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Chapter 8: Problem 69

Just before it is struck by a racket, a tennis ball weighing 0.560 \(\mathrm{N}\) has a velocity of \((20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\boldsymbol{J}}\) . During the 3.00 \(\mathrm{ms}\) that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 \mathrm{N}) \hat{\boldsymbol{\imath}}+(110 \mathrm{N}) \hat{\boldsymbol{J}}\) . (a) What are the \(x\) - and \(y\) -components of the impulse of the net force applied to the ball? (b) What are the \(x\) - and \(y\) -components of the final velocity of the ball?

Short Answer

Expert verified
(a) Impulse components: \(-1.14 \mathrm{Ns}\) and \(0.33 \mathrm{Ns}\). (b) Final velocity components: \(0.02 \mathrm{m/s}\) and \(1.78 \mathrm{m/s}\).

Step by step solution

01

Understand Impulse

Impulse \( \vec{J} \) is the change in momentum of the object and is given by the product of force \( \vec{F} \) and the time interval \( \Delta t \) during which the force acts. The formula is \( \vec{J} = \vec{F} \cdot \Delta t \).
02

Calculate Impulse Components

Given \( \vec{F} = -(380 \mathrm{N}) \hat{\boldsymbol{i}} + (110 \mathrm{N}) \hat{\boldsymbol{j}} \) and \( \Delta t = 3.00 \mathrm{ms} = 3.00 \times 10^{-3} \mathrm{s} \), the impulse components are:\[ J_x = F_x \cdot \Delta t = -380 \mathrm{N} \times 3.00 \times 10^{-3} \mathrm{s} = -1.14 \mathrm{Ns} \]\[ J_y = F_y \cdot \Delta t = 110 \mathrm{N} \times 3.00 \times 10^{-3} \mathrm{s} = 0.33 \mathrm{Ns} \]
03

Initial Velocity Components

The initial velocity of the ball is \( \vec{v}_i = (20.0 \mathrm{m/s}) \hat{\boldsymbol{i}} - (4.0 \mathrm{m/s}) \hat{\boldsymbol{j}} \) with components:\[ v_{ix} = 20.0 \mathrm{m/s} \]\[ v_{iy} = -4.0 \mathrm{m/s} \]
04

Determine Mass of the Ball

The weight \( W = 0.560 \mathrm{N} \) is given by \( W = m \cdot g \). Thus, the mass \( m \) of the ball is:\[ m = \frac{W}{g} = \frac{0.560 \mathrm{N}}{9.8 \mathrm{m/s}^2} \approx 0.0571 \mathrm{kg} \]
05

Apply Impulse-Momentum Theorem

The impulse-momentum theorem states that \( \vec{J} = m (\vec{v}_f - \vec{v}_i) \), where \( \vec{v}_f \) is the final velocity. Solving for \( \vec{v}_f \), we have: \[ m \cdot \vec{v}_f = m \cdot \vec{v}_i + \vec{J} \] \[ v_{fx} = v_{ix} + \frac{J_x}{m} \] \[ v_{fy} = v_{iy} + \frac{J_y}{m} \]
06

Calculate Final Velocity Components

Substitute to find the components:\[ v_{fx} = 20.0 \mathrm{m/s} + \frac{-1.14 \mathrm{Ns}}{0.0571 \mathrm{kg}} = 20.0 \mathrm{m/s} - 19.98 \mathrm{m/s} = 0.02 \mathrm{m/s} \]\[ v_{fy} = -4.0 \mathrm{m/s} + \frac{0.33 \mathrm{Ns}}{0.0571 \mathrm{kg}} = -4.0 \mathrm{m/s} + 5.78 \mathrm{m/s} = 1.78 \mathrm{m/s} \]

