathrm{~kg}\) hall that is travcling hurizontally at \(10.0 \mathrm{~m} / \mathrm{s}\). Your mass is \(70.0 \mathrm{~kg}\). (a) If you calch the ball. with what speed do you and the hall mowe afterward? (b) If the hall hils you and bounces off your chest, so aflerward it is moving hurizontally at \(8.0 \mathrm{~m} / \mathrm{s}\) in the opposite direction, what is your speed after the cullision?

Short Answer

Expert verified
a) The speed after catching the ball is 0.085 m/s. b) The speed after the ball bounces off is 0.154 m/s

Step by step solution

01

- Calculate combined momentum after the catch in Part a

For part a, the initial momentum (before the catch) will be the momentum of the ball since you were standing still. To calculate momentum, the mass of an object is multiplied by its velocity, so this will be \(0.600 \mathrm{~kg} \times 10.0 \mathrm{~m/s} = 6.0 \mathrm{kg \cdot m/s}\). After the catch, you and the ball move together, so the momentum will be the total mass times the final velocity, which will be \((0.6 \mathrm{kg} + 70.0 \mathrm{kg}) \times v = 70.6 \mathrm{kg} \times v\). The total momentum before the catch will be equal to the total momentum after the catch due to the conservation of momentum, therefore \(6.0 \mathrm{kg \cdot m/s} = 70.6 \mathrm{kg} \times v\). To solve for \(v\), you would divide both sides of the equation by \(70.6 \mathrm{kg}\) to get \(v = 6.0 \mathrm{kg \cdot m/s}/70.6 \mathrm{kg} = 0.085 \mathrm{m/s}\). This is rounded to three significant figures.
02

- Calculate combined momentum after the bounce in Part b

For part b, the initial momentum is still the same, \(6.0\, \mathrm{kg \cdot m/s}\). After the bounce, the football moves in the opposite direction with a velocity of \(8.0 \mathrm {m/s}\), so has a momentum of \(-0.600 \mathrm{ kg} \times 8.0\, \mathrm{m/s} = -4.8 \mathrm{kg \cdot m/s}\). The total momentum afterward is therefore \(-(4.8 + 70.0 \times v)\) \(\mathrm{kg \cdot m/s}\). Again, since momentum is conserved, the total momentum before the bounce is equal to the total momentum afterward. Therefore, \(6.0 = -4.8 -70v\). Solving for \(v\) gives \(v = (6.0 + 4.8)/70 = 0.154 \mathrm{ m/s}\). This is rounded to three significant figures.

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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, representing the quantity of motion an object possesses. It is calculated as the product of an object's mass and its velocity, given by the formula \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. When two or more objects interact, their total momentum before the interaction is equal to the total momentum after the interaction, provided no external forces are acting. This principle, known as the conservation of momentum, is crucial in analyzing collisions and other interactions.

In the given exercise, the scenario involves a ball being thrown to a person standing on ice, ensuring minimal external resistance. Before the person catches the ball, only the ball has momentum since the person is initially at rest. If we assume no outside forces (like friction or air resistance) are acting on the system, the product of the ball's mass and velocity will equal the system's total momentum after the catch. This principle aids in determining the velocity of the person after catching the ball.
Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together or deform, and kinetic energy is not conserved, though momentum is. During inelastic collisions, some kinetic energy is transformed into other forms of energy, such as thermal energy or sound. However, under the conservation of momentum, the combined momentum of the system before and after the collision remains the same.

In part (a) of the exercise, the event where the ball is caught and moves alongside the person can be considered an inelastic collision because the two masses move together after the collision at the same velocity. Calculating the combined mass and using the conservation of momentum allows us to find this final velocity. This presents an excellent example of how inelastic collisions behave in real-world applications, emphasizing the constancy of momentum despite changes in energy forms.
Recoil Velocity
Recoil velocity refers to the velocity an object acquires when it exerts a force on another object and, as a result, is propelled backward. It's encountered in various scenarios, such as when shooting a gun or when objects collide and bounce off each other. After a collision, if an object moves in one direction, another object involved in the collision must move in the opposite direction for momentum to be conserved, provided no external forces are at play.

In part (b) of the exercise, when the ball bounces off the person, we examine the recoil velocity. The ball exerts a force on the person, and by Newton's third law of motion, the person exerts an equal and opposite force on the ball. The calculation of the person's speed after the ball bounces off his chest is an example of finding the recoil velocity of the person. This exercise showcases how conservation of momentum is utilized to determine recoil velocities in a closed system.

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Most popular questions from this chapter

Two figure skaters, one weighing \(625 \mathrm{~N}\) and the other \(725 \mathrm{~N}\) push off against cach other on frictionless ice. (a) If the heavicr skater travels at \(1.50 \mathrm{~m} / \mathrm{s}\), how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this cnergy come from?

A rifle bullet with mass \(8.00 \mathrm{~g}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a frictionless, horizontal surface and is attached to an ideal spring (Fig. \(\mathrm{P} 8.79\) ). The impact compresses the spring \(15.0 \mathrm{~cm} .\) Calibration of the spring shows that a force of 0.750 \(\mathrm{N}\) is required to compress the spring \(0.250 \mathrm{~cm}\). (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullct?

Find the position of the center of mass of the system of the sun and Jupiter. (Since Jupiter is more massive than the rest of the solar planets combined, this is essentially the position of the center of mass of the solar system.) Does the center of mass lie inside or outside the sun? Use the data in Appendix F.

In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$ x_{\mathrm{cm}}=\frac{1}{M} \int x d m \quad y_{\mathrm{cm}}=\frac{1}{M} \int y d m $$ where \(x\) and \(y\) are the coordinates of the small picce of the object that has mass \(d m\). The integration is over the whole of the object. Consider a thin rod of length \(I_{4}\), mass \(M\), and cross-sectional area \(A\). I et the origin of the coordinates be at the left end of the rod and the positive \(x\) -axis lie along the rod. (a) If the density \(\rho=M / V\) of the object is uniform, perform the integration described above to show that the \(x\) -coordinate of the center of mass of the rod is at its gcometrical center. (b) If the density of the object varies linearly with \(x\) - that is, \(\rho=\alpha x,\) where \(\alpha\) is a positive constant - calculate the \(x\) -coordinate of the rod's center of mass.

Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's wcight is \(800 \mathrm{~N}\), Jane's weight is \(600 \mathrm{~N}\), and that of the sleigh is \(1000 \mathrm{~N}\). They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal (relative to the ice), and Jane jumps to the right at \(7,00 \mathrm{~m} / \mathrm{s}\) at \(36.9^{\circ}\) above the horizontal (relative to the ice). Calculate the sleigh's horizontal velocity (magnitude and direction) after they jump out.
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