Practice: You throw a 100-g ball with 30 m/s. If the ball is in your hand for 0.2 s, find the impulse you deliver to it.

EXTRA: Calculate the average force that you exert on the ball.

Subjects

Sections | |||
---|---|---|---|

Intro to Momentum | 17 mins | 0 completed | Learn |

Intro to Impulse | 42 mins | 0 completed | Learn |

Impulse with Variable Forces | 17 mins | 0 completed | Learn |

Intro to Conservation of Momentum | 20 mins | 0 completed | Learn |

Push-Away Problems | 29 mins | 0 completed | Learn |

Adding Mass to a Moving System | 14 mins | 0 completed | Learn |

How to Identify the Type of Collision | 13 mins | 0 completed | Learn |

Inelastic Collisions | 16 mins | 0 completed | Learn |

2D Collisions | 22 mins | 0 completed | Learn |

Newton's Second Law and Momentum | 11 mins | 0 completed | Learn |

Momentum & Impulse in 2D | 25 mins | 0 completed | Learn |

Push-Away Problems With Energy | 12 mins | 0 completed | Learn |

Elastic Collisions | 8 mins | 0 completed | Learn |

Collisions & Motion (Momentum & Energy) | 68 mins | 0 completed | Learn |

Collisions with Springs | 9 mins | 0 completed | Learn |

Intro to Center of Mass | 14 mins | 0 completed | Learn |

Concept #1: Impulse & Impulse-Momentum Theorem

**Transcript**

Hey guys. So, now that we have a basic understanding of what momentum is and how it works I want introduce you to a new concept that's related to momentum it's called impulse, I'll show you the impulse is and how it's tied to momentum, let's check it out. So, first I want to show you, to get started that Newton's second law, Newton's second law is f equals m, a, can be rewritten in terms of momentum, check this out. So, sum of all forces equals m, a instead of writing a, I'm going to write the definition of a, which is a change in velocity over change in time. So, I'm going to replace it or Delta V over delta t and now, I'm going to move the V inside of the Delta, you can do that it's, cool? So, the Delta now has an m, v inside of it delta t and now that you know what momentum is, hopefully you identified it, you notice that this is action momentum p equals m, v. So, I can rewrite this as Delta p over delta t, so the idea that FD equals m, a can be written here, like this or it can be written like this and this is actually Newton's original definition, when Newton figured out f equals m, a he actually didn't write it as f equals m, a, he wrote it in terms of changing momentum over change of time, we just learned it this way first because it's simpler, right? I hope you agree that this is simpler than this, okay? So, but you can write it this way. So, by rearranging this new version, f equals Delta p over delta t, we'll get a new physical quantity, let me show you, sum of all forces equals Delta p over delta t and what I'm going to do is I'm going to move the delta t over here. So, it's going to look like this, sum of all forces delta t equals Delta p, this thing here equals the change in momentum and we're going to give it a name, when you apply a force on something over a certain amount of time, right? There's a box you push it for five seconds, that is called impulse and it's given by the letter J, okay? So, the new physical quantity we're talking about is impulse, which is given by the letter J, why J? I don't know, I suspect it's because they had already assigned I to something else, I is moment of inertia the next letter is J, okay? So, J is defined as f delta t and it causes a change in momentum. So, that's sort of the full equation for J, it's going to get a little bit more, a little bit longer in just a bit but I want to make a quick point here, that J equals or is defined as, that's the definition of J, sometimes definitions are shown in terms of like a three dash equal sign, it's equal to and define as, but it causes, it causes a change in momentum. So, impulse is force times time and it causes a change in momentum, this is similar to work. Now, I have a little asterisk here, because depending on the sequence that your book takes your professor takes you may or may not seen work yet, most of you will already have two covered work and energy and then you'll be able to understand this analogy, work is Force times Delta x. Now, it's f Delta x cosine of theta but this is the simpler version f delta x and it's defined as this, and it causes a change in kinetic energy, causes a change in kinetic energy. So, I hope you can see how these are similar, if you push something with a force of 10 for 5 seconds, that's impulse, if you push with a force of 10 for 5 meters its work, one causes the change in momentum and remember, momentum is m, v and the other causes a change in kinetic energy and if you've seen this kinetic energy is half m, v squared you might even notice how these things are similar, these two guys here are similar to each