Can someone explain this formula?

Test Preparation AP Sciences
1

<p>Can someone show this formula in action?
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<p>Thanks.</p>

2

<p>rearrange it to get
integral of F(t) dt = change in momentum </p>

<p>integral of F(t) dt is impulse
change in momentum problems typically use a constant mass, so change in momentum would just be m*delta v</p>

<p>putting it in that form is a bit misleading/confusing</p>

3

<p>^
And so what’s coming next? I just want to see this formula in action, I mean, in some problem.</p>

4

<p>another way to look at it is:
F=ma and a=dv/dt so F=m*dv/dt, and Fdt=mdv, so the integral of Fdt (also called impulse) is equal to an object’s change in momentum (change in the quantity mv).
If you really wanted to see the formula in action, I guess the best way is to consider a collision, Newton’s Third Law tells us the forces are equal and opposite, and it acts for the same time interval dt, therefore each object’s change in momentum is equal and opposite delta p1= -delta p2, and this leads to the law of conservation of momentum.
Hope this helps!</p>

5

<p>Let’s say we know the mass of an object is 2 kg and the initial velocity is 2 m/s. Let’s say we also know that the force applied to the object is F=2t+6. And let’s say that you need to figure out the velocity of the object after 3 seconds.</p>

<p>What you need to do is realize that impulse and momentum are the same thing. The application of a force over a time (impulse) leads to a change in momentum (mass*velocity). The integration of the force (F) over a certain time interval (dt) gives us how much the momentum changes over that time interval. So to find how much the momentum changes from t=0 to t=3, we do fnInt(2t+6,t,0,3), which is 27.</p>

<p>27 is the change in momentum, the change in mass*velocity. Since the mass remains constant, we can substitute 2 in for the mass.</p>

<p>27=2*(change in v)</p>

<p>Change in v=27/2=13.5</p>

<p>To find the velocity after 3 seconds, we add the change in velocity to the initial velocity. Final velocity=13.5+2=15.5 m/s</p>

<p>That’s a basic example of the calculus-based impulse-momentum theory at work!</p>

6

<p>Wow! This is great. Now I have some picture what does it mean. Thanks!
But what is the difference between</p>

<p>a=dv/dt and a=d^2x/dt^2</p>

<p>Thanks!</p>

7

<p>*One thing: what force applied? If that F=mg? or normal Force?</p>

8

<p>^^dv/dt is the derivative of the velocity. d^2x/dt^2 is the double derivative of position. Think of it this way:</p>

<p>To go one down, take the derivative. To go one up, integrate:</p>

<p>Position
Velocity
Acceleration</p>

<p>So to get from position to acceleration, you need to take the derivative twice (d^2x/dt^2). You can think of it as the double derivative of position or the derivative of the derivative of the position. To get from velocity to acceleration, you only have to take the derivative once (dv/dt).</p>

9

<p>^
Why d^2x/dt^2? Isn’t the second derivative d^2y/dx^2? </p>

<p>Here I did some sketches so…are they correct? I didn’t find the proper limit :(</p>

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