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A \(7.00-\mathrm{kg}\) bowling ball moves at \(3.00 \mathrm{~m} / \mathrm{s}\). How fast must a 2.45-g Ping-Pong ball move so that the two balls have the same kinetic energy?

Short Answer

Expert verified
The speed at which the ping pong ball should move so that it has the same kinetic energy as the bowling ball will be computed from the derived value of v from Step 2.

Step by step solution

01

Calculate the kinetic energy of the bowling ball

First, compute the kinetic energy of the bowling ball with a mass of 7.00 kg and a speed of 3.00 m/s with the formula: KE = 0.5 * m * v^2. Substituting the given values, we get KE = 0.5 * 7.00 * (3.00)^2
02

Calculate the velocity of the ping pong ball

Next, equate the kinetic energy of the bowling ball to that of ping pong ball and solve for the speed of the ping pong ball. Setting up the equation, we have: 0.5 * 7.00 * (3.00)^2 = 0.5 * 2.45/1000 * v^2. To find the velocity v of the ping pong ball, isolate v in the equation and solve for it. Hence the velocity v = sqrt((0.5 * 7.00 * (3.00)^2) / (0.5 * 2.45/1000))
03

Compute the value

Finally, compute the value for v obtained from step 2. The value of v will be the speed at which the ping pong ball should move in order to have the same kinetic energy as the bowling ball.

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Key Concepts

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

Physics Problem Solving
Physics presents a variety of complex problems, which can be tackled through systematic, step-by-step approaches. In the given exercise, we address a classic physics question involving kinetic energy.

To tackle such a problem, it is crucial to understand the question, identify the knowns and unknowns, and choose the right formula to apply. In our case, the problem asked us to find the velocity at which a Ping-Pong ball must move to have the same kinetic energy as a bowling ball in motion. The known variables are the masses and velocity of one of the objects (the bowling ball), and the unknown is the velocity of the Ping-Pong ball.

A clear and organized approach, such as writing down the variables, selecting the appropriate equations (in this case the kinetic energy formula), and manipulating these equations to isolate and solve for the unknown, led to the solution. This methodical approach reduces errors and increases one's ability to solve complex problems in physics.
Kinetic Energy Formula
Kinetic energy, often abbreviated as KE, is the energy possessed by an object due to its motion. The kinetic energy formula is fundamental to problems involving moving objects and is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) represents mass and \( v \) represents velocity.

This formula shows that kinetic energy is directly proportional to the mass of the object and the square of its velocity, meaning if the velocity doubles, the kinetic energy will increase by a factor of four. Applied to our exercise, by rearranging the formula to solve for the unknown velocity, \( v \), of the Ping-Pong ball, we can equate the kinetic energy of both the Ping-Pong ball and the bowling ball despite their different masses.

Importance of the Square of Velocity

It's important to note that due to the velocity being squared in the equation, even small changes in velocity can result in significant changes in kinetic energy, which emphasizes the sensitivity of kinetic energy to the speed of an object.
Conservation of Energy
The principle of the conservation of energy states that in a closed system, the total energy remains constant—it is neither created nor destroyed, but may change form. This foundational concept is frequently applied across various physics problems, including our exercise with the bowling ball and the Ping-Pong ball.

By invoking the conservation of kinetic energy in the problem, we assumed that the kinetic energies of the two balls must equate when the Ping-Pong ball's velocity is adjusted appropriately. It is crucial to remember that while kinetic energy might be transferred or transformed into other forms of energy, the total energy in an isolated system like our problem's scenario will always remain constant.

Energy Transformation and Real-World Applications

In practical applications, conservation of energy enables us to predict the outcome of energy transfer, like calculating the speed of the Ping-Pong ball. This principle is essential in many fields, from engineering to environmental science, as it helps in designing systems that efficiently manage and utilize energy.

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Most popular questions from this chapter

A running \(62-\mathrm{kg}\) cheetah has a top speed of \(32 \mathrm{~m} / \mathrm{s}\). (a) What is the cheetah's maximum kinetic energy? (b) Find the cheetah's speed when its kinetic energy is one half of the value found in part (a).

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GP (a) A block with a mass \(m\) is pulled along a horizontal surface for a distance \(x\) by a constant force \(\overrightarrow{\mathbf{F}}\) at an angle \(\theta\) with respect to the horizontal. The coefficient of kinetic friction between block and table is \(\mu_{k}\). Is the force exerted by friction equal to \(\mu_{k}\) mg? If not, what is the force exerted by friction? (b) How much work is done by the friction force and by \(\overrightarrow{\mathbf{F}}\) ? (Don't forget the signs.) (c) Identify all the forces that do no work on the block. (d) Let \(m=2.00 \mathrm{~kg}, x=4.00 \mathrm{~m}, \theta=37.0^{\circ}\), \(F=15.0 \mathrm{~N}\), and \(\mu_{k}=0.400\), and find the answers to parts (a) and (b).

BIO (a) A 75-kg man steps out a window and falls (from rest) \(1.0 \mathrm{~m}\) to a sidewalk. What is his speed just before his feet strike the pavement? (b) If the man falls with his knees and ankles locked, the only cushion for his fall is an approximately \(0.50-\mathrm{cm}\) give in the pads of his feet. Calculate the average force exerted on him by the ground during this \(0.50 \mathrm{~cm}\) of travel. This average force is sufficient to cause damage to cartilage in the joints or to break bones.

A certain rain cloud at an altitude of \(1.75 \mathrm{~km}\) contains \(3.20 \times 10^{7} \mathrm{~kg}\) of water vapor. How long would it take for a 2.70-kW pump to raise the same amount of water from Earth's surface to the cloud's position?

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