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Push-Away Problems With Energy | 12 mins | 0 completed | Learn |

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Collisions & Motion (Momentum & Energy) | 68 mins | 0 completed | Learn |

Collisions with Springs | 9 mins | 0 completed | Learn |

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Concept #1: Collisions with Springs

**Transcript**

Hey guys so here is another example of a collision problem with motion, now this one has a spring in it so let's check it out. So I have a 15 kilograms block that is on a horizontal smooth frictionless surface and it's attached to a horizontal spring so let draw this real quick I got a 15 kilograms block on a horizontal on smooth surface and it's attached to a horizontal spring which in turn is attached to a wall so it looks something like this and the spring has spring coefficient or force constant K equal to 800 Newtons per meter so the block is at rest initially so I'm going to say that the initial velocity of the block is 0 when a 400 gram bullet strikes the block so I'm going to put a bullet here and the bullet has mass M1=0.4 kilograms and then the block has M2=15 and I wrote there, the bullet becomes embedded on the block so this is a completely inelastic collision where the final velocity would be the same because the bullet becomes embedded on the block causing it to move so the block gets hit by a bullet and moves they will actually move together, right? And this also causes the spring to be compressed so the block starts compressing the spring but then eventually it stops so the block goes from here to let's say here and this compression of the spring is X=0.7, a maximum compression implies that the final velocity is 0 because the block is going all the way over here and it stops so this is 0 maximum compression so I can say that this velocity right here is 0. Alright what is the initial speed of the bullet? So, I want to know what is this velocity here now this is going to be V1 because it's the bullet A and we want to know what is V1a? Two things that are happening here first is a collision and then there is a motion of a block compressing a spring so let's draw a little diagram for that I go from A to B, there's a collision so we're going to use momentum and then I go from B to C and in that interval, there's motion so we're going to use energy so before the initial part of momentum is pre-collision, the final part of momentum is post collision and this is the end where the system stops, OK? So, what we want we want the velocity here so let's write the momentum equation for the first part and the energy equation for the second so between A and B I have momentum so M1V1+M2V2=M1V1+M2V2 this is A-A B-B and for the second piece from B to C I have the energy equation so it's Ku, K+U+work non-conservative=K+U, this is from B so I have B-B and C so I have C-C, OK? So, I haven't done much other than just draw what's happening draw a little schematics of the different steps and write the equations now let's plug in numbers and remember I want the initial speed which is this right here, this is what we're looking for, OK? So I'm going to start here now actually before I put a numbers let's just look at what we have I have both masses, the initial velocity of the block before the collision A is pre-collision is 0 it's not moving that's typically how these bullet to block problems work the block doesn't move upfront so this goes away the bullet stays lodged inside so these velocities are the same I don't know what that velocity is yet but I know it's the same so I can write Vb instead of V1band V2b I can just write Vb and I combine the masses M1+M2, OK? And on the left here I have M1V1a, V1a is what I want so what I'm going to do is I'm going to divide both sides by M1 and I get it out of here and that's an expression for V1a I have both masses what I don't have is VB so I'm going to go over here and get Vb, Vb is going to come out of this guy here the kinetic energy at point B since kinetic energy is 1/2MV squared, but let's do it so is there a kinetic energy at point B? Point B is post collision there's always going to be kinetic energy after collision because the bullet hits and the block moves so yes. What about potential energy? Well just before the bullet hits the spring hasn't compressed at all.... Actually, I'm sorry this is after the collision but just after the collision after the block has hit it's just beginning to move the spring isn't compressed quite yet, there's also no height here, right? So, there's no heights no spring compression just yet so there is no potential energy, the work done by non-conservative forces is 0 because you're just watching and there is no friction it says it was a smooth surface, there is no kinetic energy at point C because at the end of the thing at the end of the problem of the motion the system stops, right? But there is potential energy at point C because point C is where we have full compression so we have elastic potential energy so we have Kb and Uc all the kinetic energy I at point B turns into potential energy at point C, OK? So, let's expand these 1/2MV squared and this is 1/2 KX squared, this is the elastic potential energy of the spring, the halves cancel notice that the mass does not cancel, OK? Very important in a lot of these problems mass cancel, here it doesn't so which M you're going to use matters very much in this case the bullet's embedded inside the block so we'll use both M1+M2, right? And remember what we're looking for we came to the second equation to find VB So let's get that and back go back to the first equation Vb will be Kx Squared divided by M so Vb is by itself and then I got to take the square root if you want to make this a little cuter you can pull out the X and then this is the square root of K/M, OK? But so, this is the sort of the cleanest you can make it now let's put in some numbers, X is the compression which is 0.7 square root of K, K is 800 and the mass I'm supposed to use is both masses so I guess I could have written and M1+M2 I should probably have written in M1+M2 but it's fine I'm just going to plug in the number which is 15.4, OK? 15.4. Now if you plug all of this in the calculator I have it here you get 5.05 meters per second and that's what I wanted that's Vb so now we're going to stick Vb back in here, V1a is Vb which is 5.05 the two masses added up which is 15.4 divided by the mass of the bullet which is 0.4 if you plug this in the calculator you get a velocity of 194 meters per second and that is our final answer, OK? So very straightforward collision you use momentum and then motion you use energy put the two of them together write the two equations look for what you need and almost all of these problems you're going to have to use both equations, OK? So, this is predictable do a lot of this and practice let me know if you have any questions.