other, okay? So, these two I like to put them side-by-side. So, you can see that they look very similar, might help you remember how this stuff works, but we're not going to be talking about work or energy in this chapter. So, at least not for a while. So, because impulse can be written in two ways, impulse can be written like this or like this then it's going to have two sets of units and they're going to be equivalent sets of units. So, force is measured in Newtons and time in seconds and then the other way is the change in momentum, change the momentum has units of momentum, kilograms meters per second. Now, Newton, Newton's f equals m, a is measured in Newtons, one Newton is the equivalent of a kilogram times meters per second squared. So, if you replace Newton with all of this here and then multiply by sm, I'm not going to do it doesn't matter that much, you'll see that you end up with exactly this. So, if you're curious you can check for yourself or you can just trust me that that's how it works those two sets of units are identical, not identical but they're equivalent. So, you can use whichever one, I'm going to use n s just because it's less writing to do, okay? So, let's do an example here and I have a practice problem for you guys. So, a 50 kilogram crate is, initially at rest, let's draw this 50 kilogram crate, initially at rest, on a smooth no friction horizontal surface and then you push on it with a constant horizontal 100 Newton. So, 100 Newton is a force, it's a constant force and it's a horizontal force, F equals 100, for eight seconds. So, I'm going to say that it goes from here to here and the time between these two points is 8 seconds. So, here the initial velocity is 0, here there's some sort of final velocity, which is what we want in Part B, what is the final velocity, but the first part is asking for the impulse again impulse isn't I there are a few books, very few that use I as impulse but a vast majority books are going to use J. So, what is J, the impulse that you deliver to the crate, you're giving impulse to the crate kind of like as if you were giving energy to the crate but it's not energy, it's impulse, it's force over time or force for a period of time, force for a certain period of time. So, let's do this, the way to calculate, this is J is, I'm going to show you now the expanded version of the J equation that I slightly quickly hinted at earlier. So, J is f delta t there's not much I can do here and it's also Delta p. Now, I'm going to expand Delta p, the delta operator means final minus initial. So, it's going to be p final minus p initial, I'm going to make this really long and I'm going to make it a little shorter so don't freak out p is m, v. So, p final is ,v final minus m, v initial, if the mass doesn't change, which in 99% times of the problems it won't change. So, you're good there, you can combine the masses by factoring out the m and it's going to look like this, okay? I know what you're thinking, wow, this is crazy, it's a long equation but it's really this, this, this and then this whole thing was just to get you to the simpler version, which you should know how to go from Delta p to this. So, in most of the impulse problems as you see as I solve them, I'm going to write sort of this longer version to include this and this, the reason for that is because impulse problems will give you a different combination of variables and if you have this whole thing laid out it's easier for you to see which way to go, okay? So, here I want impulse, I want J, and what do I have? Do I have f and T? Do I have delta p? Do I have mass and velocities? Well, I have the force and I have the time so this is going to be, I can already stop right there and I know that I can find the impulse by just using those two numbers. So, J equals f Delta team that nothing else matters because already got it, force is 10, time is 8, I'm delivering 800 of impulse, where the units are force Newton's time seconds, NS, cool?

For Part B, we want to know what is the final velocity and if you go back to the impulse equation right here, final velocity is right there. So, I can simply write J all the way to the left because I had J and then this stuff all the way to the right because I want to v final, right? So, when you're building equations are left and right, one side needs to be what you have and the other side what you don't have or at least you can only have one unknown. So, I have J equals m, V final minus V initial, V initial is 0. So, this becomes even simpler, J equals m, V final, so v final is j over m, j is 800, m is 50 and if you divide the two you get 16 meters per second, okay? So, these are the two final answers. Alright, so hopefully that makes sense, let me know if you're stuck in anything, I have practice problem here for you to try, give it a shot, let's see if you get it.

