Ch 11: Momentum & ImpulseWorksheetSee all chapters
All Chapters
Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
Ch 04: 2D Kinematics
Ch 05: Projectile Motion
Ch 06: Intro to Forces (Dynamics)
Ch 07: Friction, Inclines, Systems
Ch 08: Centripetal Forces & Gravitation
Ch 09: Work & Energy
Ch 10: Conservation of Energy
Ch 11: Momentum & Impulse
Ch 12: Rotational Kinematics
Ch 13: Rotational Inertia & Energy
Ch 14: Torque & Rotational Dynamics
Ch 15: Rotational Equilibrium
Ch 16: Angular Momentum
Ch 17: Periodic Motion
Ch 19: Waves & Sound
Ch 20: Fluid Mechanics
Ch 21: Heat and Temperature
Ch 22: Kinetic Theory of Ideal Gasses
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 25: Electric Force & Field; Gauss' Law
Ch 26: Electric Potential
Ch 27: Capacitors & Dielectrics
Ch 28: Resistors & DC Circuits
Ch 29: Magnetic Fields and Forces
Ch 30: Sources of Magnetic Field
Ch 31: Induction and Inductance
Ch 32: Alternating Current
Ch 33: Electromagnetic Waves
Ch 34: Geometric Optics
Ch 35: Wave Optics
Ch 37: Special Relativity
Ch 38: Particle-Wave Duality
Ch 39: Atomic Structure
Ch 40: Nuclear Physics
Ch 41: Quantum Mechanics

Concept #1: Collisions with Springs


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.