I What does the area under a net-force vs. time graph...

i What does the area under a net-force vs. time graph represent? ii Determine the total impulse felt by this object. iii If the object is known to have a mass of m = 11 kg and a velocity of 3 m/s at t = 2 s, determine its velocity at t = 8 s iv What is its velocity at t = 13 s? v Prove that if total momentum is conserved, then total impulse must be zero. vi Bonus:f I want to change the maximum positive force felt to 1 N and the minimum negative force felt to -7 N, what new values must t1, t2, t3 and t4 take such that total momentum is conserved?

Transcript

00:01 Okay, so what does the area under the net force times time graph represent? so that is going to give you the impulse, because impulse is the change in force times the time, or the force times the time interval that it's applied over.

00:24 And that's precisely the area under this graph.

00:29 So part two says, determine the total impulse felt by this object.

00:34 Well we can see that the impulse from the first non -zero force, that force has magnitude 4 .5, and it's applied over a time interval of three seconds, so we get this contribution to the impulse.

00:47 And then we have to add on the next contribution to the impulse, which this time has the forces of value minus 4 .5, and again is applied over three seconds.

00:54 So we can see that the total impulse is zero.

00:58 Part three says, the object has mass 11 kilograms, and its velocity at t equals two is three meters per second.

01:08 What's its velocity at t equals eight? well at t equals eight, the only force it's undergone is the one between t equals three and t equals six.

01:18 And that has an impulse of 4 .5 times three, or equivalently 13 .5.

01:29 And this is also equal to the change in momentum.

01:32 So that's the mass times the speed after, minus the speed before.

01:41 So we can see that the speed after is equal to 13 .5 over 11, plus the speed before.

01:53 The speed before that impulse was just three meters per second, and so we find that the speed after is equal to 4 .227 meters per second...

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