Main content

### Course: Class 11 Physics (India) > Unit 9

Lesson 14: Introduction to linear momentum and impulse# Force vs. time graphs

David explains how to use a force vs. time graph to find the change in momentum and solves an example problem to find the final velocity of a spaceship. Created by David SantoPietro.

## Want to join the conversation?

- Would it be possible to calculate the distance the rocket travelled using this graph? If so, how?(2 votes)
- I think so.

In the first 4 seconds, the acceleration is constant(the force is constant) and can be found by using F=m*a which in this case is 3=2.9*a so a = 1 m/s^2

For seconds 3 to 7, we can find the acceleration by finding the mean force, which is 3/2= 1.5 N and then, 1.5=2.9*a so a=0.5 m/s^2

For the last part, the mean force is -2/2 = -1 N and so, -1=2.9*a a=-0.34 m/s^2

Now all we have to do is calculate the distance travelled in each part using the kinematic equations, (especially S = So Vo*t+a*t^2/2) and add them.(11 votes)

- If the net impulse becomes negative what will happen to the rocket? Will it fall or just do something else?(3 votes)
- Yes! If the net impulse becomes negative, it means the change in momentum is happening in the apposite direction of the initial momentum. Here the initial momentum was away from the planet, hence the negative impulse will change the momentum of the alien rocket towards the planet. In simple terms, KABOOM!(9 votes)

- i would want to know how do the varying force-time graph when we try to estimate the aveage impulsive force is somehow similar to mean's value theorem?(1 vote)
- Yes you would take the mean value of the force if you happen so come across such a uniform slope. You can view it similar to speed and time graph where the area is the distance travelled.

So if there is a slope in such a graph then we know that the average speed is total distance travelled / total time taken. That is total area under the graph(area of traingle) / total time take(length of the base)= mean(average) value of the speed = half the value of height of the triangle.

Therefore the mean value of such a slope is the mid point of the slope or half the value of the height.

So I think

Impluse = average force * time

if force increase or decrease at constant rate then

impulse = 1/2 total change in the magnitude of force * time take.(12 votes)

- Can anyone tell me how to approach a similar problem when mass is not constant? I have a basic knowledge of integration an differentiation.(6 votes)
- you would use a "law of conservation of Momentum problem. Watch this video, (last problem)

https://www.khanacademy.org/science/physics/ap-physics-1/ap-linear-momentum/introduction-to-linear-momentum-and-impulse-ap/v/introduction-to-momentum(2 votes)

- At the end of the video, David divides by 2.9 kg, and the left side of the equation goes from Ns to m/s. Can anyone explain why?(3 votes)
- Force is equal to rate of change of momentum.

F = Δp/Δt

or Δp = F*Δt (units of N*s)

since Δp = mΔv (units of kg*m/s)

F*Δt = mΔv

Δv = F*Δt/m or units of N*s/kg is same as m/s(4 votes)

- If I have a F_g (N) vs Time (s) graph, where F_g equals the gravitational force on the person by the Earth(weight) and t = time, what would the slope represent?

Would it represent Net Force?(2 votes)- Um… not really sure whether this is a valid question. This is because F_g is a constant, that is F_g = 9.8 m/s^2 anywhere on Earth. So the slope of this graph is 0. I don’t know of any quantity that is equal to Force/time.(2 votes)

- What is the slope of at graph?(3 votes)
- When you say "at graph" are you asking about an acceleration vs time graph? The slope of an acceleration vs time graph is the rate of change of acceleration. The automatic filter for insults will not allow me to put the name of rate of change of acceleration which is "j e r k"(2 votes)

- Is there a standard name for units of impulse other than Newton-Seconds? (I might suggest 1 Ns = 1 Kirk or 1 Sulu).(2 votes)
- The metric system unit for impulse could be
*Newton-Seconds*(*Ns*), or you could also use*kg⋅m/s*. But there isn't a specific name for it, there are only two different units for impulse (in the metric system).(3 votes)

- I don't understand where he gets the 3 Newtons from?(1 vote)
- It is from the graph. The force is the y-axis. He says in the video that the rocket has sensors that gives you this readout.(1 vote)

- what is the final answer for the final velocity(1 vote)

