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this problem on the topic of momentum impulse and collisions, were told that a fullback who is running east with the speed of five m per second and has a massive 90 kg, is then tackled by their opponent, Who is 95 kg. Running north with his feet of 3m/s. We want to know why the tackle will constitute a perfectly inelastic collision. We want to then calculate the velocity of the players immediately after the tackle and then determine the mechanical energy that is lost as a result of the collision. Lastly, we want to know where the lost energy would go now over a short time interval of the collision. The external forces have no time to impart significant impulse to the players, so the two players will move together after the tackle and since the players moved together after the tackle, this constitutes a completely inelastic collision. Next for part B of the problem, we want to calculate the velocity of the players immediately after this tackle. Now we'll take the positive X. Direction to be to the east and positive Y. To be to the north. We can use the conservation of momentum in the horizontal direction which is to the east in this case. And we know at the momentum after the tackle must equal the momentum before the tackle. Which then means that in one plus him to the mass of both of the footballers after the tackle since they moved together times the combined velocity V. F. Along the horizontal direction. So you multiply that by cosign Theta must equal and one V. One I plus zero. Or simply rearranging. We get VF course I intend to to be M. One the one I over and one lesson two. So we're assuming here that the opponent or the opponent actually has no horizontal component of velocity since they're running do not Hence v. II I is equal to zero. So if we substitute our values into this, this becomes 90 kg times five meters per second, divided by the sum of the masses, 90 kg plus 95 kg. And so calculating we get VF course to India to to be two 0.43 meters per second. So now we can calculate the Y component. Bye taking the why components of the momentum before and after the collision like we did above in this case is the vertical components or northerly component. We can see That again. M one plus M two. In this case V F sign data is equal to zero since the first player has no vertical component of momentum less M two the two I which means again we can rearrange this equation and find VF signed data. When we get VFC in theater to be M two, the two I over M one plus M two. Again, we can substitute our values into this equation. We get this to be 95 kg and the speed of the opponent three meters per second, divided by the sum of their masses, which is 90 kg, Last 95 Kg. And so calculating, we get VF times sine theater to be one 0.54 meters per second. And so we get VF to simply be the square root of the sum of squares of each of the components. So if we square both the X and Y components of the final velocity and then add them And take the square root, we get this to be the square root of 8.2, 8 meter squared, the second squared which gives us the magnitude of the final velocity As they moved together after the tackle to be 2.88 meters the second. We can then calculate the the angle which will give us the vector And we know that 10 data is equal to the why component of the velocity V. F. Signed data over VF course in theater, which is simply 1.54 meters per second over 2.4, 3 meters per second. And this gives us 10 data to be zero .634. Then hence we can calculate theater to be the arc 10 of this value And we get theater to be 32.4°.. So therefore the velocity of the two players after the tackle, the F is equal to 2.88 m/s at an angle of 32.4° not of east. And so there we have the final velocity of the two players after the tackle. Next we want to determine the mechanical energy that is lost due to the collision. So to do this, we know this is will appear as lost kinetic energy in this last kinetic energy is equal to the initial kinetic energy minus the final kinetic energy. And so this is uh huh. In one V. One I squared less a half. M two V two I squared -1 into the final masses. M one plus M. Two times V. F. Squared. And so if we substitute our values into this, this becomes half into 90 Kg. Would suppress the units here 25 m/s squared as the kinetic energy of the opponent, Which is their mess 95 Kg into the meters per second. Old squared minus a half Into 185 Kg. The mass of both players times The combined speed 2.88 m, the second all squared. And so calculating, we get the lost kinetic energy or the lost mechanical energy In this tackle to be 700 and 85 jules. Well, if the mechanical energy is lost, we want to know where they lost energy would go. And so the last energy here has actually transformed into other forms of energy, such as thermal energy and sound energy.

University of Kwazulu-Natal

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