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

Conservation of Linear Momentum

The principle of linear momentum conservation states that the total momentum of a system remains constant if no external forces act on it. This principle allows us to compute the post-collision velocity of a combined object in a collision by considering the vector sum of the momentum components from each object prior to impact, irrespective of whether the collision is elastic or inelastic.

Perfectly Inelastic Collision

A perfectly inelastic collision is one in which the colliding bodies stick together after impacting. In such scenarios, while momentum is conserved, kinetic energy is not; some of the initial kinetic energy is usually converted into other energy forms such as heat or deformation. This concept is key to understanding why, despite momentum conservation, there is a loss of mechanical energy in the collision process.

Kinetic Energy and Its Loss

Kinetic energy is the energy of an object due to its motion. In inelastic collisions, such as the perfectly inelastic case, the kinetic energy after the collision is less than the kinetic energy before the collision because some of it transforms into other forms of energy. Understanding the assessment of kinetic energy before and after the collision is critical for calculating the energy lost during the impact.

Energy Dissipation and Transformation

The loss of kinetic energy in an inelastic collision is not a disappearance of energy but rather a transformation. This 'lost' energy typically converts into internal energy, sound, heat, and energy involved in structural deformations of the colliding objects, making it a fundamental concept in energy conservation and transformation processes in physics.

Vector Decomposition and Addition

In collisions involving objects moving in different directions, it is essential to break down their velocities into components and perform vector addition. This concept helps in accurately adding the momentum vectors in different axes (e.g., east and north) to determine the resultant velocity of the combined mass after the collision, ensuring a correct analysis of motion in more than one dimension.

Transcript

00:01 In this problem on the topic of momentum impulse and collisions, we are told that a fullback who is running east with a speed of 5 meters per second and has a mass of 90 kilograms is then tackled by their opponent, who is 95 kilograms running north with a speed of 3 meters per second.

00:18 We want to know why the tackle will constitute a perfectly inelastic collision.

00:23 We want to then calculate the velocity of the players immediately after the tackle and then determine the mechanical energy that.

00:30 Is lost as a result of the collision.

00:33 Lastly, we want to know where the lost energy would go.

00:37 Now, over a short time interval of the collision, the external forces have no time to impart significant impulse to the players.

00:46 So the two players will move together after the tackle.

00:58 And since the players move together after the tackle, this constitutes a completely inelastic collision.

01:13 Next for part b of the problem we want to calculate the velocity of the players immediately after this tackle.

01:25 Now we'll take the positive x direction to b to the east and positive y to be to the north.

01:33 We can use the conservation of momentum in the horizontal direction which is to the east in this case and we know that the momentum after the tackle must equal the momentum before the tackle, which then means that m1 plus m2 the mass of both of the footballers after the tackle, since they move together, times their combined velocity vf along the horizontal directions, so we multiply that by cosine theta, must equal m1 v1i plus 0 or simply rearranging we get vf cosine theta to be m1 v1 i over m1 plus m2 so we're assuming here that the opponent or the opponent actually has no horizontal component of velocity since they're running due north hence v2 i is equal to zero so if we substitute our values into this this becomes 90 kg times 5 meters per second divided by the sum of the masses 90 kg plus 95 kg and so calculating we get vf cosine theta to be 2 .43 meters per second.

03:33 So now we can calculate the y component by taking the y components of the momentum before and after the collision, like we did above.

03:52 In this case, this is the vertical components or northerly component.

04:00 We can see that again, m1 plus m2, in this case vf sine theta is equal to zero since the first player has no particle component of momentum plus m2 v2 i which means again we can rearrange this equation and find vf sine theta and we get vf sine theta to be m2 v2 i over m1 plus m2 again, we can substitute our values into this equation.

04:45 We get this to be 95 kg times the speed of the opponent, 3 meters per second, divided by the sum of their masses, which is 90 kg plus 95 kg.

05:12 And so calculating we get vf times sine theta to be 1 .54 meters per second...

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