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Chapter 9: Problem 25

A \(1.2 \mathrm{~kg}\) ball drops vertically onto a floor, hitting with a speed of \(25 \mathrm{~m} / \mathrm{s} .\) It rebounds with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\). (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for \(0.020 \mathrm{~s}\), what is the magnitude of the average force on the floor from the ball?

Short Answer

Expert verified
Impulse is 42 kg m/s; average force is 2100 N upward.

Step by step solution

01

Understand Impulse

Impulse is the change in momentum of an object. When the ball hits the ground and rebounds, its momentum changes.
02

Calculate Initial and Final Momenta

The initial momentum of the ball before hitting the ground is given by \( p_i = m \cdot v_i \), where \( m = 1.2 \, \text{kg} \) and \( v_i = 25 \, \text{m/s} \). Hence, \( p_i = 1.2 \times 25 = 30 \, \text{kg m/s} \, \text{(downwards)} \). The final momentum after rebounding is \( p_f = m \cdot v_f \), with \( v_f = -10 \, \text{m/s} \) (since the direction is opposite to the initial). Hence, \( p_f = 1.2 \times -10 = -12 \, \text{kg m/s} \).
03

Calculate the Impulse

Impulse \( J \) is the change in momentum: \( J = p_f - p_i \). Substituting the values: \( J = -12 - 30 = -42 \, \text{kg m/s} \). The impulse is 42 kg m/s upwards (since impulse changes the direction of motion).
04

Calculate Average Force

Average force can be calculated using the formula: \( J = F_{avg} \times \Delta t \). Rearrange to find \( F_{avg} \): \( F_{avg} = \frac{J}{\Delta t} \), where \( \Delta t = 0.020 \, \text{s} \). Substituting the values \( F_{avg} = \frac{42}{0.020} = 2100 \, \text{N} \). The force on the floor is 2100 N directed upwards.

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Key Concepts

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

Momentum
In the realm of physics, momentum is a key concept often described as the "oomph" of a moving object. It is the product of an object's mass and its velocity: \( p = m \cdot v \). Momentum carries both magnitude and direction, making it a vector quantity. Consider our example of a ball dropping onto the floor. When the ball moves downward toward the ground, it carries a certain amount of momentum based on its mass and speed. As it hits and rebounds from the floor, its speed鈥攁nd hence its momentum鈥攃hanges.
To understand this interaction better:
  • Initial momentum: This is calculated just before the ball makes contact with the floor, using its initial speed.
  • Final momentum: This takes into account the ball's speed just after rebounding, which is in the opposite direction, hence carrying a negative sign.
Tracking these changes helps in calculating the impulse, or the change in momentum, crucial to analyzing the effect of forces during contact.
Average Force
Calculating average force requires an understanding of impulse, which is the total change in momentum experienced over time. In our scenario, when the ball hits and bounces back off the floor, it exerts a force throughout its contact duration. The average force, which is consistent over this short time, can be determined from the impulse-momentum theorem.
Here's the breakdown:
  • Impulse-momentum theorem: This states that the impulse is equal to the change in momentum (\( J = \Delta p \)) and also equals the average force times the time duration it acts (\( J = F_{avg} \times \Delta t \)).
  • Short contact time: The ball is in contact with the floor for a mere 0.020 seconds. This brief interaction means that even a small change in momentum results in a significant average force.
Understanding this will provide insights into how forces work over time, especially in situations involving fast interactions.
Kinematics
Kinematics focuses on motion without considering the forces causing it. It's a vital concept when analyzing how the ball moves before and after interacting with the floor. By using kinematic equations, one can relate variables such as initial and final speeds, acceleration, and time without diving into the complexities of force.
  • Speed changes: In our example, kinematics helps us identify the change from an initial downward speed of 25 m/s to an upward speed of 10 m/s after the floor impact.
  • Direction matters: Kinematically, we observe that a switch in direction signifies a significant shift in momentum, requiring a calculated impulse to explain the turnaround.
Through kinematics, we connect these speed changes with the overall motion, painting a complete picture of how objects move and interact in real-world scenarios.

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Most popular questions from this chapter

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A}=(15 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) and \(\vec{v}_{B}=(-10 \hat{\mathrm{i}}+5.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\). After the collision, \(\vec{v}_{A}^{\prime}=(-5.0 \hat{1}+20 \hat{j}) \mathrm{m} / \mathrm{s} .\) What are (a) the final velocity of \(B\) and (b) the change in the total kinetic energy (including sign)?

A \(1400 \mathrm{~kg}\) car moving at \(5.3 \mathrm{~m} / \mathrm{s}\) is initially traveling north along the positive direction of a \(y\) axis. After completing a \(90^{\circ}\) right-hand turn in \(4.6 \mathrm{~s}\), the inattentive operator drives into a tree, which stops the car in \(350 \mathrm{~ms}\). In unit-vector notation, what is the impulse on the car (a) due to the turn and (b) due to the collision? What is the magnitude of the average force that acts on the car (c) during the turn and (d) during the collision? (e) What is the direction of the average force during the turn?

A cue stick strikes a stationary pool ball, with an average force of \(32 \mathrm{~N}\) over a time of \(14 \mathrm{~ms}\). If the ball has mass \(0.20 \mathrm{~kg}\), what speed does it have just after impact?

Figure 9-51 shows a 0.300 \(\mathrm{kg}\) baseball just before and just after it collides with a bat. Just before, the ball has velocity \(\vec{v}_{1}\) of magnitude \(12.0 \mathrm{~m} / \mathrm{s}\) and angle \(\theta_{1}=35.0^{\circ}\). Just after, it is traveling directly upward with velocity \(\vec{v}_{2}\) of magnitude \(10.0\) \(\mathrm{m} / \mathrm{s}\). The duration of the collision is \(2.00 \mathrm{~ms}\). What are the (a) magnitude and (b) direction (relative to the positive direction of the \(x\) axis) of the impulse on the ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from the bat?

During a lunar mission, it is necessary to increase the speed of a spacecraft by \(2.2 \mathrm{~m} / \mathrm{s}\) when it is moving at \(400 \mathrm{~m} / \mathrm{s}\) relative to the Moon. The speed of the exhaust products from the rocket engine is \(1000 \mathrm{~m} / \mathrm{s}\) relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?
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