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Lab 7 Conservation of MomentumIntroduction:Background:The momentum of a moving object is defined as the object's mass multiplied by its velocity (p mv): The la...

Question

Lab 7 Conservation of MomentumIntroduction:Background:The momentum of a moving object is defined as the object's mass multiplied by its velocity (p mv): The law of Conservation of Momentum states that the total momentum of a system does not change when there is no net external force. An air track nicely demonstrates conservation of momentum: Many small air jets provide a nearly frictionless cushion of air for the gliders to slide on: If an air track is leveled, gravitational force balances

Lab 7 Conservation of Momentum Introduction: Background: The momentum of a moving object is defined as the object's mass multiplied by its velocity (p mv): The law of Conservation of Momentum states that the total momentum of a system does not change when there is no net external force. An air track nicely demonstrates conservation of momentum: Many small air jets provide a nearly frictionless cushion of air for the gliders to slide on: If an air track is leveled, gravitational force balances the normal force. Thus, the net external force on a glider that is placed on an air track is zero. Purpose; To explore the concept of momentum; and to verify the law of conservation of momentum_ The details: moving car hits a stationary car and they couple and move together as shown below: Before collision: VR ML MR After collision: mL IR The total momentum before collision = mRVR +Okgmls = mRVr The total momentum after collision (mL+mr)v (since initially mL is at rest): If momentum is conserved, then mRVR (m+mR)VL



Answers

$\bullet$ On a frictionless air track, a 0.150 $\mathrm{kg}$ glider moving at
1.20 $\mathrm{m} / \mathrm{s}$ to the right collides with and sticks to a stationary
0.250 $\mathrm{kg}$ glider. (a) What is the net momentum of this two-
glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use
your answers in parts (a) and (b) to find the speed of the glid-
ers after the collision. (d) Is kinetic energy conserved during
the collision?

Okay for this question. We're asked to analyse two cars. Well, they call them cars are sliding objects on this frictionless surface. They asked us questions before and after a collision. So, for the sake of um, and easy and clear and to be clear, I'm going to label this first part A and the second Kirby. And the problem tells us that Car A Has a massive 0.15 kg and is moving 1.2 m/s to the right. Car B Has a massive .25 kg and is not moving. All right, so let's look at part A first for A wants us to find the net momentum of this system before this collision. So to do that, we can say that our net momentum is equal to the momentum of car A plus the momentum of Kirby, which we can break down further into Minister A times velocity of A plus must be times the last thing to be. We know that velocity B is equal to zero because it is not moving. And so that will make Carvey has zero momentum. So are all of our momentum comes from Car A Will be 0.15 kg Times 1.2 it is per second. So in that momentum, if we put that into our calculators will give us The total that momentum of 0.18 several grams times meters per second. Okay, now for part B, they want us to um identify what our momentum after our collision must be. And then they ask us why now our momentum is always conserved in these collisions. And so if our momentum of our entire system was 0.18 kg tens meters per second, then after a collision, our net momentum must be conserved. So it must be 0.18 kilograms times meters per second. That must be because the momentum, it's actually right now is concert. So that means we're not going to lose momentum or danish now for currency. They ask us to use those two answers and to come up with what our velocity must be. So we have been appearing our picture after a collision which is represented by the squiggling line. These masses are now stuck together, mess A. Msb. They want us to find what that velocity is going to be. So adding the two masses together. Okay, we can find our total mass now of our new car, as we could say, Which would just be 0.15 Plus 0.25. Okay, so let's use that to calculate the velocity. You know where momentum is equal to mass times velocity. And now our math here is going to be our total mass. So it's going to be mass A plus, mass B. And all of that time. The final velocity. Now the momentum we are given And that must be 0.18 since it's concerned. So let's put some numbers into that equation. 0.18 kill advance tens meters per second is equal to Math A. was 0.15 kilograms. Mass be with 0.25 kg and then we have that final velocity. Mm So let's work with our friend. This is 1st 0.15 Plus 0.25 will give us 0.4 0.4 kg Times the frame of velocity is equal to 0.18 kilograms meters per second. Now, if we divide both sides as you're going forward, general grams we get that our final velocity Is equal to 0.45 meters per second. We can know that it's needed per second because kilograms will cancel out on that side of our equation. So the answer to part C 0.45 m/s. Now, it's good to note that our velocity is going to greatly slow down because our math has increased. Okay, so for party they want us to ask or they want us to identify whether this kinetic energy is concerned. The best way to look at this is to look before and after. Kinetic energy Is equal to 1/2 some squared Now, since Kirby has zero Kinetic energy Where it has zero velocity, They will have zero kinetic energy. So the kinetic energy is ultra um a Which is going to be 1/2 0.15 kg. Change its velocity of 1.2 m/s, correct? So if we plug it into our calculator, we can get a number and analyze. I'll take 1/2 times 0.15 Times The Square of one Point. And we should get an answer. 0.108 jewels Now for our kinetic energy. After we've increased the mass and decrease the speed, which hard hard to tell. But we can calculate it. There will be 1/2 Now. Our new mass of A&B 0.4 kg and a new velocity, which we just calculated was 0.45 m person. Well, we will square that and we can plug that into Our new kinetic interviews for 1/2 Times 0.4 Times The Square of 0.45. And we get that our new kinetic energy is 0.0405 jules. So we can see from these two comparing kinetic energy before and after that are kinetic energy is not concerned. That is a good thing to remember about our collision. Or kinetic energy will not be conserved within this collision will lose, energy is either sound or heat, um but our momentum will always be concerned.

