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### Course: Physics archive > Unit 6

Lesson 1: Momentum and Impulse# Impulse and momentum dodgeball example

In this video, David shows how to solve for the impulse and force applied during a dodgeball collision using the impulse momentum relationship. Created by David SantoPietro.

## Want to join the conversation?

- Is this classified as an elastic collision?(17 votes)
- An elastic collision is one in which kinetic energy remains the same before and after the impact. The only way to determine whether this specific collision is elastic is the find the final and initial kinetic energy of the ball-face system. Since the formula for finding kinetic energy, KE = 1/2(m)(v^2), requires both mass and velocity, and neither the mass nor the initial velocity of the face are provided, it cannot be determined if this collision is elastic.(26 votes)

- What is basically an impulse? (Apart from its mathematical definition).(14 votes)
- Love the question dude!

anyways, this is just how I've understood impulse intuitively, and I'm just an 11th grader studying in India,

J is like the total increase/decrease in momentum over a given period of time (I know it's synonymous with J=change P, but read this through)

basically, what do we mean by a force of, say, 5N?

= The momentum is increasing by 5Ns per s, so the total increase in p over 2s = 5n*2 = 10N right?!

so over t seconds, it would be 5t Ns.

J = force acting over a period of time, what does this lead to, a change in momentum (over that period of time) hence the 2 have an equivalent formula.

so for the above case, we can say a force of 5 N exerted over a time period of 2s causes an impulse of 10Ns which is basically the total change in momentum

I know this is basic, but simplicity is (usually) the key,

I hope this appeals to your imagination as well,

if wrong, do let me know,

hope this helps :D

Onward!(7 votes)

- What about the air resistance?Doesn't air resistance affects the impulse?(8 votes)
- In most of the cases, while calculating such questions , Air resistance is usually neglected. If we consider air resistance, friction and such other non-conservative forces, we wont be able to apply conservation of energy / momentum in the problems as they make the system more complicated and prone to errors. If we are supposed to take air resistance under consideration, the value must be given in the question. Then we can calculate the force exerted by it and subtract it from the initial momentum. But still it will make the concept more complicated.(8 votes)

- Doesn't J stand for joules? What's the difference between J in impulse and J in joules?(4 votes)
- J for impulse and J for joules are used differently, like, J=Ft or 40J of energy. thus, you can always distinguish them and thus it doesn't matter. J for joules is a unit, J for impact is a vector quantity.(17 votes)

- So the final answer of the Second question (at the bottom) is +150N or -150N ?(3 votes)
- The answer would be -150 N because of the way the speaker formulated negatives and positives. The choice between which way is negative and which is positive is up to you, but just remember to be consistent. However, if the problem had asked for the magnitude impulse (hypothetically), the answer would always be 150 N, as magnitude means asking for the absolute value.(6 votes)

- What happens if the ball impacts at an angle and leaves at the same angle, do we need to consider that since Impulse is a vector?(5 votes)
- we just need to define a +ve and -ve direction on the x axis and a +ve and -ve direction on the y axis, then break down the impulse in its two perpendicular elements and then define the impulse as +ve or -ve for both axes individually(1 vote)

- At4:00shouldn't it have been -5 instead of 5, as it comes back?(2 votes)
- No, in the problem the horizontal direction to the right was chosen as positive. In the equation the 10 m/s is negative and the 5 m/s is positive.

If you consider the initial direction of the balls movement, left, as positive then the 10 m/s is positive and 5 m/s would be negative.

Either is correct but you need to keep the directions labeled the same throughout the problem.(4 votes)

- Hi I thought you multiply the net impulse force 3kg m/s by the time interval 0.02s but it is divided to give 150N, am I missing something?(3 votes)
- if you want to get an impulse given you know the net force and time interval, you can multiply them

: impulse = net_force * change_time

but here we know the net impulse (impulse is not a force, by the way) and time interval. thus we use the same formula above but with a bit of modification

: impulse = net_force * change_time

dividing both sides with change_time

impluse/change_time = net_force * change_time/change_time

which is same as

impluse/change_time = net_force

#change_time/change_time = 1

this is why we divide impulse (not a force, once again) by change_time to get a net force(1 vote)

- Couldn't you, if you had the mass of that person's head, calculate how fast their head moved away? Since momentum is conserved for collisions?(2 votes)
- If you know the change in momentum of the ball and over what time interval, then you could calculate the acceleration of the head if you know it's mass.