Practice: You throw a 100-g ball with 30 m/s. If the ball is in your hand for 0.2 s, find the impulse you deliver to it.

EXTRA: Calculate the average force that you exert on the ball.

Example #1: Intro to Impulse

**Transcript**

Alright, so here we have a baseball that is moving horizontally with 40 before coming to contact with a bat, with a bat and then it's going to move with 45 in the opposite direction right after that, So it doesn't tell us, the question doesn't say if the balls are initially moving to the left or to the right. So, we're going to pick one and I'm going to say the ball is initially moving, 0.45 is the mass, always in kilograms, I'm going to say the ball is initially moving this way with 40 meters per second then it's going to hit the bat, I'm going to draw a bat, this is probably going to come out terrible. So, it's like a cricket bat and then it's going to come back this way with 45 meters per second. Remember, please be careful, opposite directions means you have to have different signs, opposite signs. so this is, we're going to call this positive because it's going to the right and this is negative to the left, the Bat is in with the ball, with the ball for 0.03 seconds, I'm going to call that Delta T 0.03 seconds, and we want to know the impulse delivered to the ball and the impulse delivered to the wall I'm asking for the impulse for two things and the point here is to make sure you understand that there is impulse happening on both, just like when you push something, a force, that something pushes back on you, action reaction, when you impulse something, that object is also in pulsing you back because it's also applying a force. Remember, impulse is force, time, that object is also applying a force back at you for the same amount of time that you are applying a force at the object, if you want the impulse on the ball, I'm going to call this J ball, it is f, delta t but this is f on the ball equals changing p, this is the change in momentum of the ball, okay? So, this is J on the ball, force on the ball, change the momentum of the ball, this is mass of the ball, and these are the velocities of the ball. So, everything is of the ball, the only difference is that this is a force on the ball, right? When we write forces it's about, about the object that has, that is receiving the force, cool? So, that's how I can calculate this stuff, I don't know the force on the ball, I do know the time, I don't know the change in momentum, this is what we're looking for right here, I know the mass and I know the two velocities. So, I can use this to calculate it, it's going to be pretty straightforward, mass of the ball is 0.145, the final velocity of the ball. So, one way to think about this is that before the collision is your initial and then after the collision is your final, so this is a initial, final. So, final is negative 45 and initial is positive 40, when you do this, I'm going to skip a step here, when you do this you get negative 12.3 Newtons, seconds, I got a negative, does it make sense that my impulse is negative? Well, impulse is a force times the time. So, you can think of impulse as a force, if I am hitting a wall and going that way as I'm hitting the wall I'm being pushed this way, right? I'm going this way the wall pushes me. So, I end up going that way. So, this makes sense, it's because you're being pushed and you're being impulsed to the left, okay?

