# Ap physics free response practice momentum and impulse relationship

### momentum_and_impulse_practice_problems_with_solutions

General Notes About AP Physics Scoring Guidelines. 1. One exception to this practice may occur in cases where the numerical answer to a later part. View miyagi-marugoto2012.info from PHYS at Woodland Community College. AP Physics Free Response Practice Momentum and Impulse A bullet. General Notes About AP Physics Scoring Guidelines. 1. The solutions such as those given on the exam equation sheet. For a description Solutions usually show numerical answers using both values when they are For using conservation of angular momentum or momentum-impulse reasoning to conclude that.

### Learn AP Physics - Momentum

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. My initial velocity is not 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.

**AP Physics 1 review of Momentum and Impulse - Physics - Khan Academy**

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 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.

### Impulse and momentum dodgeball example (video) | Khan Academy

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.

## AP PHYSICS

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. 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?

## Reviewing Linear Momentum

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.

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. If I take three kilogram meters per second and I divide by. The students can ask general questions during this time, but I encourage them to see me after class if they have a specific question about their own work. This entire process takes about five minutes, and I encourage students to be as honest as possible in their self grading, since one goal of this activity is to see how well they'd do if this question were actually on the AP Physics 1 exam.

Students are each given a different problem from a review problem set. These problems are taken from our textbookand are problems that I feel are most representative of the test questions. Students stay at their lab tables for this activity so we don't lose time while they get into groups I give each group one of the problems with its final answer. Including the final answer with the problem is an important part of this activity. I want students to not focus so much on getting the right answer, but on being able to explain their reasoning and justify their decisions.

The problem that each group receives is random: I literally walk down the center aisle and give the group whichever problem is on the top stack of my pile. All of the students should be able to answer any of these questions. My expectation is that students take about ten minutes and actively work together to write down the solution on their papers. Also, I encourage students to use pen as they work. Using pen keeps them from erasing, so even if they make a mistake or change their thinking, I can see evidence of their entire thought and solution process.

As students are working at their lab tables, I walk around and ensure that everyone is engaged in the discussion and thinking critically about their assigned problem. I am willing to give students hints as I observe, but my feeling is that by this point in the Linear Momentum Unit students should be able to independently work through these problems. Momentum Review Problem Set Students Share Linear Momentum Solutions 20 minutes After the collaborative work time is over, I share with students that they are presenting their solutions to the rest of the class.

The goal of this activity is to show students the variety of problems that are on the unit exam. Because my students are sometimes shy, I ask if any groups volunteer to go first. There is always at least one group that wants to get the presentation out of the way, so I choose them and applaud them for being so willing.