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 |

Example #1: Inelastic Collisions

**Transcript**

Hey guys. So, here's another very basic collision question, let's check it out. So, I have a wooden block of mass 10m, I don't know what m is but I can still solve it, it's a literal question with letters not numbers. So, I have a wooden box, it doesn't matter that is wooden, mass 10m at rest on a flat surface, so the initial velocity of this object here is 0, a bullet of mass 1 m. So, instead of 10 m is 1 m. So, here's a bullet it's got mass Big M it's shot and reaches the block with speed V, okay? So, I'm going to say that the mass of the first object is big M and the initial velocity of the first object is big D, mass of the second object is 10m and the initial velocity of the second object is 0 because it's at rest and then it says, if the bullet stays inside of the block, the bullet stays lodged inside of the block find the final speed that the system will have, so the idea you got a block, the bullet goes inside of block and then it makes, it causes the block to move forward, because the bullet stays lodged inside the block, those are the key words, this is a completely inelastic collision, they stick together. Now, more important than knowing the name is to know what it means, it means that v1 final equals v2 final, which I can just write as V final and that's what we want to know, what is V final, okay? So, we do this by using conservation of momentum equation. So, m1 v1, m2 v2, m1 v1, m2 v2, the first mass is big M, the second mass is 10 big M, big M 10, big M. Alright, so what is the initial velocity? The initial velocity of m is big V to the right, positive, the initial velocity of the second object the block is 0, okay? Static block. So, this whole thing is gone, the final velocity here, I don't have it, I don't have this guy, I don't have this guy but they are the same. So, I can write V, f and V, f, which means I can then combine the two masses, V, f is m plus 10 m, which is 11 m, again, another way to think about this is that it acts as a single object of mass combined, right? And this stuff on the left here is just m, V, we're looking for V final. So, V final is m, v divided by 11 m. Notice that the masses will cancel and we get the V final is V over 11, okay? That's the answer. Now, let's talk about this thing about this real quick, the bullet had a mass of m and then moved to the velocity of V after the collision the bullet get basically added on to the block. Now, the total mass is 11 m therefore the total velocity is m over 11 the mass grew by a factor of 11 and the velocity got shrunk by a factor of eleven. So, it's a reversal effect that you've had eleven times heavier therefore you've got 11 times slower, okay? And that should make sense, that's how it works. Alright, let me know if you have any the questions.

Example #2: Inelastic Collisions

**Transcript**

Hey guys. So, here I have another example of inelastic collisions, in this problem I have two similar situations. So, we have a 50 caliber sniper rifle bullet with a mass of 800 grams and it reaches a speed, this is speed of 882 meters per second, in the first situation A here, we have an 80 kilogram person who is at rest on ice and that person is going to going to be shot by the bullet, the bullet that stays lodged inside of the person's vest, bulletproof vest. So, let me draw this real quick you got the person on ice and then there's a bullet, that's going to hit the person and stay lodged in there the mass of the bullet, we're going to call this m1 equals 0.8 kilograms the mass of the person m2 is 80 kilograms, okay? And the bullet is going to stay lodged inside the person's bulletproof vest, this means that this is a completely inelastic collision, completely inelastic but most importantly it means that the final velocity of the first object is the same as the final velocity of the second object. So, we'll refer to them as just VF, cool? And then we want to know what is the person speed after being shot, in other words, v2 final and to do this we're going to use the conservation momentum equation, very straightforward. So, for Part A I want to know on, what is v2 final so I'm going to use, m1 v1, m2 v2, m1 v1, m2 v2, this is initial and this is final, okay? So, we're going to write out the masses, the masses are 0.8 and 80, 0.8 and 80 and as always, I'm going to plug in the velocities, so the bullet moves at the velocity of 800, initially the person is not moving, right? It says, at rest at rest on ice, ice just means it's going to be frictionless and then at the end of their final velocities are the same, okay? So, I can just write v final because they're distinct, very straightforward, we can just solve this now, this is going to be, I'm sorry, this is, this is 882, okay? So, if you divide. So, we got 0.8, 882 equals 80, 0.8 V final, if you solve for V final by dividing both sides by 80.8 you get, I have it here, 8.73 meters per second and just out of curiosity I looked it up I convert it and this is approximately 19 miles per hour, if you get shot with a sniper, with this particular sniper bullet, sniper rifle bullet you would actually move back as a result at a speed of 90 miles per hour it's pretty crazy.