The second question B is to find the impulse on the wall. So, one way you could have thought about this, though this is the wrong way but I'll show you real quick, you could have tried j wall equals f wall, delta t, change momentum of the wall. and now it starts to get tricky because the doesn't really going to change velocity, right? The wall is not moving and I'll address that a little bit more in a second, so this is just the wrong way to think about this, you're not going to calculate the velocity of the wall but the way to do this is by realizing that J wall is the same as J ball because they're pushing on each other, except in the opposite direction, right? So, if the ball is being impulsed to the right, it comes and gets pushed that way then the wall is being impulsed to the, I'm sorry, the ball being impulsed to the left, right? Come this way gets pushed that way, the wall is being pushed this way, it doesn't move but it does get pushed that way. So, since it's been impulsed to the right, it's going to be not negative 12 but positive 12.3, okay? And this, I'm going to put the unit's right there, and this is what this little note talks about, right? Because of Newton's third law, action reaction, the force of A and B is the same of the force of B and A, forces are the same but opposite direction, same thing here, the impulse of A and B is opposite to the impulse of B and A, and also if you look at this equation, notice how J equals change in momentum and this also tells us that the change in momentum of one object is the same as the change in momentum of the other except in the other direction, when two objects collide, if object A loses 10 of momentum it's because object B gained 10 of momentum, cool? So, the answer is 12, negative 12 and positive 12. Now, the last thing I want to talk about real quick is this idea that the law moves or actually wall doesn't move. So, where does that impulse go if the wall doesn't move, and this is a little weird, but you may remember something like this, when you jump you push up against the earth, there's an action reaction, therefore, I'm sorry, you're pushing down against the earth so the earth pushes you up, so the earth is pushing you this way and you are pushing the earth down you go up and supposedly, technically the earth goes down but if you want to work out the numbers it turns out that the earth moves down a ridiculously irrelevant number, there's also animals jumping on the other side, which cancel that, but supposedly, technically as you jump up the earth gets pushed down? Well, what happens to a wall that is hit by an object and doesn't move is something similar. So, I'm going to put the earth here, let's build, I don't know, the Great Wall of China whatever, it's fixed on the earth, right? And then a ball is going to hit it and bounce back? Well, what happens is that because this is fixed to the earth here, it's going to act like a lever and cause the earth to turn a little bit obviously that little bit is completely relative significant and doesn't actually do anything, right? It's a tiny, tiny amount that we just say is basically 0, right? Otherwise, we could throw a bunch of balls on against the wall to make the earth spin a little bit faster, a little bit slower, cool? So, that's the idea, you can just think of it as it doesn't do anything and the J on the wall is just whatever the J on the ball is and you can't calculate J wall but you can calculate J ball and then make them the same, cool? That's it for this one, let me know if you have any questions.

Example #2: Momentum & Impulse

**Transcript**

Hey guys. So, let's solve a few more momentum impulse questions, let's check this first one out, we have a 1200 kilogram car that collides with 20 meters per second against the wall and then it stops after 1 meter of the car being compressed against the wall, so the front part of the car gets all squished up, right? So, let's draw this real quick, 1200 kilogram car, let's put some wheels, is moving with 20 meters per second when he hits a wall. Now, the car stops in a distance of 1 meter. So, from here to here is 1 meter, that's a delta x of 1 meter or d, the initial velocity over here is 20 and the final velocity here is 0, so the first question is, assuming a uniform deceleration, so the deceleration is at the same case same rate, how long is the collision last. So, we want to know what is the delta t here. So, since, we're talking about momentum on impulse you might think immediately of this equation and that's a good thing but I'm going to show that that equation is actually not going to work, another reason why you might think of this equations because of collisions. So, even though even outside of discussing this topic specifically like, if you have a test, that's mixing this chapter with a bunch of other chapters, how would you think of impulse, you would think of impulse and momentum because there is a collision going on, right? So, as long as whatever you see a collision you should thinking should be thinking of a momentum impulse, but this equation is not going to work because once you write it out, let me put here that part a, we're looking for delta t, once you write it out J, f, delta t, Delta p, m, v final, v initial, you realize that you don't have J, you're looking for T, you don't have f, you don't have Delta p and you have these three guys here, but that's not enough to solve because if you try to set this up like this, f delta t, this is what I'm looking for, equal to the right side, which is everything I have you will notice that I'm missing f here and without f I can't solve this, so this is not solvable using this equation, the only other recourse you have is to use motion equations and you will be able to use motion equations because it says, here, uniform deceleration those three or four equations of motion can only be used if you have uniform acceleration or deceleration, okay? So, we're going to use motion equations to find time, I'm going to list my five variables here, v initial, which is a 20, v final, which is a 0 acceleration I don't know, delta t is what we're looking for and Delta x is 1, hopefully remember this stuff if you don't you can go back and review, the acceleration is my unknown variable or my ignored variable rather, if you look through the four equations of motion, the equation that you're supposed to use when you don't care about the acceleration is what I call the fourth equation, which most professors will be fine with you using, some professors don't give you that equation and some professors don't even want you to use that equation, it's kind of weird but whatever, in case you know that your professor doesn't want you to use this equation and only want to use the other three you're going to have to use two out of the three equations you're first going to have to find a. So, if you can't use equation for which we'll read in a second, you're going to have to first find a then t. So, you're going to have to use two equations instead, just in case that shows up I want to remind you of this stuff because it probably been a while since you've done this. So, equation number four says that Delta x is V initial plus V final over 2 times delta t, hopefully you can use this and it would be easier. Alright, so let's move some stuff around, Delta x is 1, v initial is 20, v final is 0 and this is 2. So, that's your delta t. So, this becomes a 10. So, your delta t is 1 over 10 or 0.1, the collision is 0.1 seconds, that's it for that first part.

