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Chapter 5: Problem 10

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 Ping-Pong ball needs to move at approximately 400 m/s to have the same kinetic energy as the bowling ball.

Step by step solution

01

Calculate the Kinetic Energy of the Bowling Ball

The formula for kinetic energy is \(KE = \frac{1}{2}mv^2\). Plug in given values, where \(m = 7.00 \,kg\) is the mass and \(v = 3.00 \,m/s\) is the velocity for the bowling ball. The energy comes out as \[KE = \frac{1}{2} * 7.00 \,kg * (3.00 \,m/s)^2 = 31.5 \,J\].
02

Solve for the Ping-Pong Ball's Velocity

The kinetic energy of the ping-pong ball should be equal to the kinetic energy of the bowling ball, i.e., \(31.5 \,J\). The equation becomes \(31.5 \,J = \frac{1}{2}m_{pp}v_{pp}^2\). Where \(m_{pp} = 2.45 \,g = 0.00245 \,kg\), after converting grams to kilograms. Now solve for \(v_{pp}\) (velocity of ping-pong ball). This provides the solution \(v_{pp} = \sqrt{\frac{2*31.5 \,J}{0.00245 \,kg}} \approx 400 \,m/s\).

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Key Concepts

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

Understanding Physics Problems
When working on physics problems, a clear understanding of the given parameters and what you are solving for is vital. In this particular problem, we are dealing with kinetic energy and comparing two different objects with distinct masses and velocities. The effective approach is to first isolate the key concept 鈥� kinetic energy 鈥� and apply its formula:

Kinetic energy (KE) is defined by the equation
\[ KE = \frac{1}{2}mv^2 \]
where 'm' represents mass and 'v' represents velocity. Physics problems often require a step-by-step analysis, which in this case starts with calculating the bowling ball's kinetic energy. This initial step frames the problem and provides a target value to work towards when finding the velocity of the second object, the Ping-Pong ball.
Energy Conservation
Energy conservation is a principle stating that energy cannot be created or destroyed, only transformed from one form to another. This is a cornerstone of mechanics and is incredibly useful in solving physics problems like the one we're addressing.

With the conservation principle, we assert that the kinetic energy of the bowling ball should be equal to that of the Ping-Pong ball for them to share the same energy. It is important to remember that energy's conservation allows us to set their kinetic energies equal. This gives factual support to proceed with the calculations and solve for the unknown 鈥� the velocity of the Ping-Pong ball.
Mechanics
Mechanics is a branch of physics that deals with the motion of objects and the forces that affect them. In this scenario, we focus on a particular aspect of mechanics: finding the velocity necessary for two objects with different masses to have equivalent kinetic energies.

Mechanics involves understanding different types of energies and forces. The problem we solved uses concepts of linear kinetic energy, which is kinetic energy associated with the linear, or straight-line, motion of an object. Through the formula provided and considering the mass of each object, we can deduce how fast or slow an object should be moving to have a certain amount of kinetic energy鈥攁 fundamental concept in the field of mechanics.

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Most popular questions from this chapter

A horizontal force of \(150 \mathrm{~N}\) is used to push a \(40.0-\mathrm{kg}\) packing crate a distance of \(6.00 \mathrm{~m}\) on a rough horizontal surface. If the crate moves at constant speed, find (a) the work done by the \(150-\mathrm{N}\) force and (b) the coefficient of kinetic friction between the crate and surface.

A 2.0-m-long pendulum is released from rest when the support string is at an angle of \(25^{\circ}\) with the vertical. What is the speed of the bob at the bottom of the swing?

A shopper in a supermarket pushes a cart with a force of \(95 \mathrm{~N}\) directed at an angle of \(25^{\circ}\) below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. (a) Find the work done by the shopper as she moves down a \(50.0-\mathrm{m}\) length aisle, (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?

In a needle biopsy, a narrow strip of tissue is extracted from a patient with a hollow needle. Rather than being pushed by hand, to ensure a clean cut the needle can be fired into the patient's body by a spring. Assume the needle has mass \(5.60 \mathrm{~g}\), the light spring has force constant \(375 \mathrm{~N} / \mathrm{m}\), and the spring is originally compressed \(8.10 \mathrm{~cm}\) to project the needle horizontally without friction. The tip of the needle then moves through \(2.40 \mathrm{~cm}\) of skin and soft tissue, which exerts a resistive force of \(7.60 \mathrm{~N}\) on it. Next, the needle cuts \(3.50 \mathrm{~cm}\) into an organ, which exerts a backward force of \(9.20 \mathrm{~N}\) on it. Find (a) the maximum speed of the needle and (b) the speed at which a flange on the back end of the needle runs into a stop, set to limit the penetration to \(5.90 \mathrm{~cm}\).

A \(1.50 \times 10^{3}\) -kg car starts from rest and accelerates uniformly to \(18.0 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\). Assume that air resistance remains constant at \(400 \mathrm{~N}\) during this time. Find (a) the average power developed by the engine and (b) the instantaneous power output of the engine at \(t=12.0 \mathrm{~s}\), just before the car stops accelerating.
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