Phys 201 1nd Edition Lecture 17 Outline of Last Lecture I. Conservation of MomentumOutline of Current Lecture II. Conservation of Momentum III. ExplosionsA. From RestB. With initial VelocityIV. Momentum Conservation in Two DimensionsV. Other types of CollisionsCurrent LectureConservation of Momentum:For objects with the same momentum, the object with the fastest speed will have the greatest kinetic energy. If the same car with the same speed collides with three different objects, the most energy will belost when it collides with an object of a greater velocity, even if it has a smaller mass“Explosions” from Collisions:An explosion refers to when two objects collide and then move away from one another. An explosion canrefer to objects that being at rest, or objects that begin in motion.Explosion from Rest:Example: A 4kg rifle fires a 10g bullet at 891m/s. what is the recoil velocity of the rifle?891 4kg .01kgV1’Pi=pfThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.For this problem we are trying to find the final velocity of the coil in the rifle. Both the initial velocity of the bullet and the initial velocity of the coil are equal to 0. This means that we will rearrange our equation to solve for V1’ using m1, m2 and V2’.m1V1 +m2V2 = 0 = m1V1’+m2V2’m1V1’=-m2v2’v1’=(-m2/m1)V2’V1’=(-.01/4)891 = -2.2m/sExplosion with Initial Velocity: Example: A multi stage rocket consisting of a booster and payload takes off at an initial velocity of 900m/s. After explosion they separate from one another at 100m/s. What is the final speed?mb mpVb + Vp  vi=900m/sWe still solve for this problem using momentum conservation.(Mb+mp)Vi=mbVb +mpVpVrelative=Vp-VbSubstitute Vp=Vrel +Vb(mb+mp)Vi=mbVb+mp(Vrel+Vb)mpVrel = (mb+mp)Vb[(mb+mp)Vi/(mb+mp)]-[(mpVrel)/(mb+mp)] = [(mb+mp)Vb]/(mb+mp)Vi-[mp/(mb+mp)Vrel]=VbRemember that the mass of the booster is 4 times the mass of payload. This means that even though we don’t know the exact mass of either, we can guess using the formula; Mp/(mb+mp)=1/(1+4)=1/5Vi-(1/5)Vrel=Vb  900-1/5(100)=880m/sMomentum conservation in two dimensions:When we have a two dimensional vector, we need an x and y equation. These x and y components are conserved individually. Momentum is only conserved for the brief moment of the collision, so we want to look at collision lasting for a very short period of time. However, in most real life situation, momentumdoes not stay conserved because after the collision outside forces, such as friction and other resistive forces, will start to work on the object. Example: Car 1, traveling in the x direction, and car 2, traveling in the y direction, collide at an intersection and stick together after the collision. Car one is 1500kg and was traveling at 10m/s. Car 2 is 200kg and was traveling at 15m/s. Find the x and y components of the final velocity.Vf VfyVfx1500Kg 10m/s15m/s200kgVf describes the velocity at which the cars move after collision. To find Vf we need to find the x and y components of Vf.X directionPix=Pfxm1V1=(m1+m2)Vfx [ m1/(m1+m2)]V1=VfxVfx=[1500/(1500+200)]10=4.3m/sy-directionPiy=PfyM2V2=(m1+m2)Vfy [ m2/(m1+m2)]V2=VfyVfy=[200/(1500+200)]15=8.6m/sOnce you have these two components, you can calculate the value of Vf, or the angle at which the cars will go off at after the collision.Other ways to describe a collision:Everything we have discussed involving collisions has assumed that there is a moment in which the two objects colliding stick to one another. However, there are collisions in which objects immediately repel away from one another. Consider a bouncy ball being dropped on the ground. When the ball collides with the floor, it immediately shoots back up. This is called a rebound. Within a rebound, if the ball travels back up to the height that it was dropped at, it is considered to have a “perfect bounce”. This perfect bounce means that kinetic energy has been