## Video transcript

- [Voiceover] There's
a miniature rocketship and it's full of tiny
aliens that just got done investigating a new moon with lunar pools and all kinds of organic new life forms. But they're done investigating,
so they're gonna blast off and take their findings home
to tell all their friends. Let's say at some moment
during their ascent, they're moving at four meters per second. And they're tiny aliens, their spaceship is only 2.9 kilograms, but they need to know,
are they gonna be able to get off this moon or not? So they gotta pay
attention to their speed, but instead of using a speedometer, they clever aliens, they use
a force versus time graph. So on their dashboard, they've
got a force versus time graph and it tells them what
the net force is on them, so let's say this is the net
force, not just any force, but this is the total force
on them from rocket boosters and the force of gravity
and whatever other forces there might be, they've
got advanced force sensors. I mean come on, they can
determine their net force, let's say, and it gives them this force as a function of time. But they want to know what
is their velocity gonna be after nine seconds? So they check their force
versus time readout, and this is the graph they get. And now they can determine it. And here's how they do it. They say, "All right
there's a force, a net force "of three newtons acting
for the first four seconds." So during this entire first four seconds there's a constant force of three newtons. And every alien worth his weight knows that force, the net force
multiplied by the time duration during which that force is applied, gives you the net impulse. So this gives us the net impulse. If we take this constant
three newtons that acts, multiplied by four seconds
during which it acts, we get that there's an
impulse of 12 newton seconds. And you might be like, "Wait, who cares "about newton seconds
here, I want the velocity. "I don't want the force and the time, "I want to know the
velocity at nine seconds." But they teach you at
the alien space academy, that the net impulse is not only equal to the net force times the time, it's also equal to the change
in momentum of the object that the force was exerted on. And this is good. We know the mass of the object. We wanna know something about velocity. So we know momentum is M times V, this net impulse is
gonna help us get there. But his 12 newton seconds was only for the first four seconds. How do we figure it out
for the next three seconds? Look at this. During the next three seconds,
there's not a constant force, this force is varying, the
force is getting smaller. So how do I do this? The force isn't a constant value, so I can't just simply
take force times time, because, I mean, what force do I pick? So we're gonna use a trick. We're gonna use a trick,
because if you notice for this first section,
for the first four seconds, we took the force and multiplied by the time interval, four seconds. So what we did really is we just took the height of this rectangle times the width of this rectangle and that gives us the
area of this rectangle. So what we really did is we found the area under the force versus time graph. That gave us the impulse and
that's not a coincidence. The impulse equals the area
under a force versus time graph, and this is extremely useful to know, because now in this section,
where the force was varying, we can still use this, we
can just find the impulse by determining the area under that curve. And by area under the curve,
we mean from the line curve, in general, to the x-axis,
which, in this case, the x-axis is of the time axis. So let's do this. We found the impulse
for this first section. That was 12 newton seconds. Now we can find the impulse
for this next section by just determining the area. So this is a triangle. We'll do 1/2 base, the base
is one, two, three seconds, and the height is still three newtons. So we get a net impulse
of 4.5 newton seconds. So we've got one more section to go, but this one's a little
weird, this one's located, the area is located below the time axis, so this is still a triangle, but since the forces are negative, this is gonna count as
a negative net impulse. So when the area lies above the time axis, it counts as a positive impulse, and when the area lies
below the time axis, it counts as a negative net impulse. So how much negative net impulse? We still find the area. So the area of a triangle,
again, is gonna be 1/2, the base this time is two seconds, and the height is negative two. So negative two newtons,
which gives us a net impulse of negative two newton seconds. And now we can figure out the velocity of the spaceship at nine seconds. So, assuming that this force
readout started at this moment right over here at t equals zero seconds was the moment when it was
going four meters per second, then we could say the total
net impulse should equal the total change in
momentum of this spaceship. And we can find the total net impulse by just adding up all
the individual impulses. So during the first four seconds, there was 12 newton seconds of impulse. During the next three seconds there was 4.5 newton seconds of impulse. And during this last portion there was negative two
newton seconds of impulse, which if you add all of those up, 12 plus 4.5 plus negative
two, you're gonna get positive 14.5 newton seconds of impulse. That's good news for our
alien buddies over here, they need to get off this moon, which means they need positive
impulse, upward impulse. They got some positive impulse. Let's see what their final velocity was. We know that delta P is
the change in momentum, so this is final momentum
minus initial momentum, which we could write as mass times v final minus
mass times v initial. Now, if this were an earth
rocket, this would be hard, 'cause earth rockets
using earth technology, eject fuel at a huge rate out the backend, and that loses mass. That means this mass
isn't gonna stay constant. So earth rockets
essentially push fuel down which causes and equal and opposite force back on the rocket upward. But if you're losing mass, this
mass doesn't stay constant, and this whole process is a lot harder, because m final and m initial
aren't gonna be the same. Maybe this is where the phrase,
"It's not rocket science" comes from, 'cause rocket
science is a little harder when that mass changes. So let's just say, these
clever aliens can eject only a little bit of fuel, they do so. You might say, "How?" Well, there's gotta be a certain
amount of momentum, right, that they eject to give
themselves momentum up. But let's say they can
eject only a small mass at a huge speed, so there's not much fuel that they're losing. The fuel that they eject is
ejected at enormous speed, so that they get their momentum upward, but they lose almost no mass. And that let's us solve this problem assuming that the mass is constant. So if we do that, if we
assume the mass is constant, we get positive 14.5
newton seconds equals, we can pull the mass out,
the mass is a constant, so we could just write it as m times v final minus v initial, since I can pull out a common factor of m, which means I can write
this as 2.9 kilograms multiplied by the final
velocity after nine seconds and I know it's after nine
seconds, 'cause I added up all the impulse during the nine seconds, minus the initial velocity
was four meters per second. So if I divide both
sides by 2.9 kilograms, 14.5 over 2.9 is 5 and
that will be newton seconds over kilograms which has
units of meters per second, and it's positive. And that's gonna equal
the leftover over here is gonna be v final minus
four meters per second. And now finally if I add
four meters per second to both sides, I get the
v final somewhere up here, v final of this rocket is gonna
be nine meters per second. So after nine seconds it ended up going nine meters per second. That's just a numerical coincidence. The way you find it, so
recapping the way we did this, we found the area under the curve, 'cause the area under a curve under our force versus
time graph represents the impulse on the object. We found it for the entire trip, noting that underneath the time axis, when this curve goes under the time axis, the net impulse is gonna be negative. We added up all the net impulse, we set it equal to the change in momentum, we plugged in our values and we solved for our velocity after nine seconds.