For part B, it says, use momentum impulse, meaning this equation up here, to find the average force exerted on the car, the reason why these questions says specifically use momentum and impulse is because there are other ways you could have done this, if you wanted to find the average force in the car you could have done f equals m, a and then once you calculate the acceleration right here, then you'd be able to plug in the mass and you will know the answer but we don't want to do it that way, we want to do it using momentum and impulse but I do want to point out the fact that sometimes there's multiple ways of doing these things, right? So, find the average force, J equals f delta t, Delta p, m, v final, v initial, I want the force, I know the time and I know these three things, I don't know J, I don't know Delta p, so the best way to do it is to just set this side equal to this side, f delta t equals m, v final minus v initial, v final is 0. So, this just becomes m, V initial with a negative there, right? So, delta t is negative m, v, I, I'm sorry, f, sorry, f, Delta t, f is what we're solving for, Delta T goes under, looks like that. So, if you're plugging these numbers, I have it here, if you plug in these numbers you get really big number actually. So, 1200 initial velocity 20, this is 0.1 so the answer is negative 240,000 Newton's, okay? It's negative because it's a stopping force, the force is to the left and the reason why you have such a big number is because you're stopping a heavy car, that's moving really fast in just a meter, think about a collision, how much that hurts, that's why you get a big f there, so that should make sense, cool? That's the final answer, I got a practice problem here, let's give it a shot.

Practice: You catch a 624-g basketball originally moving with 10 m/s. Calculate the impulse delivered to the ball.

Practice: When a 300-g ball is dropped from 3 m, it hits the floor and rebounds to a height of 2 m. If the ball is in contact with the floor for 0.02 s, what is the average force exerted by the floor on the ball?

Practice: A box of unknown mass on a flat surface slows down from 12 m/s to 4 m/s after crossing a rough patch. The box-patch coefficient of friction is 0.5. Use impulse/momentum to find how long (in s) the box takes to cover the entire patch.

A block with mass 5.0 kg is initially at rest on a horizontal frictionless surface. Then a constant horizontal force F is applied to the block. After the force has been applied for 4.0 seconds the speed of the block is 12.0 m/s. What is the magnitude of the force?
A) 15.0 N
B) 3.0 N
C) 2.5 N
D) 30.0 N
E) None of the above answers

A 0.24 kg blob of clay is thrown at a wall with an initial velocity of 21 m/s. If the clay comes to a stop in 91 ms, what is the average force experienced by the clay?
A. 46 N
B. 34 N
C. 55 N
D. 67 N

A golf club exerts an average force of 1000 N on a 0.045-kg golf ball which is initially at rest. The club is in contact with the ball for 1.8 ms. What is the speed of the golf ball as it leaves the tee?A) 45 m/sB) 50 m/sC) 35 m/sD) 30 m/sE) 40 m/s

A 1000-kg car is traveling at 20.0 m/s toward the north (let that be the +y-direction). During a collision, the car receives an impulse of 1.00 x 104 N•s toward the south. What is the velocity of the car after the impulse is applied to the car?
A) 10.0 m/s south
B) 0.00 m/s
C) 20.0 m/s north
D) 10.0 m/s north
E) 30.0 m/s north

After a 0.300-kg rubber ball is dropped from a height of 1.75 m, it bounces off a concrete floor and rebounds to a height of 1.50 m. (a) Determine the magnitude and direction of the impulse delivered to the ball by the floor. (b) Estimate the time the ball is in contact with the floor and use this estimate to calculate the average force the floor exerts on the ball.

