Physics 111: Lecture 15 Today’s Agenda Elastic collisions in one dimension Center of mass reference frame Colliding carts problem Some interesting properties of elastic collisions Killer bouncing balls

Momentum Conservation: Review The concept of momentum conservation is one of the most fundamental principles in physics. This is a component (vector) equation. We can apply it to any direction in which there is no external force applied. You will see that we often have momentum conservation even when kinetic energy is not conserved.

Comment on Energy Conservation We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Energy is lost: Heat (bomb) Bending of metal (crashing cars) Kinetic energy is not conserved since work is done during the collision! Momentum along a certain direction is conserved when there are no external forces acting in this direction. In general, momentum conservation is easier to satisfy than energy conservation.

Lecture 15, Act 1 Collisions A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest. What is the ratio of initial to final kinetic energy of the system? (a) 1 (b) (c) 2

Lecture 15, Act 1 Solution No external forces in the x direction, so PX is constant. v m m m m v / 2 x

Lecture 15, Act 1 Solution Compute kinetic energies: v m m m m v / 2

Lecture 15, Act 1 Another solution We can write P is the same before and after the collision. The mass of the moving object has doubled, hence the kinetic energy must be half. m m m m

Lecture 15, Act 1 Another Question: Is it possible for two blocks to collide inelastically in such a way that the kinetic energy after the collision is zero?

Lecture 15, Act 1 Another Question Is it possible for two blocks to collide inelastically in such a way that the kinetic energy after the collision is zero? YES: If the CM is not moving! CM CM

Elastic Collisions Elastic means that kinetic energy is conserved as well as momentum. This gives us more constraints We can solve more complicated problems!! Billiards (2-D collision) The colliding objects have separate motions after the collision as well as before. Start with a simpler 1-D problem Initial Final

Elastic Collision in 1-D m2 m1 initial v1,i v2,i x v1,f v2,f final m1 m2

Elastic Collision in 1-D m1 m2 before Conserve PX: m1v1,i + m2v2,i = m1v1,f + m2v2,f v1,i v2,i x after v1,f v2,f Conserve Kinetic Energy: 1/2 m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f Suppose we know v1,i and v2,i We need to solve for v1,f and v2,f Should be no problem 2 equations & 2 unknowns!

Elastic Collision in 1-D Airtrack Collision balls However, solving this can sometimes get a little bit tedious since it involves a quadratic equation!! A simpler approach is to introduce the Center of Mass Reference Frame m1v1,i + m2v2,i = m1v1,f + m2v2,f 1/2 m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f

CM Reference Frame We have shown that the total momentum of a system is the velocity of the CM times the total mass: PNET = MVCM. We have also discussed reference frames that are related by a constant velocity vector (i.e. relative motion). Now consider putting yourself in a reference frame in which the CM is at rest. We call this the CM reference frame. In the CM reference frame, VCM = 0 (by definition) and therefore PNET = 0.

Lecture 15, Act 2 Force and Momentum Two men, one heavier than the other, are standing at the center of two identical heavy planks. The planks are at rest on a frozen (frictionless) lake. The men start running on their planks at the same speed. Which man is moving faster with respect to the ice? (a) heavy (b) light (c) same

Lecture 15, Act 2 Conceptual Solution The external force in the x direction is zero (frictionless): The CM of the systems can’t move! X X X X x CM CM

Lecture 15, Act 2 Conceptual Solution The external force in the x direction is zero (frictionless): The CM of the systems can’t move! The men will reach the end of their planks at the same time, but lighter man will be further from the CM at this time. The lighter man moves faster with respect to the ice! X X X X CM CM

Lecture 15, Act 2 Algebraic Solution Consider one of the runner-plank systems: There is no external force acting in the x-direction: Momentum is conserved in the x-direction! The initial total momentum is zero, hence it must remain so. We are observing the runner in the CM reference frame! Let the mass of the runner be m and the plank be M. m M Let the speed of the runner and the plank with respect to the ice be vR and vP respectively. vR vP x

Lecture 15, Act 2 Algebraic Solution The speed of the runner with respect to the plank is V = vR + vP (same for both runners). MvP = mvR (momentum conservation) Plugging vP = V - vR into this we find: m vR vP So vR is greater if m is smaller. M x

Example 1: Using CM Reference Frame A glider of mass m1 = 0.2 kg slides on a frictionless track with initial velocity v1,i = 1.5 m/s. It hits a stationary glider of mass m2 = 0.8 kg. A spring attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e. energy is conserved). What are the final velocities? Video of CM frame v1,i m2 v2,i = 0 m1 VCM + = CM m2 m1 x m2 v2,f v1,f m1

Step 1 Example 1... Four step procedure First figure out the velocity of the CM, VCM. VCM = (m1v1,i + m2v2,i), but v2,i = 0 so VCM = v1,i So VCM = 1/5 (1.5 m/s) = 0.3 m/s Step 1 (for v2,i = 0 only)

Example 1... If the velocity of the CM in the “lab” reference frame is VCM, and the velocity of some particle in the “lab” reference frame is v, then the velocity of the particle in the CM reference frame is v*: v* = v - VCM (where v*, v, VCM are vectors) v VCM v*