Hi. So no question were given that 16 grand mass moves in the positive X. Direction at 38 centimeters per second and in four grandmas mostly negative X. Direction at 50 cents per second. And they collide and stick together and want to find the velocity. And so the total ensure momentum musical to total final momentum and the momentum musical to the product of mass times velocity. So I went to multiply the mass of each by the by the velocity of each at each point for the initial and then for the final. So for the initial danish of last year of 16 grandmas is Travelers of six, single mass is going to be 30 cm/s times 13 plus four times ninja velocity of the for graham mass is 15 but we need to take the directions into account Because the four grandma's wasn't more than negative extraction, so it's going to be -15 and the total final final momentum they collide and stick together so you have the same velocity which is what we want to find. So let's go to R. V. And We add their masses. So this 16 plus four and we saw this, this would be South 74 ft is up 16 times 30 plus four Time final 15 over 22 mm. And actually noticed that we used our our mass and our initial velocity in grams and then centimeters per second. So this was grams. This was centimeters per second. Same thing for this. So the answer is we're going to get the on site and sent me just a second. Our velocity will be After calculation, this is 14 cm/s. And to convert this to me that's the 2nd 1 divided by by Andre. And we get physical to 0.14 m/s. That's a fallacy. And since this is positive it means that it is in the positive X. Direction. So our mass our our final collision, final velocity moves in the positive extraction to the right, right, that positive X direction.

Okay, so in this question we have to objects object is coming along the X axis east, an object is going along the Y axis north. And so they collide, they stick together and we need to find the velocity of the of the two objects together then so we can do this by finding the momentum's by finding the velocity X. Component and the Y. Component. And then we just use pythagorean theorem to sell for the velocity and to get the angle. So for the X momentum, we just use the X values of the velocity. So for object A. We know that's eight m/s, but for B it has no X. Component, so that goes to zero. And because objects stick together, we can add that we can add their masses up and we'll get the V. X. In the end. So We we sub in the numbers, so M. A. is equal to 17, V.A. is eight. This section is zero Then we add these up, their masses up. Since they stick together 17 was 29 and re x. Is here. So if we put this in our calculator, we get that V. X at the end is going to be 2.9, 6 meters per second. Now, Same similarly for the white momentum, the object A. Has no Y. Direction of the lawsuits goes to zero. So We have zero plus the massive b, which is 29 kg Times the velocity at five m/s And then it equal to the objects are sticking together. So the massacre added 17 kg 29 and the B. Y. Here. So if we put this in the calculator, We get that B. Y. is equal to three 15 meters per second. So these are the two components of our our final velocity. If we think of this as a triangle, we know that our X component Is 2.96 And our white component is 3.14, this is not the scale. And our velocity that we're trying to find here, let's call this is V. Prime is we can find this with the Pythagorean theorem. So we know that re prime square, There's going to be 2.96 squared plus 3.15 Squared. Princeton are calculator. Again, we'll get that prime is equal to 4.3, m/s. Now, if we're trying to find data here, We can use we know these two values so we can use opposite over adjacent which is 10. So you know that 10 data is equal to opposite over adjacent, just 3.15 over 2.96. We can get that data equals 10 inverse of this. And the data in the end is 40 6.78 Degrees. Now for trying to stay true to significant digits, we can only go up to we only go up to one thing if you did it for both of these. So our final loss is gonna be four m/s and her ankle, at which It will go out as 50° from the horizontal. So our final statement is going to be prime is equal to four m/s, and our direction can be, for example, it can be east 50 degrees north. Thank you that sense.

For this problem on the topic of momentum and collisions, We have two gliders moving on a horizontal frictionless air tracks, And the first glider has a mass of 160.1 g, moving to the right with a speed of 2.72 or three m/s. The second glider has mass 354.1 g, and is moving to the left at a speed of 3.515 m/s. The two gliders then undergo a totally inelastic collision, and we want to find the velocity of glider one after this collision. Now, since the masses are given in grams and the velocities meters per second, we don't have to convert any of the units in this problem, and we know for a one dimensional, totally elastic collision at the speed of the first glider, after the collision is given by the following equation. The final speed of glider, one along the X direction F the F one X is equal to The massive later one. M 1 minus the massive glider to M two over. M one plus M Two Times the initial speed of glider one the I one in the X direction less. Two M two over and one plus M two. And the initial speed of glided to the X. Direction the I two X. Now we are given the masses and velocities, so we just substitute them in and we can find the final speed of the first glider. And so this is 160 0.1 g minus 300 and 54.1 g, divided by 160.1 g plus 354 0.1 g Times The speed. The initial speed of glider one which is to .7-3 meters per second. And we'll add this to the second term, Which is two times a massive glider, to 354.1 g, Divided by the sum of mass is 160.1 g Plus 354.1 g, Multiplied by the speed of collider to which is initially in the opposite direction, so that's -3 .515 meters per second. And so during the calculation, we get the final velocity of glider one along the X direction to be minus five .869 meters per second, Or 5.869 m/s to the left.


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