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

Short Answer

Expert verified
The impulse components are \( -1.14 \text{ Ns}, 0.33 \text{ Ns} \). The final velocity components are \( 0.03 \text{ m/s}, 1.78 \text{ m/s} \).

Step by step solution

01

Understand the Impulse Formula

The impulse applied to an object is given by the product of the net force and the time interval over which the force acts. It is also equal to the change in momentum of the object. The formula is: \[ \vec{I} = \vec{F} \times \Delta t \] where \( \vec{I} \) is the impulse, \( \vec{F} \) is the net force, and \( \Delta t \) is the time interval.
02

Calculate Impulse in x-component

Given the constant net force \( \vec{F} = -(380 N)\hat{\imath} + (110 N)\hat{\jmath} \) and the time of contact \( \Delta t = 3.00 \times 10^{-3} \text{s} \), calculate the impulse in the \( x \)-direction. \[ I_x = F_x \times \Delta t = (-380 N) \times 3.00 \times 10^{-3} \text{s} = -1.14 \text{ Ns} \]
03

Calculate Impulse in y-component

Calculate the impulse in the \( y \)-direction using the \( y \) component of the force: \[ I_y = F_y \times \Delta t = (110 N) \times 3.00 \times 10^{-3} \text{s} = 0.33 \text{ Ns} \]
04

Analyze Change in Momentum

The impulse is equal to the change in momentum \( \Delta \vec{p} \): \[ \Delta \vec{p} = \vec{I} \]Since impulse is the change in momentum, we can express this as:\[ m\Delta\vec{v} = \vec{I} \]where \( m \) is the mass of the ball and \( \Delta\vec{v} \) is the change in velocity. It is critical to express the mass of the ball as the weight divided by the acceleration due to gravity: \[ m = \frac{0.560 \text{ N}}{9.81 \text{ m/s}^2} \approx 0.0571 \text{ kg} \]
05

Calculate Change in Velocity \(x\)-component

Use the impulse-momentum relation in the \( x \)-direction: \[ I_x = m \Delta v_x \Rightarrow \Delta v_x = \frac{I_x}{m} = \frac{-1.14 \text{ Ns}}{0.0571 \text{ kg}} \approx -19.97 \text{ m/s} \] Next, find the final \( x \)-component of velocity:\[ v_{fx} = v_{ix} + \Delta v_x = 20.0 \text{ m/s} + (-19.97 \text{ m/s}) \approx 0.03 \text{ m/s} \]
06

Calculate Change in Velocity \(y\)-component

Apply the same logic to the \( y \)-direction: \[ I_y = m \Delta v_y \Rightarrow \Delta v_y = \frac{I_y}{m} = \frac{0.33 \text{ Ns}}{0.0571 \text{ kg}} \approx 5.78 \text{ m/s} \] Then find the final \( y \)-component of velocity:\[ v_{fy} = v_{iy} + \Delta v_y = (-4.0 \text{ m/s}) + 5.78 \text{ m/s} \approx 1.78 \text{ m/s} \]

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Key Concepts

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

Impulse Formula
Impulse is a pivotal concept when studying dynamics, particularly when it comes to how forces influence the motion of objects. Essentially, impulse quantifies the effect of a force acting over a period. The formula for impulse is expressed as \( \vec{I} = \vec{F} \times \Delta t \), where:
  • \( \vec{I} \) represents impulse, a vector quantity that reflects how the force changes the object's momentum.
  • \( \vec{F} \) is the net force applied to the object.
  • \( \Delta t \) signifies the time duration for which the force is applied.
By merging force and time, impulse provides the magnitude and direction of the change in momentum. It's crucial to recognize that while force might be large, a minimal time span reduces the overall impulse, potentially leading to small changes in velocity. Conversely, prolonged application of even a small force can result in significant motion changes. Understanding the impulse formula is critical for calculating how external forces can alter the motion of objects in a specified direction.
Net Force
Net force is the total force acting on an object, resulting from the combination of all individual forces. It plays a crucial role in determining how an object's motion changes. In this context, the net force acting on the tennis ball has two components: \( -(380 N)\hat{\imath} \) in the x-direction and \( (110 N)\hat{\jmath} \) in the y-direction.
This concept essentially sums up all forces to give a single resultant force that influences the object's state of motion.
  • If the net force is non-zero, the object experiences an acceleration, reflecting a change in its speed or direction.
  • Alternatively, a net force of zero means the object maintains its present state of motion, be it rest or uniform motion.
Net force influences the object's momentum, and hence tying it directly to the impulse experienced due to the force applied over a given duration.
Change in Velocity
Change in velocity is a direct consequence of the impulses exerted on an object, as dictated by the net force and time. This results in an alteration of the object's speed or direction. The relationship between impulse and this change is mathematically described using impulse-momentum theorem: \( m\Delta\vec{v} = \vec{I} \) where:
  • \( m \) is the mass of the object, calculated by dividing the weight of the tennis ball (0.560 N) by gravity (9.81 m/s²), yielding approximately 0.0571 kg.
  • \( \Delta\vec{v} \) is the change in velocity, derived from calculating the impulse in both x and y components and dividing by the mass.
Through changes in the net force or its application duration, velocity components along different axes of the object can be distinctly altered.
For the tennis ball, as calculated, a change of \( -19.97 \text{ m/s} \) in the x-direction and \( 5.78 \text{ m/s} \) in the y-direction occurs. Understanding how velocity changes underpins much of classical mechanics by describing the motion resulting from external forces applied to an object over time.

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Most popular questions from this chapter

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring \((\textbf{Fig. P8.79})\). The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

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