/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 \(\cdot\) You are standing on a ... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\cdot\) You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.400 \(\mathrm{kg}\) ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s} .\) Your mass is 70.0 \(\mathrm{kg} .\) (a) If you catch the ball, with what speed do you and the ball move afterwards? (b) If the ball hits you and bounces off your chest, so that afterwards it is moving horizontally at 8.00 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?

Short Answer

Expert verified
(a) 0.0568 m/s, (b) 0.1029 m/s.

Step by step solution

01

Set Up the Conservation of Momentum for Catching the Ball

When you catch the ball, the principle of conservation of momentum states that the total momentum before catching is equal to the total momentum after catching. The momentum is calculated as the mass multiplied by the velocity. Since only the ball is moving initially, we have:\[ m_b \cdot v_b = (m_b + m_p) \cdot v_f \]where \( m_b \) is the mass of the ball, \( v_b \) is the initial velocity of the ball, \( m_p \) is your mass, and \( v_f \) is the final velocity when both you and the ball are moving together.
02

Substitute Values and Solve for Final Velocity (a)

Substitute the known values into the momentum equation:\[ 0.400 \cdot 10.0 = (0.400 + 70.0) \cdot v_f \]Simplify and solve for \( v_f \):\[ 4.0 = 70.4 \cdot v_f \]\[ v_f = \frac{4.0}{70.4} \approx 0.0568 \text{ m/s} \]Thus, the speed after catching the ball is approximately 0.0568 m/s.
03

Set Up the Conservation of Momentum for Ball Bouncing Off

For the scenario where the ball bounces off you, apply the conservation of momentum principle again. The initial momentum is the same as before, but now the ball's final momentum must be subtracted:\[ m_b \cdot v_b = m_p \cdot v_f + m_b \cdot v_{b'} \]where \( v_{b'} \) is the new velocity of the ball after bouncing.
04

Substitute Values and Solve for Final Velocity (b)

Substitute the known values into the momentum equation:\[ 0.400 \cdot 10.0 = 70.0 \cdot v_f + 0.400 \cdot (-8.0) \]This simplifies to:\[ 4.0 = 70.0 \cdot v_f - 3.2 \]\[ 4.0 + 3.2 = 70.0 \cdot v_f \]\[ 7.2 = 70.0 \cdot v_f \]\[ v_f = \frac{7.2}{70.0} \approx 0.1029 \text{ m/s} \]So, your speed after the ball bounces off is approximately 0.1029 m/s.

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Key Concepts

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

Understanding Collisions
Collision refers to the event where two or more objects come into contact in such a way that momentum and sometimes energy are transferred between them. It is more common in physics to deal with perfectly elastic or inelastic collisions. When solving problems involving collisions, what we focus on is how objects interact by either sticking together or bouncing off each other.

The core principle in collision problems is the conservation of momentum. Regardless of the nature of the collision, the total momentum of the system before the collision is equal to the total momentum after the collision. This is a powerful principle because it helps us find unknown speeds and movements of the objects involved.
  • Elastic collisions: objects bounce off with no loss of kinetic energy.
  • Inelastic collisions: objects may stick together, and kinetic energy is not conserved.
In the given exercise, the ball could either be caught (a scenario of inelastic collision) or it bounces off (an elastic collision). Understanding these nuances helps in approaching problems systematically and figuring out the velocities involved.
Solving Physics Problems
Physics problems, especially those involving collisions, can be tricky. However, with systematic steps, they become manageable. Start by understanding the problem setup and identifying known and unknown variables. For instance, take note of masses and initial velocities of the objects involved, as these play a crucial role in momentum calculations.

Once the setup is clear, apply relevant physics equations like the conservation of momentum. This involves equating the initial and final momentum to solve for unknown variables such as final velocity. Create clear momentum equations to depict the scenario correctly.
  • Formulate momentum conservation equations based on problem requirements.
  • Substitute known values into the equations for solving.
Working through each step and double-checking your calculations ensures precision and enhances understanding, contributing to a better grasp of physics as a whole.
Velocity Calculations In Detail
Calculating velocity, especially after collisions, involves carefully considering how momentum is shared between the objects involved. The exercise demonstrates two different setups: catching the ball and when it bounces off.
In the first case, the ball is caught, leading to combined mass moving together. Use the equation:\[ m_b \cdot v_b = (m_b + m_p) \cdot v_f \]By substituting the values, you find the final velocity of both.
  • Calculate initial total momentum using ball's mass and velocity.
  • Equate it to the final combined momentum of both the ball and the person.
For the scenario where the ball bounces, adjust the equation to account for its new velocity going in the opposite direction. Here, momentum direction matters, hence use:\[ m_b \cdot v_b = m_p \cdot v_f + m_b \cdot v_{b'} \]Again, substitute known values diligently and solve for your speed after the bounce. These detailed calculations highlight the pivotal role velocity plays in momentum conservation problems.

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Most popular questions from this chapter

. Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 \(\mathrm{ms}\) contact time throughout this problem. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) .(a) What impulse and what force did Big Bill exert on the ten- nis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

\(\bullet\) Bird defense. To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a 600 gram falcon flying at 20.0 \(\mathrm{m} / \mathrm{s}\) ran into a 1.5 \(\mathrm{kg}\) raven flying at 9.0 \(\mathrm{m} / \mathrm{s}\) . The falcon hit the raven at right angles to its original path and bounced back with a speed of 5.0 \(\mathrm{m} / \mathrm{s}\) . (These figures were estimated by one of the authors \((\mathrm{WRA})\) as he watched this attack occur in northern New Mexico.) By what angle did the falcon change the raven's direction of motion?

A machine part consists of a thin, uniform \(4.00-\mathrm{kg}\) bar that is 1.50 \(\mathrm{m}\) long, hinged perpendicular to a similar vertical bar of mass 3.00 \(\mathrm{kg}\) and length 1.80 \(\mathrm{m} .\) The longer bar has a small but dense 2.00 -kg ball at one end (Fig. 8.42\()\) By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through \(90^{\circ}\) to make the entire part horizontal?

A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of 6000 \(\mathrm{kg}\) and ejects gas at a relative velocity of magnitude 2000 \(\mathrm{m} / \mathrm{s}\) , how much gas must it eject in the first second to have an initial acceleration of 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) .

. Accident analysis. A 1500 \(\mathrm{kg}\) sedan goes through a wide intersection traveling from north to south when it is hit by a 2200 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one there after. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pave- ment is \(0.75,\) and the cars slide to a halt at a point 5.39 \(\mathrm{m}\) west and 6.43 \(\mathrm{m}\) south of the impact point. How fast was each car traveling just before the collision?

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