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

Short Answer

Expert verified
The x and y components of the impulse are \(-1.14 N.s\) and \(0.33 N.s\) respectively. The final x and y components of the velocity are \(-20 m/s\) and \(2 m/s\) respectively.

Step by step solution

01

Calculate the Impulse

Impulse is given by the product of the force and the time it acts on, thus use the formula Impulse (J) = Force (F) x Time (t). For 'x' component, the force is \(-380 N\) and time is \(3.0 ms = 3 x 10^{-3} s\), therefore Impulse on 'x', \(Jx = -380 N * 3 x 10^{-3} s = -1.14 N.s\). For 'y' component, the force is \(110 N\) and time remains the same, therefore Impulse on 'y', \(Jy = 110 N * 3 x 10^{-3} s = 0.33 N.s\).
02

Calculate the Final Velocity

The change in momentum equals the impulse applied, and momentum is the product of mass and velocity. Thus, Impulse = mass x change in velocity. We rearrange this formula to find the final velocities: change in velocity = Impulse / mass. The weight of the ball provided (0.56 N) is converted to mass by dividing by acceleration due to gravity (9.8 m/s^2). So, mass 'm' = 0.56 N / 9.8 m/s^2 = 0.057 kg. For x-component, initial velocity \(Vxi = 20 m/s\), final velocity \(Vxf = Vxi + Jx / m = 20 m/s - 1.14 N.s / 0.057 kg = -20 m/s\). For y-component, initial velocity \(Vyi = -4 m/s\), final velocity \(Vyf = Vyi + Jy / m = -4 m/s + 0.33 N.s / 0.057 kg = 2 m/s\).

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Key Concepts

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

Impulse Calculation
The concept of impulse is crucial when analyzing situations involving forces and motion over short periods. Impulse measures the impact of a force applied over time. Calculating impulse involves the formula:
  • Impulse (\( J \)) = Force (\( F \)) \( \times \) Time (\( t \))
In this exercise, we have forces provided in two perpendicular directions, the x- and y-components. For the x-component, we use the force of \(-380 \text{ N}\) and a time of \(3.00 \text{ ms} = 3 \times 10^{-3} \text{ s}\). Thus, the impulse in the x-direction, \(J_x\), is calculated as:
  • \( J_x = -380 \text{ N} \times 3 \times 10^{-3} \text{ s} = -1.14 \text{ Ns}\)
In the y-direction, the force is \(110 \text{ N}\), maintaining the same time period. Hence, the impulse, \(J_y\), is:
  • \( J_y = 110 \text{ N} \times 3 \times 10^{-3} \text{ s} = 0.33 \text{ Ns}\)
Impulse provides a direct measure of how a force changes the momentum of an object. This concept ties directly into momentum through Newton's laws of motion.
Final Velocity Determination
Determining the final velocity requires understanding the relationship between impulse and change in momentum. Momentum is the product of mass and velocity, and changes in momentum relate to the impulse:
  • Change in momentum = Impulse
Given that mass can be expressed through weight and gravitational acceleration, i.e., mass (\( m \)) = weight / \( 9.8 \text{ m/s}^2 \), the mass of the tennis ball is \( 0.056 \text{ kg} (\text{from}\, 0.560 \text{ N})\).For the x-component, starting with an initial velocity \( V_{xi} = 20 \text{ m/s}\):
  • Final velocity: \( V_{xf} = V_{xi} + \frac{J_x}{m} = 20 \text{ m/s} - \frac{1.14 \text{ Ns}}{0.056 \text{ kg}} = -20 \text{ m/s}\)
For the y-component, initial velocity is \( V_{yi} = -4 \text{ m/s}\):
  • Final velocity: \( V_{yf} = V_{yi} + \frac{J_y}{m} = -4 \text{ m/s} + \frac{0.33 \text{ Ns}}{0.056 \text{ kg}} = 2 \text{ m/s}\)
The final velocities reflect how external force changes the trajectory and speed of the ball in both directions.
Force and Motion
In physics, motion is often defined by forces, and the relationship between these is outlined by Newton's second law of motion. This core law explains how a net force acting on an object results in acceleration, affecting its velocity and position over time. Here, the ball experiences a net force during its contact with the racket, influencing its speed.The exercise shows that the force into components states how differing directions of force can distinctly change motion. The understanding of motion affected by force is:
  • Force (\( F \)) in x- and y-direction affects motion based on \( F = ma \) (where \( a \) is acceleration).
The constant forces provided alter the ball’s speed and direction:
  • In x-direction: \(-380 \text{ N}\) pulls back, \( \text{updating direction}\).
  • In y-direction: \(110 \text{ N}\) directs upwards, affecting elevation.
The application and effect of net external forces on an object's motion can be seen practically in sports like tennis, for predictive motion analysis.
Component Vectors
Understanding component vectors is fundamental in physics when dealing with multidimensional motion. Vectors have both magnitude and direction, making them applicable in describing forces and velocities in distinct directions. In this problem: The tennis ball's velocity and forces applied are split into x and y components:
  • Initial Velocity Vector: \((20 \hat{i}) - (4 \hat{j})\,\text{m/s}\)
  • Net External Forces:\(-380 \hat{i} + 110 \hat{j}\,\text{N}\)
The importance of components lies in:
  • Having clarity on how each separate force impacts direction.
  • Ability to handle vector addition or subtraction, thus solving for final states like velocity.
Grasping the separation of vectors into components simplifies the solution of many physics problems, ensuring precise calculation and understanding of multi-directional forces.
Newton's Laws of Motion
Newton's Laws of Motion are the foundation for understanding mechanics. They describe the relationship between an object's motion and the forces acting upon it. Primarily:- **First Law**: An object in motion stays in motion unless acted upon by an external force.- **Second Law**: The net force on an object is equal to the mass of the object multiplied by its acceleration (\( F = ma \)).- **Third Law**: For every action, there is an equal and opposite reaction.In this exercise, the Second Law is pivotal. It relates how the constant net force during the ball's contact period results in acceleration, directly influencing the change in velocity:
  • Provides a predictive understanding of force implications on the tennis ball's motion.
  • Gives insight into explaining how speed decreases or increases based on direction and magnitude of force.
Using these laws simplifies analyzing everyday phenomena such as playing tennis, where forces cater to motion patterns and provide a deeper insight into the mechanics involved.

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Most popular questions from this chapter

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.

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