Alright, so for Part B, it's different situation but very similar, it says, Now, suppose that the bullet, same bullet goes through an 80 kilogram wooden crates also on ice but it emerges out of the crate quakes instead of staying lodged inside of the crate the bullet comes out from the other side. So, I can do something like this, m1 is the same 0.8 kilograms, m2 is the same. Now, instead of shooting to a person we're shooting to a box, and velocity of one, which is a bullet initial is 882 and the velocity of the bullet, velocity one final after the collision is going to be 300, cool? The block has initial velocity, v2 initial of 0, it's static, and what we want to know what is the block's final velocity, same thing as before, the only difference the bullet's going through now, we can use the same equation conservation momentum equation, cool? So, to save time I'm just going to go ahead and write the masses already. So, 0.8, 80, 0.8, 80 and we're going to plug in the velocities, the initial velocity of the 0.8, which is the bullet is 882, this is 0 because it doesn't move, after the collision the bullet still moves but now with 300, right? And we want to know what is V2 final, okay? So, this is 705.6 plus 0 equals 240 plus 80 v2 final and if you move everything out of the way and leave v final, V2 final by itself you get 5.82 meters per second, okay? I want to quickly talk about this, the speed of the block was a little bit less, why is that? In the first case the bullet got fully absorbed, which means it gave all of its momentum into their vest, into you, in the second case the bullet went in with 882 and came out with 300. So, you can think of it as where as the first problem the bullet lost all of its speed, we went from 882 to 8, here it only lost about 500 instead of losing 800 of speed, it lost 500 speed. So, he lost less momentum therefore the block gains less momentum as well compared to the person, okay? So, that's it for this too that's it, let me know if you have any questions.

Example #3: Inelastic Collisions

**Transcript**

Hey guys. So, here we have another example of an inelastic collision problem. Now, in this case, we have two people throwing the a to each other and when you catch a ball is essentially a collision. So, let's check it out. So, we have two people in ice, person over here is holding a ball, it says, they're facing each other, let's make this a straight line here, and the mass of this person A is 7 kilograms, the mass of B is 60 kilograms and the mass of the ball, I'm going to call this m c just because I'm already using m b, is 5 kilograms, right? And this ball is going to get thrown with 10 meters per second and I want to know, what is the speed of A after throwing the ball. So, there's two events here, first you throw the ball and then for Part B I want to know what is the speed of B after catching the ball, and then you catch the ball, these are two different events, one is throwing, is a push away type of problem and you can think of catching as a collision type of problem. So, I'm going to be able to write the conservation momentum equation twice. Now, that being said, when you throw, there are two objects that are interacting with each other it's a and C, okay? B has nothing to do with it while you're throwing, when you're catching it's B and C because this person be catching the ball, which is C, and person A has nothing to do with this, okay? So, let's do this real quick, I'm going to write the momentum equation, conservation momentum equation twice, one for each situation. So, I got ma va initial plus mB VB initial equals, it's actually A and C, right? So, this is C, and then ma Va final plus mC vC final, the masses in question are 70 and 5, so I'm going to do 70, 5, 70, 5 and all I got to do is put these the velocities, the initial velocity is before you throw, right? Since, this is a throw you're talking about before the throw, after the throw, for the catch we can talk about before and attach and after the catch. So, before you throw you're here, you're not moving, the ball is not moving, remember how this works, right? So, both of these are 0. Now, the ball goes to the right with 10. So, let's call it positive 10, that's the 5 kilogram and we want to know what is the final velocity of a in this case final means after you throw, this number is going to go over here, negative 15 over 70 equals VA therefore VA final is, I have it here, negative 0.71 I got a negative, which means it's going to the left, it makes sense that it should go to the left because if I throw something this way, I'm going to get pushed over this way. Alright, so that's part A, very straightforward.

Now, let's do Part B, actually do it over here, same setup but now, I'm going to have mb vb i, initial, plus mc vc initial equals mb vb final plus mc vc final, the masses are now going to be 60 and 5. So, 60, 5, 60, 5 and I want to know, here I was looking for VA final, which is this guy here and here, I'm looking for Vb final, which is this guy here, cool? So, remember you're no longer talking about the throw, you're talking about the catch. So, initial means before the catch, right? So, before the catch the 60, the person has no velocity at all, they're static, but the ball is moving with 10, after the catch we don't know the person's final velocity, that's what we're looking for, and we don't know the ball's final velocity but since you caught it these two velocities are the same. So, we're going to write them as VF just VF instead of B and C, we're going to write just VF, which means I can combine the two masses and I get 65, okay? So, this is 0, I have 50, I can divide by 65 and that is my final velocity, my final velocity is, I have it here, 0.77 meters per second, okay? Now, before I end the video, let's talk about this real quick, one velocity is bigger than the other, is greater than the other or rather one speed this has higher speed, does that make sense? It should make sense because this person is lighter, okay? So, this person will recoil a certain amount but the amount of momentum that the second person gains is the same but because it has a smaller mass it picks up more speed okay, if you're lighter you're typically faster and that works almost all the time in momentum problems, okay? So, it's the same momentum coming from A ends up transferring to the ball and then transferring to B and you end up with a higher speed since the mass is smaller, cool? That's it, let me know if you have any questions.