F = Δp/Δt = ma

a = (Δp/Δt)/m(2 votes)

- Could we answer the question"What was the average force on the person's face by the ball?" by means other than Newton's Third Law? I tried several ways but none worked.(2 votes)

## Video transcript

- [Instructor] This person right here is about to play dodgeball. They're just unfortunately
not gonna dodge the ball. It's gonna fly in, it's
gonna bounce off their head. This may or may not have happened to you. I think this probably happened to me. It's been a long time
since I played dodgeball. And although, unfortunate for this person, it's a wonderful opportunity,
scientifically speaking, to talk about the impulse, momentum, force, time relationship, so let's do that. Let's put some numbers on here. So, we're gonna need to
know the mass of the ball. Let's say this is a .2 kilogram ball, and we're gonna need to
know some other numbers. Let's say the ball comes in at a speed of about 10 meters per second. So let's say it's comin'
in at 10 meters per second and let's say it leaves at a speed of five meters per second. So it's probably gonna recoil with a little less speed
than it came in with. It comes in with 10, leaves with five, and let's say that time, time right here, the time period that it's actually in contact with the person's face, let's say the time, when the ball is getting
kind of compressed and then recoils and expands again, let's say the time that it's actually in contact is about .02 seconds,
or about 20 milliseconds. So, knowing this information, we can ask all kinds of questions. One of them is: What was the impulse on
the ball from the person? Now, the definition of impulse, we use the letter J for impulse, that always seemed a little weird to me. There's no J in impulse. I end up calling it jimpulse, just so I can remember that it's impulse, and there's a J for it. So the jimpulse, or the impulse, is defined to be the force
acting on the object, multiplied by the time duration during which that force is acting. In other words, the impulse, from a force,
is equal to that force, multiplied by how long that
force was acting on the object. So, if we knew the force on this ball, we could use this formula
to get the impulse, but we don't. I don't know the force that this person's face
is exerting on the ball, so I can't use this formula
to solve for the impulse. But, there's an alternate
formula for impulse. If you're talking about the net impulse, in other words, the impulse from all forces
on an object, like this ball, that should just equal
the change in momentum of that object, like the change in momentum of this ball. So if we can figure out the
change in momentum of this ball, we can figure out the
net impulse on this ball. And since it's the net impulse, and this formula appears also true, this is equivalent, which is saying that it's the net force, multiplied by the time duration, during which that net force is acting. This is hard for people to remember, sometimes my students like
to remember it as Jape Fat. So, if you look at this, it looks like J-A-P, this
kinda looks like an E, F-A-T. So if you need a way, a pneumonic
device, to remember this, Jape Fat is a way to remember how impulse, change in momentum, force, and time, are all related. So let's do it. We can't use force because
we don't know it yet, but I can figure out
the change in momentum 'cause I know the velocities. So, we know that the change in momentum is gonna be P final, the final momentum, minus the initial momentum. What's my final momentum? My final momentum is M times V, so it's gonna be mass times V final, minus mass times V initial, and my mass is .2, so I've got a mass of 0.2 kilograms. My final velocity is five, because the ball recoiled to
the right with positive five. Positive five 'cause
it's moving to the right. I'm gonna assume rightward is positive. Then minus, the mass is .2 again, so 0.2 kilograms. My initial velocity is not 10. This is 10 meters per second to the left, and momentum is a
vector, it has direction, so you have to be careful
with negative signs here. This is the most common mistake. People just plug in positive
10, then get the wrong answer. But this ball changed directions, so the two velocities here have
to have two different sides, so this has to be a negative
10 meters per second, if I'm assuming rightward is positive. This leftward velocity, and
this leftward initial velocity, has to be negative 10. And, if you didn't plug that in, you'd get a different answer, so you gotta be careful. So, what do I get here if
I multiply this all out? I'm gonna get zero, no, sorry, I'm gonna get one kilogram
meters per second, minus a negative two
kilogram meters per second, and that's gonna give me positive three kilogram meters per second is the impulse, and that should make sense. The impulse was positive. The direction of the
impulse, which is a vector, is the same direction as
the direction of the force. So, which way did our face
exert a force on the ball? Our face exerted a force
on the ball to the right. That's why the impulse on
the ball is to the right. The impulse on this person's
face is to the left, but the impulse on the
ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and
bounce back to the right. That's why this impulse has
a positive direction to it. Now, if you've been paying attention, you might be like, wait a minute, hold on. What we really did was we found the change