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Key Concepts

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

Newton's Second Law
One of the fundamental principles of physics is Newton's Second Law of Motion. It states that the net force acting on an object is equal to the product of its mass and acceleration. In equation form, this is written as \( F = m imes a \). This law explains how the velocity of an object changes when subjected to an external force.
For instance, if you know the mass of a tennis ball and the forces applied to it, you can determine its acceleration. For our sphere, the force components are given: \(-380 \mathrm{N}\) in the \(\hat{\boldsymbol{i}}\) direction and \(110 \mathrm{N}\) in the \(\hat{\boldsymbol{j}}\) direction. The fact that the force is constant simplifies our calculations.
By knowing the mass, which we find using the weight and gravitational pull, you can apply Newton鈥檚 Second Law to determine how the motion of the ball changes when it interacts with external forces like the racket.
Impulse-Momentum Theorem
The Impulse-Momentum Theorem is a key concept in understanding how forces affect motion over time. Impulse \( \vec{J} \) is defined as the product of the force \( \vec{F} \) and the time period \( \Delta t \) over which the force acts, or \( \vec{J} = \vec{F} \cdot \Delta t \). This results in a change in momentum of an object and it shows how powerful forces are applied over time.
In our example, the force is applied for \(3.00 \mathrm{ms}\) and we can calculate the impulse components using the given force and time. Once we have the impulse, we can apply the Impulse-Momentum Theorem: \( \vec{J} = m(\vec{v}_f - \vec{v}_i) \). This enables you to calculate the change in velocity of the ball, helping you identify its final velocity components after the interaction.
Kinematics
Kinematics explores the motion of objects without considering the forces that produce this movement. It describes motion in terms of displacement, velocity, and acceleration.
In our problem, the initial velocity components of the tennis ball were \( (20.0 \mathrm{m/s}) \hat{\boldsymbol{i}} \) and \(- (4.0 \mathrm{m/s}) \hat{\boldsymbol{j}} \). With kinematics, we can further predict the motion of the ball after forces are applied.
Using the Impulse-Momentum theorem alongside Kinematics, the relationship between initial velocity, impulse, and mass can help you determine the new velocity after the racket impacts the ball. This understanding is essential in solving both components of velocity effectively.
Vector Components
When dealing with physics problems, forces and motion often occur in multiple dimensions. This requires the use of vector components, allowing us to break down vectors into perpendicular components.
In the tennis ball exercise, both the force exerted and the ball鈥檚 velocity are given in vector form: for the force \( -(380 \mathrm{N}) \hat{\boldsymbol{i}} + (110 \mathrm{N}) \hat{\boldsymbol{j}} \) and for velocity \( (20.0 \mathrm{m/s}) \hat{\boldsymbol{i}} - (4.0 \mathrm{m/s}) \hat{\boldsymbol{j}} \).
By resolving these into components, you can handle each direction independently, using tools like the Impulse-Momentum theorem to explore how each direction behaves when forces are applied. This process simplifies solving complex physics problems and enhances understanding of motion in two or more dimensions.

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Most popular questions from this chapter

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less damage. In one such accident, a 1750 -kg car traveling to the right at 1.50 \(\mathrm{m} / \mathrm{s}\) collides with a \(1450-\mathrm{kg}\) car going to the left at 1.10 \(\mathrm{m} / \mathrm{s} .\) Measurements show that the heavier car's speed just after the collision was 0.250 \(\mathrm{m} / \mathrm{s}\) in its original direction. You can ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic energy of the two-car system during this collision.

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 piece of the object that has mass \(d m .\) The integration is over the whole of the object. Consider a thin rod of length \(L,\) mass \(M,\) and cross-sectional area \(A .\)Let 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 geometrical 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.

Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an elastic collision with the floor and then rebounds to make an elastic collision with the still-descending small. Just before the collision between the two balls, the large ball is moving upward with velocity \(\vec{\boldsymbol{v}}\) and the small ball has velocity \(-\vec{\boldsymbol{v}}\) . (Do you see why? \()\) Assume the large ball has a much greater mass than the small ball. (a) What is the velocity of the small ball immediately after its collision with the large ball? (b) From the answer to part (a), what is the ratio of the small ball's rebound distance to the distance it fell before the collision?

Changing Mass. An open-topped freight car with mass \(24,000 \mathrm{kg}\) is coasting without friction along a level track. It is raining very hard, and the rain is falling vertically downward. Originally, the car is empty and moving with a speed of 4.00 \(\mathrm{m} / \mathrm{s}\).(a) What is the speed of the car after it has collected 3000 \(\mathrm{kg}\) of rain- water? (b) since the rain is falling downward, how is it able to affect the horizontal motion of the car?

A Ricocheting Bullet. 0.100 -kg stone rests on a frictionless, horizontal surface. A bullet of mass 6.00 g, traveling horizontally at 350 \(\mathrm{m} / \mathrm{s}\) , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 \(\mathrm{m} / \mathrm{s}\) . (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?
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