0 of 3 completed

Example #1: Inelastic Collisions

Example #2: Inelastic Collisions

Example #3: Inelastic Collisions

A 5 kg mass moves with an initial speed of 1 m/s and collides with a 10 kg mass at rest. After the collision, the 5 kg mass is at rest and the 10 kg mass is in motion, with some speed v. (a) What type of collision is this? (b) What is the speed v? (c) IF the collision is inelastic, how much kinetic energy was lost during the collision?

An important quantity to know during an inelastic is known as the coefficient of restitution, which essentially measures how "ellastic" a collision is. The coefficient of restitution, e, is defined as
e = ( v1,f - v2,f ) / ( v1,i - v2,i )
where the v's are speeds, not velocities. If a ball were to drop from a height of 5 m, and only bounce up to a height of 4 m, what is the coefficient of restitution for the collision between the ball and the floor?

Suppose two asteroids strike head on. Asteroid A ( exttip{m_{
m A}}{m_A}8.1×1012 kg) has velocity 3.6 km/s before the collision, and asteroid B (exttip{m_{
m B}}{m_B}1.6×1013 kg) has velocity 1.3 km/s before the collision in the opposite direction.(a) If the asteroids stick together, what is the magnitude of the velocity of the new asteroid after the collision?(b) What is the direction of the velocity of the new asteroid after the collision?

Blocks A and B are moving toward each other on a horizontal frictionless surface. Block A has mass 5.0 kg and is moving to the right with speed 8.0 m/s. Block B has mass 15.0 kg and is moving to the left with speed 4.0 m/s. The two blocks collide and stick together. After the collision, the combined object (mass 20.0 kg) has velocity A) 0.50 m/s to the left B) 0.50 m/s to the right C) 1.0 m/s to the left D) 1.0 m/s to the right E) 2.5 m/s to the left F) 2.5 m/s to the right G) none of the above answers

Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth, throwing up so much dust that the sun was blocked out for a period of many months. Suppose an asteroid with a diameter of 2.0 km and a mass of 0.80*1013 kg hits the earth (6.0*1024 kg) with an impact speed of 4.1*104 m/s^{4} m m/s. What is the earth’s recoil speed after such a collision? (Use a
reference frame in which the earth was initially at rest.)

A 10.0-g bullet is fired into a stationary block of wood having mass m = 5.00 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?

A 7600-kg boxcar traveling 15 m/s strikes a second car. The two stick together and move off with a speed of 6.6 m/s. What is the mass of the second car?

If all three collisions in the figure are totally inelastic, which cause(s) the most damage (deformation of objects, thermal energy increase, etc.)? Assume that the wall is stationary and the car is completely stopped by it in the first diagram.
1. I, III
2. I
3. all three
4. I, II
5. II, III
6. III
7. II

In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic.You have probably learned that "momentum is conserved" in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement "momentum is conserved."1) What physical quantities are conserved in this collision?a) the magnitude of the momentum onlyb) the net momentum (considered as a vector) onlyc) the momentum of each object considered individually2) Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are V1 and V2. What is the speed of the two-car system after the collision?a) V1+V2b) V1-V2c) V2-V1d) (V1V2)^(1/2)e) (1/2)(V1+V2)f) (V1+V2)^(1/2)g)The answer depends on the directions in which the cars were moving before the collision.3) Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are P1 and P2. After the collision, what is the magnitude of their combined momentum?a) P1+P2b) P1-P2c) P2-P1d) (P1P2)^(1/2)e) (1/2)(P1+P2)f) (P1+P2)^(1/2)g) The answer depends on the directions in which the cars were moving before the collision.4) Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are p1 and p2. After the collision, their combined momentum is vector p. Of what can one be certain?a) Vector P= p1+p2b) Vector P= p1-p2c) Vector P= p2-p1

A bullet of mass mb is fired horizontally with speed viat a wooden block of mass mw resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling at speed.vfA. Which of the following best describes this collision?B. Which of the following quantities, if any, are conserved during this collision?a. kinetic energy onlyb. momentum onlyc. kinetic energy and momentumd. neither momentum nor kinetic energyC. What is the speed of the block/bullet system after the collision?Express your answer in terms of vi, mw, and mb.

A car with mass mc = 1134 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck of mass mt 1752 kg traveling south at vt = 8.8 m/s fails to yield and collides with the car. The vehicles become stuck together and slide on the asphalt, which has a coefficient of friction of μk = 0.5.A) Write an expression for the velocity of the system after the collision, in terms of the variables given in the problem statement and the unit vectors i and jB) How far, in meters, will the vehicles slide after the collision?

A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet.If the bullet has a mass of 26.8g and a speed of 230m/s , how high will the block rise into the air after the bullet becomes embedded in it?Express your answer to three significant figures and include the appropriate units.

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.