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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 |

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Example #1: 2D Collisions

**Transcript**

Hey guys. So, in every collision problems so far would have two objects moving along the same direction, typically the x-axis, the horizontal and after colliding that would continue to move in the x-axis. So, because of the thing was along the same direction these were one-dimensional collision or one-dimensional momentum problems, which are simpler. Now, we're going to deal when you're looking to two-dimensional collisions, two-dimensional mental problems, where a ball hits another one and then they might go off at an angle like that. So, instead of having just x or just Y, we're going to have a combination of those, of the two, let's check it out, so it says here, as with every two-dimensional problem in physics whether it is a velocity force problem momentum collision problem, we're going to treat the x and y-axis separately, okay? So, we're going to do is we're going to write the momentum equation because this is a collision. So, we write the momentum equation but we are going to do it twice, we're gonna do it one for the x axis, one for the y axis and we're going to, we're going to treat the x and y axis separately, as if they had nothing to do with each other. Notice that this is just m1 v1, m2 v2, our long momentum equation except this has a bunch of x's, this is x only and this is y only, we don't mix them, okay? Because things are going to be moving in two dimension, we also have to, I have to quickly remind you how to decompose vectors. So, let's say you have velocity this way and it goes that angle. So, I can, if the angle is right here, this is where we want to have our angle, that's our standard position for the angle, I can decompose this into VX and VY and if the angle is at the right place the x will always be V cosine of theta and VY will always be V sine of theta. So, make sure the angle is there, okay? That's how you do that, also remember, if we have VX and VY you can get V back, it's just using the Pythagorean theorem, okay? So, quick review of vectors. So, here's an example I picked out V one final, V1 initial in the x-axis and I picked out V2 final in the y-axis, which is this one and I have them here as examples to decompose. So, v1 I x is v1 I cosine of theta because there's an x. Alright, so this first piece repeats and then the little x gets replaced with cosine theta. So, what would this be? I want you to take a few seconds to do this, v2, f y, it's a lot of letters, if you expand it, right? It's it sine or cosine, right? It's, I hope you said sine because of the Y and it's going to be v2 f, sine of theta. And notice that the angle is the angle for the second object and its final because this one is final right there, okay? There's a lot of little subscripts a lot of little letters down there and you got to be careful. So, let's do a practice problem, let's do an example real quick to see how this works. So, I have a white billiard ball mass of white is 1.57, let's put it right here, and ball is going to go this way with 4 meters per second and then it's going to hit a black ball and after the collision the ball, the black ball is going to, black ball is originally at rest, I'm going to put 0 meters per second here, after the collision the black ball is going to go this way, it says here, that black, I'm doing the dotted lines to signify that this happened later, right? So, this is sort of before to the left and after to the right, actually I'm going to write that here, before the collision and then this is after the collision, okay? So, it moves this way with 3 meters per second at 37 below, below the x-axis, it looks like that. Now, which way would the white ball go, the ball, if the black ball goes down, I hope you're thinking that the white ball would go somewhere in this direction. Now, it doesn't mean it goes the same speed, it doesn't mean it goes with 37 degrees, in fact, usually it doesn't turn out to be that nice but that's what we want to know, I want to know what is v white final and I want to know what is Theta white final, in other words, what is the magnitude and the direction of the white balls velocity after, okay? And we're going to write simply, we're going to write the two momentum equations, one for the x, one for the Y, plug in a bunch of numbers. So, let's do that real quick. So, m1 v1 initial, m2 v2 initial, m1 v1 final plus m2 v2 final, I'm writing it tiny because I want to write the other one here. So I want to make sure I have space, this is for the x axis. So, I should have a bunch of little X's here. Now, I do have the masses but before I plug them in I want to point out to you that mass of white and mass of black are the same, the neat thing about that, is that this m is the same as this m, which is the same as this m this m. So, every term is the same thing. So, once, if the masses are the same to simplify I can to just cancel all the M's that it makes life a little bit easier, same thing is going to happen for the y axis, I'm really only going to have v1 initial plus v2, v1, v2 initial equals v1 final plus v2 final, if you want you can write the whole equation with all the M's and then cancel them out, this is going to be why, why, why, why, cool? Let's plug in numbers. Now, what is the initial so initial is this, right? What's the initial velocity of the first ball in the x-axis? So, the first ball in the x-axis is moving with 4, I'm going to call this positive 4, what about the second ball, what's the initial velocity of the second ball in the x-axis? So, I hope you're thinking 0, right? What is the final velocity of the first ball, which is white, what's the final velocity of the ball here? we don't know. So, I'm going to write V White's final in the x-axis, I don't have it, that's fine, what about this guy here? The final velocity of the black ball, final velocity of the black ball is this in the x-axis, I actually have that, right? So, in the x-axis it's going to be, I can decompose this, this is going to be 3 cosine of 37. Remember, if you decompose along the x axis it's cosine as long as the angle is between the vector, the arrow and the y, the x axis, the angle is like right here, which it is, right? So, that's a good angle just like how this over here is the good angle. always between the arrow and the x axis, okay? So, if you plug this into your calculator you get 2.4, Let me just check here, yep, you got 2.4, okay? So, that's what this number is going to be, it's 2.4 to the right. So, it's positive 2.4, okay? Let's try to simplify this, I basically get 4 minus 2.4 equals V WF x. So, V WF in the x axis of the white ball is 1.6 meters per second. Now, I'm not going to rewrite this as v w f cosine of theta just yet and let me show you why, don't write this, the reason I'm not going to do, this is because in this case I have one unknown, which is VX, here I now have two unknowns, V and the cosine of theta, I don't know V and I don't know theta. So, it's getting uglier, I'm going from one unknown to two unknowns. So, we're going to leave it alone for now and I'll show you what happens in the end.

