(a) Kicking a soccer ball. (b) Finding the average force on the ball from its components.
(a) Before-and-after diagram
(b) Average force on the ball

Answer

(a) $\Delta p_{x}=p_{2 x}-p_{1 x}=m\left(v_{2 x}-v_{1 x}\right)=(0.145 \mathrm{kg})[55.0 \mathrm{m} / \mathrm{s}-(-45.0 \mathrm{m} / \mathrm{s})]=14.5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ $J_{x}=\Delta p_{x},$ so $J_{x}=14.5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .$ Both the change in momentum and the impulse have magnitude $14.5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$
(b) $\left(F_{\mathrm{av}}\right)_{x}=\frac{J_{x}}{\Delta t}=\frac{14.5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}}{2.00 \times 10^{-3} \mathrm{s}}=7250 \mathrm{N}$
The force is in the direction of the momentum change.

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Physics 101 Mechanics

University Physics

Chapter 8

Momentum, Impulse, and Collisions

Section 1

Momentum and Impulse

Discussion

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Zobia M.

February 26, 2022

Example 8.13

Zobia M.

February 26, 2022

Top Physics 101 Mechanics Educators

Christina K.

Rutgers, The State University of New Jersey

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Video Transcript

once again welcome to a new problem at this time will given a baseball that has a mass. So this is This is the information that's given. We have a baseball with the mask off zero point one four five kilograms, and it's moving towards the right with an initial velocity, which happens to be, you know, going to say positive forty five minutes per second because it was the right is positive. Um, so that's the peach. That's the baseball peach, and it's going to move all the way up until a specific point where there's a bat, um, and the the batted ball. So this is the Peach ball, and this is the Barton Bowl. So the bartered bowl happens to have a final, which we're going to call the final velocity off negative fifty five meters to second. So that's what we're given S O in part A. We want to find the change in momentum. Okay, we want to find the change in momentum. That's the magnitude of the magnitude off the change in momentum and the impulse Lenny in Part B. We want to find the average force, the average force that's responsible for that specific change in momentum. So that's what we're looking for. S o, you know, starting on part, eh? We see that to get the change in momentum. It's just the final of the final momentum, minus the initial momentum. Remember, momentum is the product off the mass and Exxon under velocity. So it's just m of the final minus V initial, which is m. The final happens to be negative. Fifty five meters per second minus V Initial is positive forty five minutes per second and so way have to plug in the mass. The mass happens to be zero point one for five kilograms, so zero point one four five kilograms. And so if you do the math, you end up getting a change in momentum off negative fourteen point five kilogram meters per second. The highlight off this is that the magnitude off the change in momentum happens to be fourteen point five kilogram meters per second. So that's just the fast part. You know, want to find the magnitude You can recall that the second part and we'll do that on the page. Are you interested in finding the the impulse, the impulse? Remember impulse, which is J the vector sobs superscript on top of it. Defected line direction on top of it happens to be if they are really force times change in time. So this is, um that happens to be a our impulse. The well, we have to backtrack a little bit. That happens to be Howard. Change in momentum. Okay, so this is this happens to be our change in momentum and this is exactly the same as our imports. So you know, if if we take it, we'll take the change in momentum and really mean compare that to to the to the problem. So if you go back, you find that the change in momentum was negative fourteen point five kilogram meters per second, a negative fourteen point five kilogram meters per second. So that happens to be to be our impulse. You know, this is the impulse And so the magnitude ofthe the impulse which I'll just put the the absolute value sign to show that this is the magnitude of the impulse that happens to be equivalent to fourteen point five kilogram meters per second. You know, that's the magnitude of the impulse in part B, which we can pick up in the next stage. We supposed to find the average force that's required. I remember you know you had the budget ball right there on the time it takes for that contact happened is two point zero zero milliseconds, which we can change two seconds by multiplying by one second off a thousand milliseconds. And that gives us two times ten to the negative three seconds. That's the time it takes to off contact between the bat on the ball. So we have the last e way because it goes in that direction so they have red force times. Delta t happens to be the impulse. And so the average force is the summers, the impulse of adult titty. Going back to the previous page, you can see that we are an impulse off negative fourteen point five kilogram kilogram meters. Second, I wanted to fight that, by the way, time it takes for that contact happened. And so our final average force F what you call F average calling average eyes going to be the same as seven, you know, roughly seven point two times ten to the three Newton. That's our problem right there. I hope you enjoyed it Feel feet share really comments or questions? Um and, uh hope you have a wonderful day a Sze Yu up. Who are the problems? I want to see you soon. Okay, Thanks. Bye.

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