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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 \(-42 \, \mathrm{Ns}\); average force is \(2100 \, \mathrm{N}\).

Step by step solution

01

Understand the Impulse-Momentum Theorem

The impulse (J) on an object is equal to the change in momentum (Δp) of the object. The momentum of an object is given by the product of its mass (m) and its velocity (v). Mathematically, it is expressed as:\[ J = Δp = m(v_{f} - v_{i}) \]where \( v_{f} \) is the final velocity and \( v_{i} \) is the initial velocity.
02

Identify Given Data and Write Equations

For the falling ball, the mass (m) is \( 1.2 \mathrm{~kg} \). The initial velocity \( v_{i} \) (as it strikes the floor) is \( 25 \mathrm{~m/s} \). As it rebounds, the final velocity \( v_{f} \) is \( -10 \mathrm{~m/s} \) (it goes in the opposite direction to the initial velocity). Substitute these values into the impulse equation:\[ J = 1.2 ( -10 - 25 ) \]
03

Calculate the Impulse

Substitute the values into the equation to find the impulse:\[ J = 1.2 \times (-35) = -42 \, \mathrm{Ns} \]The impulse is \(-42 \, \mathrm{Ns}\) indicating a direction opposite to the initial motion.
04

Use Impulse to Determine Average Force

The impulse is also the product of the average force (F) exerted during the contact time (Δt). Thus:\[ J = F \cdot Δt \] Given that \( Δt = 0.020 \mathrm{~s} \), we can solve for \( F \) by rearranging the equation:\[ F = \frac{J}{Δt} = \frac{-42}{0.020} \]
05

Calculate the Average Force

Plug in the impulse and time of contact to find the average force:\[ F = \frac{-42}{0.020} = -2100 \, \mathrm{N} \]The magnitude of the force is \( 2100 \, \mathrm{N} \). This large negative force means the direction is opposite to the initial direction of motion.

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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 that describes the motion of an object. It is the product of an object's mass and velocity. The formula for momentum (\(p\)) is expressed as:
  • \(p = m \cdot v\)
Here, \(m\) stands for mass and \(v\) is velocity. For example, in the case of the ball, its momentum before hitting the floor is its mass times its velocity: \(1.2 \, \text{kg} \times 25 \, \text{m/s}\).

Momentum is a vector quantity, which means it has both magnitude and direction. When an object changes direction, as the ball does when it rebounds, its momentum also changes. This ability to quantify both mass and velocity simultaneously helps quantify how hard it is to stop a moving object. High momentum objects, like a speeding car, need a lot more force to be stopped than low momentum objects, like a slowly rolling tennis ball.
Impulse
Impulse is a concept closely linked with momentum. It's the change in momentum of an object when a force is applied over a certain time interval. The Impulse-Momentum Theorem provides the relation, stating that:
  • Impulse (\(J\)) = Change in momentum (\(Δp\))
  • \(J = m(v_f - v_i)\)
Where \(v_f\) is the final velocity and \(v_i\) is the initial velocity. For the ball in the exercise, the impulse is calculated by multiplying its mass by the difference in its final and initial velocities, which results in \(-42 \, \text{Ns}\).

The negative impulse indicates that the force direction is opposite to the initial motion (the downward direction of the ball). Impulse helps us understand how forces change the motion of objects; it is not merely about having a force, but how long that force is applied.
Average Force
The average force is related to impulse and provides insight into the strength of the force applied during the motion change. It can be described mathematically by the formula:
  • \(F = \frac{J}{Δt}\)
Where \(J\) is the impulse, and \(Δt\) is the time duration. In the exercise, the impulse is \(-42 \, \text{Ns}\), and the contact time is \(0.020 \, \text{s}\).

By substituting these into the equation, we find the average force to be \(-2100 \, \text{N}\). The negative sign shows the force acts in the opposite direction to the initial motion. This large force over a short time period reflects a quick and significant change in the motion of the ball. Understanding average force is crucial in scenarios like car crashes or sports, where we need to minimize or maximize force impacts effectively.

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Most popular questions from this chapter

A man (weighing \(915 \mathrm{~N}\) ) stands on a long railroad flatcar (weighing \(2415 \mathrm{~N}\) ) as it rolls at \(18.2 \mathrm{~m} / \mathrm{s}\) in the positive direction of an \(x\) axis, with negligible friction. Then the man runs along the flatcar in the negative \(x\) direction at \(4.00 \mathrm{~m} / \mathrm{s}\) relative to the flatcar. What is the resulting increase in the speed of the flatcar?

A stone is dropped at \(t=0\). A second stone, with twice the mass of the first, is dropped from the same point at \(t=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t=300 \mathrm{~ms} ?\) (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

A railroad freight car of mass \(3.18 \times 10^{4} \mathrm{~kg}\) collides with a stationary caboose car. They couple together, and \(27.0 \%\) of the initial kinetic energy is transferred to thermal energy, sound, vibrations, and so on. Find the mass of the caboose.

ILW In Fig. \(9-63\), block 1 (mass \(2.0 \mathrm{~kg}\) ) is moving rightward at \(10 \mathrm{~m} / \mathrm{s}\) and block \(2(\mathrm{mass} 5.0 \mathrm{~kg})\) is moving rightward at \(3.0 \mathrm{~m} / \mathrm{s}\). The surface is frictionless, and a spring with a spring constant of \(1120 \mathrm{~N} / \mathrm{m}\) is fixed to block \(2 .\) When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

A \(0.15 \mathrm{~kg}\) ball hits a wall with a velocity of \((5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(6.50\) \(\mathrm{m} / \mathrm{s}) \hat{\mathrm{j}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). It rebounds from the wall with a velocity of \((2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(-3.20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). What are (a) the change in the ball's momentum, (b) the impulse on the ball, and (c) the impulse on the wall?

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