So, we're done here, let's go to the y-axis and the same thing, the initial velocity of the first object in the y axis, so the first object is initially is this one, right? In the beginning, this is the white ball, it's going to the right, If it's going flat to the right it doesn't have any velocity in the y axis so this is 0. Now, what about the second object the second object before the collision isn't even moving, okay? Second object, this guy, before the collision isn't even moving. So, it's just 0 as well. Now, what about the, one is white two is black, what about the white ball after the collision, so the white ball after the collision is here, do I know that final velocity? I don't because I don't know either one of these. So, I don't know the final velocity. So, let me erase this and since I don't know it, I'm just going to write V white final y, what about the final velocity, the post collision velocity of the black ball in the y axis, do I have that? I can get it by decomposing this here. So, if this was 3 cosine of 37, this guy is going to be, let's call this VB or v2 final and the y axis is going to be 3 sine of 37 and if you put this in your calculator, this is 1.8. Now, keep in mind that it's going down. So, I got to put a negative in front of it, okay? 1.8 and that's what this is going to be, negative 1.8. So, look, what happens I can just move this around cancel the negatives and I get that the final velocity in the y axis of the white Bowl is 1.8, okay? One unknown. Now, look, what happens, I have, I have the final velocity in the x, and I have the velocity in the why, this looks like this 1.6 and this looks like this 1.8, I can combine the two, okay? I can combine the two. So, let's do that real quick. So, essentially you have a 1.6 this way and a 1.8 this way but what I really want to know is what is V white final, I'm not looking for the x or the Y I'm looking for the total vector but I can find the magnitude of the total vector if I know the components, this forms a little triangle and I hope you're thinking Pythagorean theorem, right? So, I can do square root of 1.6 squared plus 1.8 squared and if you combine this, I have it here, the answer is 2.4 meters per seconds, okay? So, that's the first answer, the second answer is pretty quick, it's theta, theta, I hope you remember, if you have the two sides of the triangles theta is just the arc tangent of the Y side divided by the x side and if you, so this is going to be the arc tangents Y is the height 1.8, this is 1.6, make sure your calculator is in degrees and the answer you get is 48, let me disappear, 48. where do I have it? 48.4 degrees, okay? So, a little bit long just because there's a lot of variables and this is the first time of doing it but I hope you realize that it's actually pretty straightforward, you write this equation once, twice, you plug in all the numbers and you end up getting the components of the final velocity for the unknown ball and then you put them together to get the total vector on length, okay? So, that's it for this one, let me know if you guys have any questions.