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Concept #1: Collisions with Springs

A block with mass 4.95 kg is on a horizontal frictionless surface. It rest against a horizontal spring that has force constant k = 300 N/m. Initially the spring is not compressed. A bullet with mass 0.0500 kg is travelling horizontally with speed v0 = 400 m/s. The bullet strikes the block and remains embedded in it. What is the maximum distance that the spring is compressed after the collision with the bullet?

A 1300 kg car rolling on a horizontal surface has a speed of exttip{v}{v}71 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.3 m. What is the spring constant of the spring?

A 7 kg mass, moving at 10 m/s, is approaching a 5 kg mass, at rest. The 7 kg mass has a light spring, with a force constant of 1000 N/m, oriented towards the 5 kg mass so that when the masses meet, the spring will compress along its axis. When the masses collide, what distance does the spring compress by? Hint: this system doesn't loose energy during the collision.

A 5.00-kg block is moving at 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant k=500 N/m that is attached to a wall. The spring has negligible mass.(a) Find the maximum distance the spring will be compressed.(b) If the spring is to compress by no more than 0.400 m, what should be the maximum value of v0?

In the figure, a block sitting on a frictionless horizontal surface is attached to a rigid wall on the right through a spring (whose axis is horizontal). A bullet is shot at the block from the left and gets embedded in it, causing the block to move to the right, thus compressing the spring. (Assume the bullet is travelling perfectly horizontally, along the axis of the spring, before hitting the block). Which of the following are true?
A. The initial kinetic energy of the bullet is completely converted to spring potential energy when the spring reaches its maximal compression.
B. The initial momentum of the bullet is equal to the momentum of the bullet+block system just after the bullet enters the block.
C. Part of the momentum of the bullet+block system is lost during the collision (i.e. before the spring-compression starts).
D. Part of the energy of the bullet+block system is ”lost” (no longer present as macroscopic kinetic energy) during the collision, before the spring-compression starts.
E. If we are given the masses of the block and the bullet, the initial speed of the bullet and the spring constant, it is possible to find the maximum compression of the spring.
1. A, E
2. A, B, D
3. B, D
4. A, B
5. A, C
6. D
7. B, D, E
8. A, C, E
9. A, B, E

A 3.0-kg block slides along a frictionless tabletop at 8.0 m/s toward a second block (at rest) of mass 4.5 kg. A coil spring, which obeys Hookes law and has spring constant k exttip{k}{k}= 810 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block.(a) What will be the maximum compression of the spring?(b) What will be the final velocities of the blocks after the collision?(c) Is the collision elastic? Ignore the mass of the spring.

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