A baseball of mass 0.190 kg moving at 30.0 m/s strikes the glove of a catcher. The glove recoils a distance of 8.00 cm. The magnitude of the average force applied by the ball on the glove is
A) 71.3 N
B) 1.07 x 103 N
C) 0.731 N
D) 10.7 N
E) 2.14 x 103 N

A rubber ball of mass m is released from rest at height h above the floor. After its first bounce, it rises to 90% of its original height. What impulse (magnitude and direction) does the floor exert on this ball during its first bounce?

In all parts, consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70×10-3 seconds.A) Assuming that this force is a constant, what is the magnitude J of the impulse on the ball?B) What is the shape of the net force vs. time graph?C) Why is the graph that shape?

It is correct to say that impulse is equal toA. momentum.B. the change in momentum.C. the force multiplied by the distance the force acts.D. all of the aboveE. none of the above

In the impulse approximation, __________.a. the forces between colliding objects can be neglectedb. no external forces act on the systemc. the true impulse is approximated by a rectangular pulsed. external forces can be neglected during the time the impulsive force acts

What represents the impulse of the force in a graph of force versus time?a) The impulse is equal to length of the curve.b) The impulse is equal to the slope of the curve.c) The impulse is equal to the area under the curve.d) The impulse is equal to the product of the maximum force times the maximum time.e) Impulse cannot be determined from this type of graph.

True/False(a) As a tile falls from the roof of a building to the ground, its momentum is conserved.(b) When a baseball player hits a home run, the baseball receives a greater impulse from the bat than the bat does from the ball.(c) In a ballistics test, a bullet traveling horizontally strikes and embeds itself in a large block that is free to move. During this process, the bullet transfers all of its momentum to the block.

1. Find the magnitude of the impulse J delivered to the particle. Express your answer in terms of m and v. Use three significant figures in the numerical coefficient. 2. Which of the vectors below best represents the direction of the impulse vector J?

A glass soda bottle is emptied of soda and filled to the very top with water. A cork is carefully fitted into the top of the bottle, leaving no air between the cork and the water. The top of the bottle has a diameter of Dtop = 2.00 cm and the bottom of the bottle has a diameter of Dbot = 6.50 cm. The glass breaks when it is exposed to pmax = 70.0 MPa of pressure.A student hits the cork sharply with her fist and the bottom of the bottle breaks. The student's fist has a mass of m = 0.480 kg and moves downward at a speed of vi = 5.00 m/s. It collides elastically with the cork and rebounds with the same speed. The collision lasts for t = 1.20×10-4 s. In this problem, the positive direction is upward.What is the force that the fist exerts on the top of the bottle?Express your answer in newtons.

A 57g tennis ball is served at 45m/s. If the ball started from rest, what impulse was applied to the ball by the racket? Express your answer using two significant figures.

A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. Assume constant acceleration of the hammer-nail pair. a. Calculate the duration of the impact.b. What was the average force exerted on the nail?

Learning Goal:To learn about the impulse-momentum theorem and its applications in some common cases.Using the concept of momentum, Newton's second law can be rewritten as∑F? =dp? dt, (1)where ∑F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives?t2t1?F? dt=?t2t1dp? dtdt. (2)The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten asJ? =p2??p1?.This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written asF(t2?t1)=mv2?mv1. (3)Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x axis. The net force F is acting on the particle along the x axis. F is a constant force.)A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.C) The particle has momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?tD) The particle has momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?

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