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Chapter 9: Problem 29

A ball of mass \(1.00 \mathrm{~kg}\) is attached to a loose string fixed to a ceiling. The ball is released from rest and falls \(2.00 \mathrm{~m}\), where the string suddenly stops it. Find the impulse on it from the string.

Short Answer

Expert verified
The impulse on the ball from the string is 6.26 kg路m/s upward.

Step by step solution

01

Understanding the Problem

The ball is initially at rest and falls freely under gravity until the string stops it. We need to calculate the impulse, which is the change in momentum of the ball when it stops.
02

Calculate Final Velocity Before Stopping

Since the ball falls a height of 2.00 meters, we can find the final velocity using the equation for free fall: \[ v^2 = u^2 + 2gh \] where \( v \) is the final velocity, \( u = 0 \) since it starts from rest, \( g = 9.81 \mathrm{~m/s^2} \) is the acceleration due to gravity, and \( h = 2.00 \mathrm{~m} \) is the height. Solving for \( v \), we get:\[ v = \sqrt{0 + 2 \times 9.81 \times 2} = \sqrt{39.24} \approx 6.26 \mathrm{~m/s} \]
03

Determine Initial and Final Momentum

The initial momentum \( p_i \) of the ball when released is \( 0 \) because it starts from rest. The final momentum \( p_f \) just before the string stops it is given by:\[ p_f = mv = 1.00 \times 6.26 = 6.26 \mathrm{~kg \cdot m/s} \]
04

Calculate Impulse from the String

Impulse \( J \) is the change in momentum, calculated as:\[ J = p_f - p_i = 6.26 - 0 = 6.26 \mathrm{~kg \cdot m/s} \]
05

Direction of Impulse

Since the ball was moving downward and is stopped by the string, the direction of the impulse is upward, opposite to the ball's 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 which describes the quantity of motion an object possesses. It is defined as the product of an object's mass and its velocity:
  • Momentum formula: \[ p = mv \] where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
  • Momentum is a vector quantity, meaning it has both magnitude and direction.
  • A higher momentum means an object is harder to stop.
In the case of the falling ball, its momentum changes as it falls. Initially, the momentum is zero because the ball is at rest. As it gains speed due to gravity, its momentum increases until it is stopped by the string. The impulse, which is the change in momentum, is a critical part of this problem, representing the force applied by the string to bring the ball to a stop.
Free Fall
Free fall refers to the motion of an object under the influence of gravitational force only. During free fall, the only force acting on the object is gravity, resulting in a uniform acceleration. Here are some key points:
  • In free fall, the initial velocity \( u \) is often zero, as the object starts from rest.
  • The acceleration due to gravity \( g \) is constant, generally \( 9.81 \text{ m/s}^2 \) on Earth.
  • The velocity of the object increases as it falls.
For the ball in the exercise, it falls freely for \( 2.00 \) meters, accelerating due to gravity. The velocity it reaches just before the string stops it (calculated as \( 6.26 \text{ m/s} \)) is crucial for determining its change in motion, or impulse.
Acceleration due to Gravity
The acceleration due to gravity is the rate at which any object speeds up as it falls toward the Earth's surface. On Earth, this acceleration is approximately \( 9.81 \text{ m/s}^2 \). Here's what you need to know:
  • This acceleration is constant for objects in free fall, meaning it remains the same regardless of the object's mass or the distance fallen.
  • The formula used to calculate final velocity in a free-falling object is: \[ v^2 = u^2 + 2gh \] where \( v \) is final velocity, \( u \) is initial velocity, \( g \) is acceleration due to gravity, and \( h \) is the height fallen.
  • Understanding this concept allows us to predict how quickly an object will gain speed as it falls under Earth's gravitational pull.
In the original problem, knowing \( g \) enabled us to calculate the ball's final velocity before the string intervened, critical for finding the ball's impulse.

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Most popular questions from this chapter

Block 1 of mass \(m_{1}\) slides along a frictionless floor and into a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=3 m_{1}\). Prior to the collision, the center of mass of the twoblock system had a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Afterward, what are the speeds of (a) the center of mass and (b) block \(2 ?\)

A \(15.0 \mathrm{~kg}\) package is moving at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\) vertically upward along a \(y\) axis when it explodes into three fragments: a \(2.00\) \(\mathrm{kg}\) fragment is shot upward with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\) and a \(3.00 \mathrm{~kg}\) fragment is shot in the positive direction of a horizontal \(x\) axis with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\). Find (a) the speed of the third fragment right after the explosion and (b) the total kinetic energy provided by the explosion.

After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 \(\mathrm{m}\). During the collision at the bottom of the elevator shaft, a \(90 \mathrm{~kg}\) passenger is stopped in \(5.0 \mathrm{~ms}\). (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

Two bodies of masses \(m=0.30 \mathrm{~kg}\) and \(2 m\) are connected by a long string of negligible mass. The string is looped over a pulley and, with the string taut, the bodies are released at time \(t=0\) so that the heavier one descends and the lighter one ascends. At time \(t=4.0 \mathrm{~s}\), the lighter one undergoes a fully inelastic collision with a third body of mass \(m\). Because the first two bodies move in rigid fashion, the collision is effectively between the third body and the system of the first two bodies. (a) Just after the collision, what is the speed of the three bodies? (b) By how much was the kinetic energy of the descending body decreased because of the collision?

A rocket that is set for a vertical launch has a mass of \(50.0 \mathrm{~kg}\) and contains \(450 \mathrm{~kg}\) of fuel. The rocket can have a maximum exhaust velocity of \(2.00 \mathrm{~km} / \mathrm{s}\). What should be the minimum rate of fuel consumption (a) to just lift it off the launching pad and (b) to give it an acceleration of \(20.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) If the consumption rate is set at \(10.0 \mathrm{~kg} / \mathrm{s}\), what is the rocket speed at the moment when the fuel is fully consumed?
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