Impulse with Variable Forces: Study with Video Lessons, Practice Problems & Examples

Impulse is calculated using the area under a force versus time graph, where positive areas indicate positive impulse and negative areas indicate negative impulse. The impulse-momentum relationship states that impulse equals the change in momentum, expressed as $J^{=}(mv\_f-mv\_i)$. For a toy car, calculating the areas under the graph yields an impulse of 12.5 N·s, leading to a final speed of 6.25 m/s when starting from rest.

Impulse with Variable Forces

Guys, in some problems, you're going to be given force versus time graphs like this graph over here, and you're going to be asked to calculate the impulse. So I'm going to show you how to calculate impulse by using these graphs here, but the idea is going to be very straightforward. It's going to be very similar to something we've already seen before. Now when we talked about work, we said that one way to calculate work was if you're given an f versus x graph, the area under this graph was going to be equal to the work, and we said that areas above the x axis were going to be positive work. So if you have an area like this, this is going to be positive work because your force is positive here on this side of the axis. Areas below the x-axis were negative works because the force is negative. It's the same idea here when you have f versus t graphs. Right? It's except instead of calculating the area under this graph and that being the work, the area underneath these f versus t graphs is really just going to be the impulse. Alright. But the rule is actually going to be the same, right? So areas above the x-axis are actually going to be positive impulses because the force is going to be positive. Right? So you have a positive force, a positive impulse, and then areas below are just going to be negative impulses like this. So the idea is very similar. It's just really the letter that's different f versus delta t. Alright? So let's take a look at an example here. So we have a remote controlled car that is moving forwards and backwards, and it's sort of due to this changing force here that we have. So we want to calculate the impulse that's delivered to the toy car. So remember that your momentum, or sorry, that your impulse in part a is going to be f×Δt. Except we can't really use this formula because this force is constantly changing over time. So we say here that this is actually just equal to the area that's underneath this curve right here. So what I'm going to do is say that the impulse I'm just basically going to break up this graph into 2 different sections here. So I'm going to call this guy right here, the area from 2 to 4. I'm going to call this a1. And this area over here from 4 to 9 seconds. Right? So this is like this is t equals 9 here, is I'm going to call this a2. And really to calculate the impulse, all I have to do is just add together these areas here, break some stuff up into a bunch of triangles and rectangles, and then I just, you know, just add everything up together. Alright. So let's check this out. So the area 1. So, really, if I sort of break this up using this triangle right here, using this line, this is going to be a triangle. So the base is equal to 2 and the force is equal to 10. So remember that the area of a triangle is 12bh, this is going to be 122×10, and that this is going to be the area of this guy right here, which is the area of the rectangle, which is just the base, which is 2, and then the area and the height, which is 10. So if you go ahead and work this out, what you're going to get is, you're going to get 30. You're going to get 30 right here. Alright. So now we're going to do the exact same thing for area 2. We're going to calculate the area of all the negative stuff that's happening on the right side of the graph. So this is a big trapezoid over here. I'm going to break this up into a triangle, a smaller square, and then a smaller triangle. So let's get started here. This is going to be a triangle 12 base times height. The base of this is 2, and the height of this is actually negative 5. So it's going to be 122×(-5). You always have to plug in the negative signs here on the y-axis to get your negative impulses. Then we're going to add the square. The square is just 2 times 5 base times height. So, this is going to be 2×(-5). And then finally, this last piece over here, just be careful because the base of this is actually 1, not 2 because we're going from 8 to 9, but the height of this is still negative 5. So 121×(-5). And if you work all this out, you're actually going to get negative 17.5 Newton seconds. Alright? So now that we have the 30 and the negative 17.5, our impulse is just going to be putting those two things together. So it's 30 plus negative 17.5 for a grand total of 12.5 Newton seconds. And that's your answer. So that's the answer to part a. That's the impulse. It's still a positive impulse because the area of the yellow is bigger than the area of the green. Now let's move on to the second part over here. So in the second part, we're told that if the car has a mass of 2 kilograms and starts from rest, we're going to calculate the final speed. So, in other words, we're going to calculate v final. Alright? How do we do that? Remember our equation here for impulse? It's that j is equal

An object experiences a force given by the graph below. What value of F_max would give an impulse of 6 N⋅s?

