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