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A \(0.70 \mathrm{~kg}\) ball moving horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) strikes a vertical wall and rebounds with speed \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in its linear momentum?

Short Answer

Expert verified
The magnitude of the change in momentum is 4.9 kgâ‹…m/s.

Step by step solution

01

Calculate Initial Momentum

The initial momentum of the ball can be calculated using the formula for linear momentum, \( p = m imes v \), where \( m \) is mass and \( v \) is velocity. The initial momentum is \( 0.70 \, \text{kg} \times 5.0 \, \text{m/s} = 3.5 \, \text{kg} \cdot \text{m/s} \).
02

Calculate Final Momentum

The final momentum of the ball, after it rebounds, is calculated using the same formula but with the rebound speed. The final momentum is \( 0.70 \, \text{kg} \times 2.0 \, \text{m/s} = 1.4 \, \text{kg} \cdot \text{m/s} \).
03

Determine Change in Momentum

The change in momentum is given by the difference between final and initial momentum. Because the ball reverses direction, the final velocity is considered negative for the calculation. So the change in momentum is \( -1.4 \, \text{kg} \cdot \text{m/s} - 3.5 \, \text{kg} \cdot \text{m/s} = -4.9 \, \text{kg} \cdot \text{m/s} \).
04

Calculate Magnitude of Change in Momentum

The magnitude of the change in momentum, which is always positive, is \( \left| -4.9 \, \text{kg} \cdot \text{m/s} \right| = 4.9 \, \text{kg} \cdot \text{m/s} \).

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Key Concepts

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

Impulse
Impulse is a core concept in motion dynamics, closely related to momentum. It describes how the force acting over a specific time interval can change an object's momentum. This is expressed mathematically as \[ \text{Impulse} = F \times \Delta t \]where - \( F \) is the force applied, and - \( \Delta t \) is the time duration during which the force acts.
The exciting part about impulse is that it doesn't just depend on the force applied but also on the time the force acts. For instance, a small force applied over a long time can have the same impulse as a large force applied briefly. What's important to note is that impulse has the same unit as momentum, \( \text{kg} \cdot \text{m/s} \). This makes sense because impulse is essentially changing the momentum of an object.
In the exercise presented, impulse would be the effect of the force exerted by the wall on the ball, causing it to reverse its direction and change its velocity. The force that the wall applies during the brief collision with the ball gives the ball a new velocity, effectively changing its momentum.
Conservation of Momentum
The conservation of momentum is a fundamental principle of physics that states the total momentum of a closed system remains constant if no external forces are acting on it. Let's put it simply: in the absence of external forces, the momentum before an event will equal the momentum after.
In our exercise, the concept applies well. Before the collision, the ball was moving with an initial momentum of \( 3.5 \, \text{kg} \cdot \text{m/s} \). After the collision, its momentum was \( -1.4 \, \text{kg} \cdot \text{m/s} \). No external forces besides the wall's force act horizontally, so the momentum is conserved. The difference in the momentum values before and after ensures the effectiveness of conservation.
  • This principle is pivotal in collisions, explosions, and similar interactions.
  • It helps explain phenomena where multiple objects interact, and energy might seem lost, but momentum remains constant.
Understanding the conservation of momentum allows us to solve for unknown factors post-collision, such as speeds or mass.
Collision Dynamics
The dynamics of collisions provide insight into how objects interact upon impact. These dynamics can tell us whether a collision is elastic or inelastic, and they greatly affect the final velocities and energy states of the involved objects.
In collisions, momentum is always conserved, but kinetic energy might not be. An elastic collision is one where both momentum and kinetic energy are conserved. Meanwhile, an inelastic collision only conserves momentum, meaning some kinetic energy is converted into other forms, like sound or heat.
For our exercise scenario: - The collision between the ball and the wall is likely inelastic since the ball does not bounce back to its original speed. - Although momentum is conserved, some kinetic energy is lost due to sound, heat, or deformation.
  • Analyzing collision dynamics is essential for understanding real-world events like car crashes or sports impacts.
  • By studying these dynamics, we can implement safety measures and improve materials to better withstand impacts.
Through studying collision dynamics, we gain valuable insights into the interaction of forces and movements in various environments.

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Most popular questions from this chapter

A \(140 \mathrm{~g}\) ball with speed \(7.8 \mathrm{~m} / \mathrm{s}\) strikes a wall perpendicularly and rebounds in the opposite direction with the same speed. The collision lasts \(3.80 \mathrm{~ms}\). What are the magnitudes of the (a) impulse and (b) average force on the wall from the ball during the elastic collision?

A completely inclastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass \(3.0 \mathrm{~kg},\) is moving upward at \(20 \mathrm{~m} / \mathrm{s}\) and the other ball, of mass \(2.0 \mathrm{~kg}\), is moving downward at \(12 \mathrm{~m} / \mathrm{s}\). How high do the combined two balls of putty risc above the collision point? (Ncglect air drag.)

A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\). of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The rocket engine is then fired for \(250 \mathrm{~s}\) while fuel is consumed at the rate of \(480 \mathrm{~kg} / \mathrm{s}\). The speed of the exhaust products relative to the rocket is \(3.27 \mathrm{~km} / \mathrm{s}\). (a) What is the rocket's thrust? After the \(250 \mathrm{~s}\) firing, what are (b) the mass and (c) the speed of the rocket?

block 1 of mass \(m_{1}\) slides from rest along a frictionless ramp from height \(h=2.50 \mathrm{~m}\) and then collides with stationary block \(2,\) which has mass \(m_{2}=2.00 m_{1}\). After the collision, block 2 slides into a region where the coefficient of kinetic friction \(\mu_{k}\) is 0.500 and comes to a stop in distance \(d\) within that region. What is the value of distance \(d\) if the collision is (a) elastic and (b) completely inelastic?

A space vehicle is traveling at \(4300 \mathrm{~km} / \mathrm{h}\) relative to Earth when the exhausted rocket motor (mass \(4 m)\) is disengaged and sent backward with a speed of \(82 \mathrm{~km} / \mathrm{h}\) relative to the command module (mass \(m\) ). What is the speed of the command module relative to Earth just after the separation?

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