in momentum of the ball, and when we do that, what we're finding is the
net impulse on the ball. In other words, the impulse
from all forces on the ball. But what this question was asking for was the impulse from a single force. The impulse from just the person's face. Now, aren't there other
forces on this ball? Isn't there a force of gravity? And if there is, doesn't that mean what
we really found here wasn't the impulse from just our face, but the impulse from the person's face and the force of gravity
during this time period? And the answer is no, not
really, for a few reasons. Most important reason being that, what I gave you up here was the
initial horizontal velocity. This 10 meters per second
was in the X direction, and this five meters per second, I'm assuming is also in the X direction. 'Cause if I'm taking the
initial velocity in the X, and the final velocity in the X, and I take the difference in momentum, what I really found was
the change in momentum in the X direction. When I do that, I'm finding the net
impulse in the X direction, and there was only one X
directed force during this time and that was our face on the ball, pushing it to the right. There was a force of gravity. That force of gravity was downward. But what that force of gravity does, it doesn't add or subtract any
impulse in the X direction. It tries to add impulse
in the downward direction, in the Y direction, so it tries to add vertical
component of velocity downward, and so we're not even
considering that over here. We're just gonna consider
that we're lookin' at the horizontal components of velocity. How much velocity does it
add vertically, gravity? Typically, not much during the situation, because the time period during which this collision acted is very small and the weight of this ball, compared to the force that our face is acting on the ball with, the weight is typically much smaller than this collision force. So that's why, in these
collision problems, we typically ignore the force of gravity. So, we don't have to
worry about that here. That's not actually
posing much of a problem. We did find the net
impulse in the X direction since our face was the
only X directed force, this had to be the impulse
our face exerted on the ball. Now, let's solve one more problem. Let's say we wanted to know: What was the average force on this person's face from the ball? Well, we know the net impulse on the ball, that means we can figure out
the net force on the ball, because I can use this relationship now. Since I know that the net impulse on the ball in the X direction should just equal the
net force on the ball in the X direction, multiplied by the time interval during which the force was applied, I can say that the net impulse on the ball was three kilogram meters per second, and that should equal
the net force on the ball in the X direction, which was supplied by,
unfortunately, this person's face, multiplied by the time interval, which is 0.02, 20 milliseconds. So, now I can solve. The force, the net force on the ball, during this time interval
in the X direction, was three, divided by .02. If I take three kilogram meters per second and I divide by .02 seconds, I'm gonna get 150 Newtons was
the net force on the ball. We got a positive number,
and that makes sense, because this person's face exerted a positive force on this ball, 'cause the force was exerted to the right. So, these were positive, and the impulse from the face on the ball should be going the same direction as the force from the face on the ball. So, this is the force on the
ball by the person's face, but notice this question is asking: What was the average
force on the person's face from the ball? Not on the ball by the face. You might think, oh no,
we gotta start all over, we solved for the wrong question, but we're in luck. Newton's third law says that the force on the face from the ball
should be equal and opposite. So, this force on the face from the ball has got to be equal and opposite to the force on the ball from the face, so that's what we found here, the force on the ball from the face, that means the force on
the face from the ball is gonna have the same size. It's gonna be 150 Newtons. It's just gonna be directed
in the leftward direction, that means it's gonna be a negative force, so technically, you could say, this would be negative
150 Newtons on the face from the ball. So, to recap, the impulse,
from an individual force, is defined to be that force, multiplied by the time interval during which that force is applied. And if you're talkin' about the net force in a given direction, multiplied by the time interval, you'd be finding the net
impulse in that direction, and this also happens to
equal the change in momentum in that direction. So, in other words, if there is a net impulse
in a given direction, there's gotta be a change in momentum in that direction by the same amount. And one convenient way to
remember how are these related, is you could use the
pneumonic device, Jape Fat. I have no idea what Jape Fat means, but it helped me remember
that the net impulse equals the change in momentum, and that also equals the net force, multiplied by the time interval, during which that force was applied. And finally, remember that
during these collisions, there's always an equal
and opposite force exerted on the two objects
participating in the collision.