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$\cdot$ The speed of the fastest-pitched baseball was $45 \mathrm{m} / \mathrm{s},$ and theball's mass was 145 $\mathrm{g}$ . (a) What was the magnitude of the mo-mentum of this ball, and how many joules of kinetic energy didit have? (b) How fast would a 57 gram ball have to travel to havethe same amount of (i) kinetic energy, and (ii) momentum?

a) 6.525 $\mathrm{kgm} / \mathrm{s}$146.8 $\mathrm{J}$b) 72 $\mathrm{m} / \mathrm{s}$114.5 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum

Physics Basics

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Cornell University

University of Washington

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Lectures

03:23

In physics, mechanical energy is the sum of the kinetic and potential energies of a system.

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

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The speed of the fastest-p…

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A baseball player throws a…

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(a) A baseball weighs $145…

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A baseball with a mass of …

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A 145 -g baseball is throw…

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(a) A 145-g baseball trave…

And that's question were given that a baseball with a velocity, a baseball with a mass of 145 grams which I've converted to 0.145 kilograms travels with a velocity of 45 meters per second. In part A were asked to find the kinetic energy of this baseball, which is pretty easy to dio. We just need to remember that, um, kinetic energy is equal to 1/2 times the mass of the ball times the velocity of the ball squared and we have both of those given to us. So 1/2 times 0.145 kilograms times 45 meters per second squared and plugging that into a calculator refines that the kinetic energy of this baseball is 146 0.8. Jules, we also have to find theme o mentum of this baseball, and we do that just using our equation. Um, momentum is equal to mass times velocity. So in this case, um, we have both of those again. So zero point, um, 145 kilograms times a velocity of 45 meters per second is equal to 6.5 to 5 kilogram meters per second. Non part B. Um, we have to set a later ball to have the same kinetic energy in part B one or momentum and part B two, and finds what velocity it must have to achieve those same values. So, um, let's do part one first. So we need our, um, kinetic energy of the ball to be equal to, um, the kinetic energy we found earlier of 1 46.8 jewels. So, um, putting that on the left hand side of the kinetic energy equation you have 1 46.8 is equal to 1/2 times the mass of the ball is 0.57 kilograms times the velocity squared. Now we want to multiply both sides by two and divide both sides, binds your points your 57 kilograms and take the square root of both sides to isolate the velocity on its own. So philosophy is then equal to the square root of two times 1 46.8 jewels over 0.57 kilograms and plugging that into a calculator, we find that the velocity necessary for this later Balto, have the same kinetic energy is about 72 meters per second. In part B. Two, we need to do the same thing, but with momentum. So um, the momentum were setting equal to 6.5 to 5 kilogram meters per second. On the left hand side is equal to set that equal toothy equation for momentum mass times velocity which is 0.57 kilograms times velocity. So now all we need to do is divide both sides by 0.57 kilograms. So velocity is equal to 6.5 to 5 kilogram meters per second over 0.57 kilograms. And plugging that into a calculator, we find that the velocity necessary to achieve the same Mo mentum is 114 point five meters per second. And that's it for this question.

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