• A. 1000 N
• B. 4000 N
• C. 2000 N
• D. 6000 N

Impulse of a Baseball Bat

Hey, everybody. So in this problem, we have a baseball bat that's hitting a bat, and we have this graph here that shows the force over time. So it's a massive amount of force, but it happens over a very, very small amount of time. This scale is in milliseconds here. Let's jump into the first part here. We want to calculate the impulse that is delivered to the baseball. So, in other words, what we want to calculate is j. Now, remember, j, whenever you're given a force versus time graph, is going to be the area that is under the curve. So in other words, you just take this graph here, you split this up into a bunch of triangles and squares and stuff like that, and you just calculate the area that's under the curve. So this shape is kind of like a triangle, but it's kind of weird because this side here is a little different than the right side. But what we can do is we can break it up into 2 smaller right triangles. And I'm going to call this one A1 and A2. So to calculate the impulse, we're going to calculate the area and that's really just the area of A1 + A2. Alright? So let's go ahead and write out these formulas here. Well, A1 is going to be this has this is a base. I'm going to call this base 1, and this has a height of h. The other triangle has a base I'm going to call base 2, and it also has the same height. So it's the h this is the same h for both of the triangles. So that means that the area formula for area 1 is going to be 1 half of base 1 times height plus 1 half of base 2 times height. And what we can do here is we can just do a little math or geometry trick, which is we're going to combine these 2 bases together. This is going to be 1 half base 1 plus base 2 times height. So, really, all you need to calculate, for any triangle, doesn't have necessarily have to be a right triangle, is you need the base, the whole entire thing, times the height, and then you multiply it by 1 half. Alright? That's just a, you know, cool little shortcut. But anyways, so we're just going to go ahead and plug this in. So we've got the base one, which is going to be what? Are we going to plug in 4? Or remember, this scale here is in milliseconds, and we have to convert it to the right units. 4 milliseconds is really just 0.004, so be careful here because this 4 here is going to be 0.004 seconds. Likewise, what happens here, is we have to do the base 2. This is going to be 8 milliseconds which is going to be 0.008. So we add that in, 0.008, and then you have to multiply by the height of the whole triangle, which is 1500. When you work this out, what you're going to get is an impulse of 9 newton seconds. Alright? So even though the force is massive, it's 1500, it acts over a very small amount of time, and so you end up with something, you know, like 9, which isn't a super huge number. Alright? That's the answer. That's the impulse. Let's move on to the second part here. Now we want to calculate the final speed of the baseball right after the impulse. Right? So after the bat hits the ball and it goes off, we want to calculate what is the speed after that impulse is delivered. So how do we do that? Well, remember that the impulse can only always be related to the change in the momentum. Another formula that we have here is that j is equal to force times time, but it's also equal to basically the change in the momentum. So we have mv_final minus mv_initial. What we really want to calculate here is v_final. So if you look through, what happens is that the second term, the mv_initial, this is the initial momentum, is going to be 0. What happens here is that there is no velocity. There's no initial velocity because the baseball is initially at rest. Right? So there's no initial speed here. So that whole entire term goes away. So what's the j? Well, we just calculated it. It's the 9 newton seconds. So this is going to be 9 is equal to, and then we've got the mass. The mass is 200 grams, so this is going to be 0.2 kilograms. This is going to be 0.2 v_final. So if you work this out, what you're going to get here is 9 divided by 0.2, and that's a final speed of 45 meters per second. And that's pretty reasonable because this is about, you know, let's say, a 100 miles an hour, and that's pretty reasonable for a baseball. Right? You get you know, you hit the baseball and it goes off, at a 100 miles an hour. That's somewhat reasonable. Alright? So that's the answer. Let me know if you have any questions.

Here’s what students ask on this topic:

How do you calculate impulse from a force vs. time graph?

To calculate impulse from a force vs. time graph, you need to find the area under the curve. The impulse (J) is the integral of the force (F) over the time interval (Δt). For a graph with varying force, break the area into geometric shapes like triangles and rectangles, calculate their areas, and sum them up. Positive areas above the x-axis contribute to positive impulse, while areas below the x-axis contribute to negative impulse. The total impulse is the net area under the curve.

What is the relationship between impulse and momentum?

The relationship between impulse (J) and momentum (p) is given by the impulse-momentum theorem, which states that the impulse applied to an object is equal to the change in its momentum. Mathematically, this is expressed as:

$J=\mathrm{\Delta p}=mv$_f-mv_i

where m is the mass, v_f is the final velocity, and v_i is the initial velocity. This means that the impulse is equal to the product of mass and the change in velocity.

How do you find the final velocity of an object using impulse?

To find the final velocity (v_f) of an object using impulse, you can use the impulse-momentum theorem. First, calculate the impulse (J) from the force vs. time graph. Then, use the equation:

$J=mv$_f-mv_i

If the object starts from rest (v_i = 0), the equation simplifies to:

$J=mv$_f

Solve for v_f by dividing the impulse by the mass (m):

$v$_f=Jm

What is the significance of positive and negative areas in a force vs. time graph?

In a force vs. time graph, the areas above the x-axis represent positive impulse, while the areas below the x-axis represent negative impulse. Positive impulse indicates that the force is acting in the positive direction, increasing the object's momentum in that direction. Negative impulse indicates that the force is acting in the negative direction, decreasing the object's momentum or changing its direction. The net impulse is the sum of these positive and negative areas, determining the overall change in momentum.

Why can't you use the formula J = FΔt directly for variable forces?

The formula J = FΔt assumes that the force (F) is constant over the time interval (Δt). For variable forces, the force changes over time, so this formula is not applicable. Instead, you need to calculate the impulse by finding the area under the force vs. time graph, which accounts for the varying force. This method involves breaking the graph into geometric shapes, calculating their areas, and summing them to find the total impulse.