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On a very muddy football field, a 110-kg linebacker tackles an 85-kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.8 m/s north and the halfback is sliding with a velocity of 7.2 m/s east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

$\tan \theta=\frac{v_{2 y}}{v_{2 x}}=\frac{4.96 \mathrm{m} / \mathrm{s}}{3.14 \mathrm{m} / \mathrm{s}}$ and $\theta=58^{\circ}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Section 3

Momentum Conservation and Collisions

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

McMaster University

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problem. 8.37. We're told of a football tackle that's occurring on a very muddy field. And so the players are able to slip around without much friction that we need to worry about, at least for the duration of time. Immediately, before and after the tackle. That's in question. And so we're told that a linebacker is moving, uh, north at this speed and that this mass towards the halfback was moving east with the given speed mass. And then we'll have what we call T, I guess, for the tackled pair of them moving off some direction north and east with the combined mass of 195 kg. We want to know the magnitude and direction of this. So because the any forces acting external to these things are going to be negligible for the situation in question, he conservation of momentum applies. So before the tackle, we have the let's do this one first because we'll call this direction X this direction, why we can just call them East and West, you know, So the X component first is going to be the mass of H times. The initial speed of age and the exhale direction and then l is going to be moving in the Y direction. Oh, initial. Why direction? I didn't. This is going to be equal to the final momentum where the Mass was going to factor out it here and we'll have the X component of the final velocity. Unsurprisingly enough in the extraction bus, white component of the final velocity, the Y direction. So then we can we can solve this. This is technically because X hats and white hats don't add together and the amazing uniform we could have written This is a system of two equations and in fact, we can make found the final student the extraction going to be the mass of halfback times. So both the half backs and it'll speed and provide component. They're like carbon farm, the linebacker with no created on his otherwise right. So this works out 3.14 m per second and this is 4.96 Yes. Now, to find the magnitude of this, of course, this is always going to be just the some of the squares of the components and then take the square root of that and that works out to be 59 m per second. And then if we call, we take the direction the angle north. It is from the east because that theater that will be the arc tangent of the ratio of the Y component of the velocity to the extent and to remember how to do this. Imagine we have any vector you like, which, in this case, we could be for velocity or for vector the X Yeah, data. And then we know that the tangent of data is opposite over the adjacent. And then just take the arc tangent of both sides and you get to the with our tangent installation and then putting these numbers we had before in we get that the net direction is 58 degrees north of the east.

Cornell University

Numerade Educator

University of Winnipeg

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