2. Topic 2.2 Extended E – Elastic and inelastic collisions

3.  We begin with a few definitions... A collision in which the total kinetic energy is conserved is called an elastic collision. Topic 2.2 Extended E – Elastic and inelastic collisions  What this means is K f = K 0. A collision in which the total kinetic energy is not conserved is called an inelastic collision.  What this means is K f  K 0.  Usually this means is K f < K 0. A collision in which the colliding objects stick together is called a totally inelastic collision.  What this means is that the final velocities are identical.

4. Topic 2.2 Extended E – Elastic and inelastic collisions  An inelastic collision is a collision where energy is lost.  For example, suppose we drop a billiard ball, as shown.  Note that potential energy never regains its previous level.  Each bounce loses a little more energy.  If instead of a billiard ball we drop a piece of clay, the clay will stick.  We say that the collisions are inelastic.  We say that the collision is completely inelastic. We define as completely inelastic any collision where the particles stick together. The kinetic energy of the system is not conserved.

5. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS FYI: In any collision, momentum is conserved.  If the particles stick together, their final velocities are equal.  Thus m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f v 1f = v 2f ≡ v f  Thus so that m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f completely inelastic collision m 1 v 1i + m 2 v 2i = m 1 v f + m 2 v f or  Completely inelastic collision problems are the easiest collisions to solve, because...

6. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS Suppose two train cars are moving down the track as shown, and they hitch together. If v 1 = 10 m/s and v 2 = 8 m/s, and the cars’ masses are equal, what is the speed of the cars after they collide? v1v1 v2v2 vfvf vfvf  Since the cars hitch together, their final speeds are identical, and we can use the completely inelastic collision equation: m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f m(10) + m(8) = (m + m)v f m(18) = 2mv f v f = 9 m/s

7. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS If the mass of each car is 2000 kg, find the change in kinetic energy that occurs during the collision. = 164000 J K 0 = m 1 v 1i + m 2 v 2i 1212 1212 2 2 = 2000·10 + 2000·8 1212 1212 2 2 K f = m 1 v 1f + m 2 v 2f 1212 1212 2 2 = 2000·9 + 2000·9 1212 1212 2 2 = 162000 J  K = K f - K i = 162000 - 164000 = - 2000 J FYI: Note the loss of kinetic energy. The 2000 J were converted to sound and heat energy.

8. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS  A ballistic pendulum in its simplest form is a hanging block of wood which has a bullet shot into it. The amount the block rises can be used to find the original speed of the bullet. h M m M m M m vivi m M m M

9. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS  We divide the whole event into two phases. M m vivi m M v Collision phase – Energy not conserved, momentum is conserved Initial Final  The first phase is the collision phase, where mechanical energy is not conserved. FYI: During this phase MOMENTUM IS CONSERVED. mv i + M(0) = mV + MV mv i = (m + M)V

10. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS  We divide the whole event into two phases. m M v Energy phase – Mechanical energy is conserved Initial  The second phase is the energy phase, where mechanical energy is conserved. m M h Final K – K i + U - U i = 0 U = K i V 2 = 2gh (m + M)gh = (m + M)V 2 1212 FYI: During this phase MECHANICAL ENERGY IS CONSERVED.

11. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS  Now we combine the two results into a useable formula that relates the bullet speed to h: mv i = (m + M)V Ballistic pendulum equation V 2 = 2gh v i = 2gh (m + M) m

12. Topic 2.2 Extended E – Elastic and inelastic collisions C OMPLETELY I NELASTIC C OLLISIONS Suppose a 12-g bullet striking the 3-kg block raises the block 8 cm. How fast was the bullet traveling? v i = 2(10)(0.08) (0.012 + 3) 0.012 v i = 317.5 m/s

13. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS  During a typical collision some of the energy of motion (kinetic energy) is lost due to heat and permanent mechanical deformation.  If the objects which collide are hard and resilient (like billiard balls or steel balls) there is no energy loss. In this case, we say that kinetic energy is conserved.  We call a collision in which kinetic energy is conserved a perfectly elastic collision, or in short, an elastic collision.  Since during any collision (elastic or otherwise) momentum is conserved, the following two equations are used to solve elastic collision problems: P i = P f K i = K f Elastic collision equation set

14. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS  Let’s apply our equations to a typical 1D collision, as shown below: From our equations we have m2m2 m1m1 p 1i + p 2i = p 1f + p 2f K 1i + K 2i = K 1f + K 2f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f m 2 (v 2i -v 2f ) = m 1 (v 1f -v 1i ) m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 1212 1212 1212 1212 2 2 2 2 m 2 (v 2i -v 2f ) = m 1 (v 1f -v 1i ) 2 2 2 2 m 2 (v 2i -v 2f )(v 2i +v 2f ) = m 1 (v 1f -v 1i )(v 1f +v 1i ) m 2 (v 2i -v 2f )(v 2i +v 2f ) m 1 (v 1f -v 1i )(v 1f +v 1i ) = m 2 (v 2i -v 2f ) m 1 (v 1f -v 1i ) v 2i + v 2f = v 1i + v 1f Elastic velocity equation

15. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS  The previous equation is, of course, only valid if the collision is perfectly elastic. We now use it to substitute v 2f out of our momentum equation: v 2i + v 2f = v 1i + v 1f → v 2f = v 1i + v 1f - v 2i m 2 (v 2i - v 2f ) = m 1 (v 1f - v 1i ) m 2 [v 2i - (v 1i + v 1f - v 2i )] = m 1 (v 1f - v 1i ) m 2 [2v 2i - v 1i - v 1f ] = m 1 v 1f - m 1 v 1i 2m 2 v 2i - m 2 v 1i - m 2 v 1f = m 1 v 1f - m 1 v 1i 2m 2 v 2i + (m 1 - m 2 )v 1i = (m 1 + m 2 )v 1f 2m 2 v 2i + (m 1 - m 2 )v 1i m 1 + m 2 v 1f = 2m 1 v 1i + (m 2 - m 1 )v 2i m 2 + m 1 v 2f = Elastic final velocities FYI: From SYMMETRY we can obtain the corresponding equation for v 2f: FYI: You won't see these formulas on any formula sheet. Enter them in your calculator for future reference!

16. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS Consider the following 1D perfectly elastic collision:  What are the final speeds of the masses? 3 kg 10 m/s 4 kg 6 m/s 4 kg v 2f 3 kg v 1f m 1 = 3 kg m 2 = 4 kg v 1i = + 10 m/s v 2i = - 6 m/s 2m 2 v 2i +(m 1 -m 2 )v 1i m 1 + m 2 v 1f = 2·4· - 6 +(3-4)10 3 + 4 = = - 8.2857 m/s 2m 1 v 1i +(m 2 -m 1 )v 2i m 2 + m 1 v 2f = 2·3· + 10 +(4-3) - 6 4 + 3 = = + 7.7143 m/s v 2i + v 2f = v 1i + v 1f  As a check, we can use the elastic velocity equation: - 6 + + 7.7143 = 10 + - 8.2857 1.7143 = 1.7413

17. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS  Suppose we have a perfectly elastic 1D collision between two equal masses: Then since m 1 = m 2 = m our formulas simplify somewhat: 2m 2 v 2i +(m 1 -m 2 )v 1i m 1 + m 2 v 1f = 2m 1 v 1i +(m 2 -m 1 )v 2i m 2 + m 1 v 2f = = v 2i v 2i + v 2f = v 1i + v 1f  As a check, we can use the elastic velocity equation: v 2i + v 1i = v 1i + v 2i 2mv 2i +(m-m)v 1i m + m = 2mv 1i +(m-m)v 2i m + m = = v 1i

18. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS  Suppose we have a perfectly elastic 1D collision between two equal masses, as shown: 1 2  Mass 2 is originally stationary, and mass 1 is originally moving at 5 m/s.  After the collision mass 2 is now moving to the right at 5 m/s, and mass 1 is stops. FYI: The effect of equal masses colliding elastically in 1D is that the masses SWAP VELOCITIES.

19. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS  Suppose we have a perfectly elastic 1D collision between two very unequal masses, as shown: m M  The masses could be an elephant and a fly. The important thing here is that M >> m.  M >> m means that M “is very much greater than” m.  Since M >> m we can simplify the following: M - m = M m - M = - M m + M = M m / M = 0 M + m = M  Thus: 2m 2 v 2i +(m 1 -m 2 )v 1i m 1 + m 2 v 1f = 2Mv 2i +(m-M)v 1i m + M = 2Mv 2i + - Mv 1i M = v mf = 2v Mi + - v mi 2m 1 v 1i +(m 2 -m 1 )v 2i m 2 + m 1 v 2f = 2mv 1i +(M-m)v 2i M + m = = v 1i + v 2i MMMM 2mM2mM v Mf = v Mi

20. Topic 2.2 Extended E – Elastic and inelastic collisions E LASTIC C OLLISIONS Suppose an elephant is traveling at 5 m/s to the left, and a fly is traveling at 10 m/s to the right, when they collide. What are the final speeds of both the elephant and the fly? for the fly: v mf = 2v Mi + - v mi v mf = 2(-5) + - 10 v mf = - 20 m/s for the elephant: v Mf = v Mi m M v Mf = - 5 m/s