A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent runnin

A 90.0-kg fullback running east with a speed of 5.00 m/s is...

Key Concepts

Perfectly Inelastic Collision

This describes a type of collision in which the colliding objects stick together after impact, moving as one combined mass. In such collisions, while the momentum is conserved, kinetic energy is not conserved due to energy losses resulting from factors like deformation, heat, and sound.

Conservation of Momentum

Momentum is a vector quantity that remains constant in a closed system with no external forces. In collision analyses, the total momentum before the collision is equal to the total momentum after the collision, regardless of the conversion of kinetic energy into other forms.

Kinetic Energy Dissipation in Inelastic Collisions

Unlike elastic collisions where kinetic energy is conserved, in inelastic collisions some of the kinetic energy is transformed into other forms of energy such as heat, sound, or internal energy. This loss is a key characteristic of inelastic collisions and explains why energy calculations must account for energy not present in the post-collision kinetic energy.

Vector Decomposition in Momentum Analysis

Momentum is a vector quantity having both magnitude and direction, which requires breaking down the momentum into its components along mutually perpendicular axes (such as east and north). This approach allows for the proper application of the conservation of momentum law in two dimensions, ensuring that momentum is properly conserved in each independent direction.

Transcript

00:01 Here you have a body that's moving with a velocity of 5 meters per second.

00:14 And this body, you're told, it weighs 90 kilograms.

00:19 And there's another body that's moving up that's 95 kilograms, and that's moving up at 3 meters per second.

00:30 And there's a collision that happens here in the form of a tackle.

00:34 And they tell you that afterwards, so after this collision happens, these two bodies stick together and they move with some velocity at some angle that's equal to vf in this case.

00:52 So the first question they ask here is, is this, so what makes this a perfectly an elastic collision? and the answer to that is every time the two objects stick together here, they're that is what constitutes you perfectly inelastic collision because you're going to lose some of your mechanical energy as these things get stuck together and move with the common velocity here.

01:16 So they don't rebound off each other but they actually stick together and become one body as they move.

01:23 So that's a perfectly inelastic collision.

01:27 So now the second question they want you to ask you is the velocity of this thing after the collision happens.

01:33 So the way to do that is again, since this is steward collision, we want to always remember that pi is equal to pf.

01:43 So in this case, again, since it's a two -dimensional problem and these are vectors that we're working with, we have to split it up into two different components as well.

01:53 So if i start by doing the x direction, i will realize that the only body that's moving in an x direction initially is this first mass here.

02:01 So essentially that would just be m1 times v1i should be because this is now zero in x, that should be equal to m1 plus m2, all of this multiplied by the final velocity, but in the x direction.

02:23 So you notice now that you have the m1, you have the v1, you have m2.

02:30 So essentially your solution is going to be m1 v1 i over m1 plus m2 here.

02:43 And all of that will give you the v final in the x.

02:48 So if you plug in the number, 90 here and 95 and 5, you see that your final velocity in the x direction is somewhere around 2 .4 .3 meters.

03:04 Second so then you go ahead and do the same thing in the y again this is the only body that's moving in the y so now it's actually m2 v2 i is equal to now m1 plus m2 this multiply by v final but in the y direction and again you do the same thing you realize that m2 v2 i over m1 plus m2 is equal to v final in the y direction and you plug in again plug in 95 here 90 here in 3 meters to second you find that that is something around 1 .54 meters per second so these are the components of the velocity but you want the vector itself so to find the final velocity vector magnitude in this case you're just going to do the square root of vf x squared plus v f f y squared and when you plug in these numbers in this formula here you find that your final velocity comes up to something like 2 .87 meters per second and then for the angle because again these these bodies are troubling in two different essentially one is going up one is going sideways so this we expect that maybe this will have like a final velocity along that direction so then the angle that this makes with the horizontal now is what you want to calculate and to calculate that if you know the x velocity you know the y velocity and you know the x velocity and you know the x velocity so you want the result in velocity here and notice to get this, what you call the hypatna side on this right triangle, if you were to move this over here, you have the adjacent opposite side.

05:21 You have to use a tangent.

05:22 So the tangent of theta will give you vy over vx.

05:27 But we want to find theta.

05:30 So to find data, we're just going to do the inverse tangent vy over vx.

05:42 Should give us the angle here.

05:45 And if you take vy to be 1 .54, and vx to be 2 .43, and you put in an inverse sign in formula, you find that your theta is around 32 .4 degrees.

06:07 So this is your angle, that's the speed...

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