Example #2: 2D Collisions

**Transcript**

Hey guys. So, let's do another example of two-dimensional collision. So, here we're going to have two cars or a car in a pickup truck hitting each other. So, I have a 1100 kilogram car. So, it's moving east. So, East is to the right, let's call this m1 equals 1100, it's moving this way with v1 initial of 30 and it collides at an intersection with a pickup truck, that's moving north. So, looks sort of like this, pickup truck is going north, I'm going to draw the pickup truck like this, I can put the mass inside m2 is 2200. Notice that it's double, exactly double the mass, okay? And this truck is moving with a V2 initial of V2 initial of 20, it says here that after the collision the two vehicles are going to become are going to stay stuck together, stay stuck and move together. So, they're going to become entangled and just sort of move together, so the first thing is, let's say that this ball represents the vehicles together after the collision, I want you to think about, which way this thing would go and I hope you're thinking this way, right? Because these two things are going like this so they're going to collide and they're going to go like that. The next thing I want you to think about is whether this angle is going to be 45 degrees or if it's going to be less than 45 or more than 45 and I want you to look at the numbers and see if you can figure that out, and the answer is it depends not on the masses or the velocities but it depends on the momentum, the momentum here, if you multiply these two numbers you get 3300 and then there's an extra 0 here and then this momentum here is 4400 and there's an extra 0 here. So, the pickup truck is a little slower but it's much heavier therefore it has more momentum, because it has more momentum I can already tell you that this is not going to be, right down the middle as a 45 but in fact, you're going to get an angle that is greater than 45, okay? So, we are going to want to know what is this final angle, let me write theta final equals question mark, we're going to want to know what is the final velocity of the entangled cars but I'm going to make a prediction here, that theta will be greater than 45 degrees, right? Sometimes you can do this sometimes you can't, I just want to talk a little bit about that. Now, how do we do this? Two dimensional collisions so write the momentum equation for the x axis and the y axis, here the masses are different. So, I'm not going to be able to cancel them, let's go. So, m1 v1, m2 v2, m1 v1, m2 v2, and then this, this is initial, initial, final, final these are in the x axis. So, X, X, X, X, it's kind of a pain to do this, the other one is in the y axis same think, m1 v1, m2 v2, m1 v1, m2 v2, initial, final, final Y, Y, Y, Y, cool? I haven't done anything yet I just wrote the equation. So, let's plug in the numbers. So, here I have 1100 and 2200 equals 1100, 2200, okay? The other one, we'll figure out a bit the other one next, let's start plugging numbers, this is the x-axis, okay? So, what is the initial, initial is pre-collision, what is the initial velocity of the first car in the x-axis. So, I hope you this right away it's 30, okay? Now, work with me here, what's the initial velocity of the second car in the x-axis? And I hope you're saying 0 because it's moving straight in the y-axis, it doesn't move sideways so it's 0. Remember, everything here is x, it's not 20, it's 0. Now, what about the final velocity, I don't know the final velocity for either one of these guys but I do know that they're the same because they get, they're stuck together. So, instead of writing v1 final x and v 2 final x, I'm just going to write v final x for both because they are together, okay? This cancels, this is going to be 3, 3, 0, 0, 0 and these guys add up. So, it's going to be 3, 3, 0, 0 v final x and I can quickly find v final x is going to be 10 meters per second, okay? 10 meters per second. Notice that m1 right here, was moving with 30 and once I collide with this guy here, it's going to go that way but the x velocity goes from 30 to now that there's more mass to 10, should make sense, you're going this way then you gain mass so you slow down in that direction from 30 to 10, okay? So, that's it for that one, let's continue the Y equation right here, I don't have a lot of space, we want to do some, some weird stuff here, let me do this, cool? So, I'm going to continue, 1100, 2200, 1100, 2200, and the velocity, okay? Alright, so what is the initial velocity? this is the most important part, is figuring out, which velocities go where, what's the initial velocity in the y axis for the first object and I hope you said 0 because it's moving in the x, it doesn't move in the y axis at all, what about the second one? the second object, the car is, has initial velocity of 20 on the y axis, going up so we're going to put plus 20, for the final velocity is the same thing, I don't know but they're stuck together. So, it's the same, so i's v final Y, V final Y and I can calculate as well, this is 4, 4, 44,000 and then this adding up here is 3,300 V final. So, I can calculate v final to be, I can calculate v final to be 13.33 meters per second, okay? Whoops coming out of the way 13.33 meters per second, same idea the car was, the truck was moving up with 20 it hits something and then its upward velocity goes down a little bit from 20 to 13 because the system now has more mass, okay? Alright, so now that I have this and this, this is v, Fy, guess what? I can combine them together again, I can say, well, if my v final x of the whole thing together is 10 and my v final y of the whole thing together is 13.33, I can then look for this guy v final of both by just doing the Pythagorean theorem 10 squared, 13.33 squared and if you do this you get 16.67 meters per second, we want to know the magnitude and the direction, direction means theta. So, theta again is the arc tangent of y over x. So, it's the arc tangent of 13.33 divided by 10, calculator should be in degrees, and the answer you get is 15, 53 rather 0.1 degrees, these are the two final answers, okay? These are probably the most common versions or sort of the most simple but also the most common versions, introductory two-dimensional collision problems, let me know if you have any questions.

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Two bags of sand are sliding on a horizontal frictionless surface. Bag A has mass 2.0 kg and bag B has mass 5.0 kg. Initially bag A is traveling east with speed vAi and bag B is traveling south with speed vBi.The two bags collide and after the collision bag A is sliding at 1.50 m/s in a direction 30.0° north of east and bag B is sliding at 2.50 m/s in a direction 37.0° south of east. What was the initial speed of each bag?

On a horizontal frictionless surface block A (mass 2.00 kg) is sliding toward the east with speed vAi and block B (mass 4.00 kg) is sliding north at speed vBi. The blocks collide and stick together. After the collision the combined object (mass 6.00 kg) is moving at 36.9° north of east at vf = 8.00 m/s.
What was the initial speed vAi of block A?

