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In a certain men's track and field event, the shotput has a mass of 7.30 $\mathrm{kg}$ and is released with a speed of 15.0 $\mathrm{m} / \mathrm{s}$ at $40.0^{\circ}$ above the horizontal over a man's straight left leg. What are the initial horizontal and vertical components of the momentum of this shotput?

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$83.9 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ ; $70.4 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

University of Washington

Simon Fraser University

McMaster University

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

06:46

In a certain track and fie…

02:01

In a certain men's t…

03:15

In a certain track and fic…

02:26

04:50

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04:08

A $2.00-\mathrm{m}$ -tall …

04:58

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in discussion. We have given moss, whoa softwood, that is I am equal to 7.3 energy. Also the speed of short for that is the equal to 15 m/s and angle good I mean thought was and origin that is Changed up equals to 40 digging. Now comes to are mm So here we had a blink flying go. The momentum of a short word for horizontal component. So we know that initial moment down of a short board can be expressed as a P. I. Equal to M marty club I V. No. Yeah. And is the mass of the sort. We is the spirit of the sort. And pia is actually the initial momentum of the shop. So we substitute 7.3 Kz for M and 15 m per second. Four we therefore we get P I equal to 7.3 Many people are by 15. So we get the I equals two 109 why kilogram meter. Part second. No, the or a gentle component of the initial momentum of a short book can be expressed as B of X. Equal to be I My people are by course cheap now. Here we substitute 109.5 kilometer per second for P I. And 14 2040. To therefore we get P of X equality. 109.5 million people have I almost 40 degree. So from about we get P of X equal to 83. 1 89. Hello Graham later. But six. And the horizontal component of the initial momentum of a short book. It was 83.89 kg meter, but second now similarly comes to part the Here we are going to calculate a vertical component of the initial momentum of a short. It's going to be expressed as P of Y equal to E I sign Cheetah. Now, here we substitute. 109.5 kilogram meter per second for P and 1484 Tita. They're falling at P or five equals 2 109.5 multiple. Have I signed 40 digging. So from above we get p of y equals two 75. So mine kilogram meter sex and the vertical component of the nation, momentum of a short book equals to 70.39 kg meter per second. Or you can say 70.4 kg meter per second approximately.

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