Example 1... Calculate the initial velocities in the CM reference frame (all velocities are in the x direction): Step 2 v*1,i = v1,i - VCM = 1.5 m/s - 0.3 m/s = 1.2 m/s v*2,i = v2,i - VCM = 0 m/s - 0.3 m/s = -0.3 m/s v*1,i = 1.2 m/s v*2,i = -0.3 m/s

Example 1 continued... Movie Now consider the collision viewed from a frame moving with the CM velocity VCM. ( jargon: “in the CM frame”) v*1,i m2 v*2,i m1 m1 m2 x m2 v*2,f v*1,f m1

Energy in Elastic Collisions: Use energy conservation to relate initial and final velocities. The total kinetic energy in the CM frame before and after the collision is the same: But the total momentum is zero: So: (and the same for particle 2) Therefore, in 1-D: v*1,f = -v* 1,i v*2,f = -v*2,i

Step 3 Example 1... v*1,f = -v* 1,i v*2,f = -v*2,i v*1,i m2 v*2,i m1 v*1,f = - v*1,i = -1.2m/s v*2,f = - v*2,i =.3 m/s m2 m1

Example 1... v* = v - VCM Step 4 So now we can calculate the final velocities in the lab reference frame, using: v = v* + VCM v1,f = v*1,f + VCM = -1.2 m/s + 0.3 m/s = -0.9 m/s v2,f = v*2,f + VCM = 0.3 m/s + 0.3 m/s = 0.6 m/s v1,f = -0.9 m/s v2,f = 0.6 m/s Four easy steps! No need to solve a quadratic equation!!

Lecture 15, Act 3 Moving Between Reference Frames Two identical cars approach each other on a straight road. The red car has a velocity of 40 mi/hr to the left and the green car has a velocity of 80 mi/hr to the right. What are the velocities of the cars in the CM reference frame? (a) VRED = - 20 mi/hr (b) VRED = - 20 mi/hr (c) VRED = - 60 mi/hr VGREEN = + 20 mi/hr VGREEN = +100 mi/hr VGREEN = + 60 mi/hr

Lecture 15, Act 3 Moving Between Reference Frames The velocity of the CM is: = 20 mi / hr 20mi/hr CM So VGREEN,CM = 80 mi/hr - 20 mi/hr = 60 mi/hr So VRED,CM = - 40 mi/hr - 20 mi/hr = - 60 mi/hr The CM velocities are equal and opposite since PNET = 0 !! 80mi/hr - 40mi/hr x

Lecture 15, Act 3 Aside 20mi/hr CM As a safety innovation, Volvo designs a car with a spring attached to the front so that a head on collision will be elastic. If the two cars have this safety innovation, what will their final velocities in the lab reference frame be after they collide? 80mi/hr - 40mi/hr x

Lecture 15, Act 3 Aside Solution v*GREEN,i = 60 mi/hr v*RED,i = -60 mi/hr v*GREEN,f = -v* GREEN,i v*RED,f = -v*RED,i v*GREEN,f = -60 mi/hr v*RED,f = 60 mi/hr v´ = v* + VCM v´GREEN,f = -60 mi/hr + 20 mi/hr = - 40 mi/hr v´RED,f = 60 mi/hr + 20 mi/hr = 80 mi/hr

Summary: Using CM Reference Frame (m1v1,i + m2v2,i) : Determine velocity of CM : Calculate initial velocities in CM reference frame : Determine final velocities in CM reference frame : Calculate final velocities in lab reference frame Step 1 VCM = Step 2 v* = v - VCM Step 3 v*f = -v*i Step 4 v = v* + VCM

Interesting Fact v*1,i v*2,i We just showed that in the CM reference frame the speed of an object is the same before and after the collision, although the direction changes. The relative speed of the blocks is therefore equal and opposite before and after the collision. (v*1,i - v*2,i) = - (v*1,f - v*2,f) But since the measurement of a difference of speeds does not depend on the reference frame, we can say that the relative speed of the blocks is therefore equal and opposite before and after the collision, in any reference frame. Rate of approach = rate of recession v*1,f = -v*1,i v*2,f = -v*2,i This is really cool and useful too!

Basketball Demo. Drop 2 balls Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M). Drop these from some height. The height reached by the small ball after they “bounce” is ~ 9 times the original height!! (Assumes M >> m and all bounces are elastic). Understand this using the “speed of approach = speed of recession” property we just proved. 3v m v v v v M v (a) (b) (c)

More comments on energy Consider the total kinetic energy of the system in the lab reference frame: v*1 v*2 but so v* (same for v2) v* = KCM = PNET,CM = 0 = KREL

More comments on energy... Consider the total kinetic energy of the system in the LAB reference frame: v* = KREL = KCM So ELAB = KREL + KCM KCM is the kinetic energy of the center of mass. KREL is the kinetic energy due to “relative” motion in the CM frame. This is true in general, not just in 1-D

More comments on energy... ELAB = KREL + KCM Does total energy depend on the reference frame?? YOU BET! KREL is independent of the reference frame, but KCM depends on the reference frame (and = 0 in CM reference frame).

Recap of today’s lecture Elastic collisions in one dimension (Text: 8-6) Center of mass reference frame (Text: 8-7) Colliding carts problem Some interesting properties of elastic collisions Killer bouncing balls Look at textbook problems Chapter 11: # 63, 67, 71