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Momentum & Impulse

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Question 1 0:19
  • Question 2 2:17
  • Question 3 3:25
  • Question 4 3:56
  • Question 5 4:28
  • Question 6 5:04
  • Question 7 6:18
  • Question 8 6:57
  • Question 9 7:47

Transcription: Momentum & Impulse

Hello everyone and welcome back to Educator.com. 0000

In this mini-lesson we are going to go over the first page of the APlusPhysics worksheet on momentum and impulse and you can find that worksheet by clicking on the link down below.0002

Take a minute or two, print it out, try and go through it, see if you can solve them and then check here and we will see if you got them right. 0012

Problem 1 -- We have a 1200 kg car traveling at 10 m/s, so mass is 1200 kg traveling at 10 m/s and it is brought to rest in 0.1 s, so this must be our initial velocity; our final velocity is 0, and the time it took is 0.1 s. 0019

What is the magnitude of the average force acting on the car to bring it to rest? 0042

In order to solve this one, I am going to have to use the relationship that impulse, which is a change in momentum, is equal to force times time, but our change in momentum -- well that is going to be change in mass times velocity equals force times time. 0047

But mass is not changing, so on the left hand side -- this will be mass times change in velocity equals force times our time interval and δv -- our change in velocity -- that is final minus initial...0060

...so that will be mass times final velocity minus initial velocity equals Fδt. 0079

Finally I am actually looking for force, so let us rearrange that to say that force then is mass times change in velocity divided by our time interval, where our mass then is 1200 kg, our change in velocity is going to be our final velocity, 0 minus our initial, 10 m/s all over our time interval of 0.1 s. 0086

As I do this, I am going to have 1200 × -10/0.1, or -120,000 N, and since we are only looking for magnitude...0111

...that is 120,000 N or 1.2 × 105 N, so the best answer there is Number 4. 0126

Looking at Number 2 -- A 50 kg student -- m = 50 kg -- threw a 0.4 kg ball -- so the mass of the ball is 0.4 kg -- with a speed of 20 m/s (v = 20 m/s). 0137

What was the magnitude of the impulse that the student exerted on the ball? 0155

If we are focused on the ball, it started at rest, so V-initial must have been 0 m/s and we want to know impulse. 0159

Impulse is change in momentum, which is going to be our final momentum minus our initial momentum, which implies then that our impulse is going to be... 0166

...well our final momentum is just mass times final velocity, so that will be 4 kg × 20 m/s (final velocity) - 0 (initial velocity) because the ball started at rest, therefore our impulse is going to be 20 × 0.4 or 8 N-s.0178

Our best answer there is Number 1. 0200

Looking at Number 3 -- In the diagram below a 60 kg roller skater exerts a 10 N force on a 30 kg roller skater for 0.2 s. 0204

What is the magnitude of the impulse applied to the 30 kg roller skater? 0213

Again, impulse is also force times time interval, so our force was 10 N and that was applied for 0.2 s, so that is just going to be 2 N-s or answer Number 2. 0218

Number 4 -- Two carts are pushed apart by an expanding spring as shown below. 0237

If the average force on the 1 kg cart is 1 N, what is the average force on the 2 kg cart? 0242

This takes us back to Newton's Third Law. 0248

We know the force of object 1 on 2 is equal in magnitude and opposite in direction to the force of 2 on 1, so if 1 N is applied to the 1 kg cart, you must have 1 N applied to the 2 kg cart -- Newton's Third Law. 0250

Number 5 -- What is the speed of a 1,000 kg car that has a momentum of 2 × 104kg-m/s East? 0268

Momentum is mass times velocity and we are looking for speed velocity, therefore velocity will be momentum divided by mass...0278

...or 2 × 104kg-m/s/103 (mass) is going to be 2 × 101 m/s, so the answer is Number 2. 0287

Moving on to Number 6 -- A motorcycle being driven on a dirt path hits a rock. 0305

Its 60 kg cyclist is projected over the handle bars at 20 m/s into a haystack. 0309

If the cyclist is brought to rest in 0.5 s, the magnitude of the average force exerted on the cyclist by the haystack is...?0315

Our impulse is change in momentum, which is force times time. 0323

Now our initial momentum of our cyclist, 60 kg × 20 m/s is going to be 1200 kg-m/s and the final momentum is 0 because the cyclist is brought to rest. 0329

The time interval is 0.5 s, so to solve for force -- that is going to be change in momentum divided by time. 0345

Our change in momentum is our final momentum minus our initial momentum, 1200/0.5 (time interval) or -2400 N and since we are only looking for magnitude of the force, 2400 N is our answer, so choice 4, 2.4 × 103. 0355

Number 7 -- We have a 70 kg hockey player skating East on an ice rink and the hockey's player hit by a 0.1 kg puck moving toward the West. 0379

The puck exerts a 50 N force toward the West on the player. 0388

Determine the magnitude of the force that the player exerts on the puck during this collision. 0391

Well if the puck exerts a 50 N force toward the West, then the player must exert that exact same magnitude of force back on the puck to the East, so that must be 50 N back, again Newton's Third Law. 0396

Number 8 -- Which situation will produce the greatest change in momentum for a 1 kg cart? 0416

Remember change in momentum is impulse, which is force times time interval. 0424

Accelerating it from rest to 3 m/s -- well that is going to be a change in momentum of 0-3, so 3 kg-m/s.0432

Accelerating it from 2-4, that will be 2 × 1 or 2 kg-m/s. 0441

Applying a force of 5 N for 2 s -- force times time -- that will be 10 N-s or applying a force of 10 N for half a second will be 5. 0447

Without a doubt, the largest impulse or change in momentum will be Number 3, 10 N-s or 10 kg-m/s. 0457

Let us try one more. A 0.149 kg baseball initially moving at 15 m/s is brought to rest in 0.04 s by a baseball glove on a catcher's hand. 0466

The magnitude of the average force exerted on the ball by the glove is...?0477

Once again impulse is change in momentum, which is force times time and we are looking for force, so force will be change in momentum divided by time. 0481

Now the initial momentum of our baseball is its mass times its initial velocity or 0.149 kg × 15 m/s, which is going to be about 2.235 kg-m/s. 0494

If it is brought to rest, its final momentum is going to be 0, so change in momentum is final minus initial, which will be 0 - 2.235 or - 2.235 kg-m/s...0514

...so the force is δp/δt or -2.235 kg-m/s/0.04 s (time), so that will give me about -56 N...0530

but because we are only looking for the magnitude, the correct answer is 56 N. 0550

Hopefully, those went pretty well for you. If they did -- Great! -- Go ahead and move on, if they did not, now would be a great time to go review the longer lesson on momentum and impulse. 0557

Thanks everyone for your time and make it a great day!0566