On a horizontal frictionless surface block A (mass 2.00 kg) is sliding toward the east with speed vAi and block B (mass 4.00 kg) is sliding north at speed vBi. The blocks collide and stick together. After the collision the combined object (mass 6.00 kg) is moving at 36.9° north of east at vf = 8.00 m/s.
What was the initial speed vBi of block B?

On a horizontal frictionless surface block A (mass 3.00 kg) is sliding toward the east with speed VA and Block B (mass 5.00kg) is sliding toward the north with speed VB. After the collision block B is sliding toward the north with speed 2.00 m/s and block A is sliding with a speed of 6.00 m/s in a direction of 30.0° north of east, as shown in the sketch. What are the speeds VA and VB of the two blocks before the collision?
Ans. VA =
Ans. VB =

A rock with mass 2.0 kg is at rest on a horizontal frictionless surface. A bullet with mass 0.100 kg is traveling due east with a speed of 400 m/s just before it strikes the rock. Just after the collision between the bullet and the rock, the bullet is moving south with a speed of 300 m/s. What is the speed of the rock just after the collision with the bullet?
A) 5.0 m/s
B) 15.0 m/s
C) 25.0 m/s
D) 59.2 m/s
E) None of the above answers

A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to move off at an angle of 45°. The neutrons initial speed is 5.1×105 m/s{
m {
m m/s}}.(a) Determine the speeds of the two particles, vn and vHe, after the collision.(b) Determine the angle of the neutron after the collision.

An object with an initial momentum represented by the vector below, strikes an object that is initially at rest. Which of the following sets of vectors may represent the momenta of the two objects after the collision? Note carefully: The original vector below and the following vectors are ALL DRAWN TO THE SAME LENGTH SCALE.

Two asteroids of equal mass in the asteroid belt between Mars and Jupiter collide with a glancing blow. Asteroid Aexttip{A}{A}, which was initially traveling at vA1 = 40.0 m/s with respect to an inertial frame in which asteroid was at rest, is deflected 30.0° from its original direction, while asteroid B exttip{B}{B}travels at 45.0° ^circ to the original direction of Aexttip{A}{A}, as shown in the figure.(a) Find the speed of asteroid A after the collision.(b) Find the speed of asteroid B after the collision.(c) What fraction of the original kinetic energy of asteroid exttip{A}{A} dissipates during this collision?

Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 14.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of 54.1° from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.(a) What is the magnitude of Daniels velocity after the collision?(b) What is the direction of Daniels velocity after the collision?(c) What is the change in total kinetic energy of the two skaters as a result of the collision?

A stone with mass 0.800 kg rests on a horizontal frictionless surface. A bullet with mass 0.0200 kg traveling horizontally at speed VAi strikes the stone and rebounds horizontally at a speed of VAf in an direction perpendicular to its original motion. After being hit by the bullet, the stone is moving at 6.00 m/s in a direction 30.0° from the original direction of motion of the bullet. What are the speeds VAi and VAf of the bullet before and after the collosion?Ans. VAi = VAf =

A(n) 7 kg object moving with a speed of 6.5 m/s collides with a(n) 18 kg object moving with a velocity of 8.2 m/s in a direction 21° from the initial direction of motion of the 7 kg object. What is the speed of the two objects after the collision if they remain stuck together?
1. 9.10604
2. 7.63104
3. 8.34111
4. 9.58276
5. 7.78221
6. 9.34713
7. 8.24562
8. 6.89647
9. 9.73592
10. 10.3766

On a horizontal frictionless surface block A (mass 4.0 kg) is sliding toward the east with a speed of 6.0 m/s and block B (mass 8.0 kg) is sliding toward the north with a speed of 9.0 m/s. The blocks collide. Immediately after the collision block A is sliding toward the north with speed 7.0 m/s and block B is sliding with a speed of VBf at an angle θ north of east. Calculate the speed VBf and the angle θ.Ans. VBf = Ans. θ =

A particle with mass m moving at speed v collides with a particle with mass 2m initially at rest. After the collision, the two particles have velocities that are directed at equal angles of θ on either side of the original line of motion of the m particle. What is the speed v' of the 2m particle after the collision?
1. v' = v/4sinθ
2. v' = v/cosθ
3. v' = vsinθ/4
4. v' = v/4cosθ
5. v' = vcosθ
6. v' = v/2cosθ
7. v' = vsinθ/2
8. v' = v/sinθ
9. v' = vcosθ4

Let two objects of equal mass m collide. Object 1 has initial velocity v, directed to the right, and object 2 is initially stationary.A. If the collision is perfectly elastic, what are the final velocities v1 and v2 of objects 1 and 2?Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.B. Now suppose that the collision is perfectly inelastic. What are the velocities v1 and v2 of the two objects after the collision?Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.C. Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the final velocities v1 and v2 of objects 1 and 2?Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.D. Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision?Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.

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