1. PHY 231 1 PHYSICS 231 Lecture 13: Collisions Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom

2. PHY 231 2 Previously: momentum and impulse Momentum p=mv F=  p/  t Impulse (the change in momentum)  p= F  t Demo: the ‘safe’ bungee jumper

3. PHY 231 3 Conservation of Momentum F 21  t = m 1 v 1f -m 1 v 1i F 12  t = m 2 v 2f -m 2 v 2i Newton’s 3rd law: F 12 =-F 21 (m 1 v 1f -m 1 v 1i )=-(m 2 v 2f -m 2 v 2i ) Rewrite: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f p 1i +p 2i =p 1f +p 2f CONSERVATION OF MOMENTUM CLOSED SYSTEM!

4. PHY 231 4 Moving in space An astronaut (100 kg) is drifting away from the spaceship with v=0.2 m/s. To get back he throws a wrench (2 kg) in the direction away from the ship. With what velocity does he need to throw the wrench to move with v=0.1 m/s towards the ship? Initial momentum: m ai v ai +m wi v wi =100*0.2+2*0.2=20.4 kgm/s After throw: m af v af +m wf v wf =100*(-0.1)+2*v wf kgm/s Conservation of momentum: m ai v ai +m wi v wi = m af v af +m wf v wf 20.4=-10+2*v wf v wf =15.7 m/s

5. PHY 231 5 Types of collisions Inelastic collisionsElastic collisions Momentum is conserved Some energy is lost in the collision: KE not conserved Perfectly inelastic: the objects stick together. Momentum is conserved No energy is lost in the collision: KE conserved

6. PHY 231 6 Perfectly inelastic collisions Conservation of P: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f After the collision m 1 and m 2 form one new object with mass M=m 1 +m 2 m 1 v 1i +m 2 v 2i =v f (m 1 +m 2 ) v f =(m 1 v 1i +m 2 v 2i )/ (m 1 +m 2 ) before after Demo: perfect inelastic collision on airtrack

7. PHY 231 7 Perfect inelastic collision: an example A car collides into the back of a truck and their bumpers get stuck. What is the ratio of the mass of the truck and the car? (m truck =c*m car ) What is the fraction of KE lost? 50 m/s20 m/s25 m/s m 1 v 1i +m 2 v 2i =v f (m 1 +m 2 ) 50m c +20c*m c =25(m c +c*m c ) so c=25m c /5m c =5 Before collision: KE i =½m c 50 2 +½5m c 20 2 After collision: KE f =½6m c 25 2 Ratio: KE f /KE i = (6*25 2 )/(50 2 +5*20 2 )=0.83 17% of the KE is lost (damage to cars!)

8. PHY 231 8 Ballistic Pendulum M block =1 kg M bullet =0.1 kg V bullet =20 m/s h How high will the block go? There are 2 stages: The collision The Swing of the block The collision The bullet gets stuck in the block (perfect inelastic collision). Use conservation of momentum. m 1 v 1i +m 2 v 2i =v f (m 1 +m 2 ) so: 0.1*20+1*0=v f (0.1+1) v f =1.8 m/s The swing of the block Use conservation of Mechanical energy. (mgh+½mv 2 ) start of swing = (mgh+½mv 2 ) at highest point 0+½1.1(1.8) 2 =1.1*9.81*h so h=0.17 m Why can’t we use Conservation of ME right from the start??

9. PHY 231 9 Elastic collisions Conservation of momentum: m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f Conservation of KE: ½m 1 v 1i 2 +½m 2 v 2i 2 =½m 1 v 1f 2 +½m 2 v 2f 2 Rewrite conservation of KE: a)m 1 (v 1i -v 1f )(v 1i +v 1f )=m 2 (v 2f -v 2i )(v 2f +v 2i ) Rewrite conservation of P : b)m 1 (v 1i -v 1f )=m 2 (v 2f -v 2i ) Divide a) by b): (v 1i +v 1f )=(v 2f +v 2i ) rewrite:(v 1i -v 2i )=(v 2f -v 1f ) Use in problems

10. PHY 231 10 elastic collision of equal masses Given m 2 =m 1. What is the velocity of m 1 and m 2 after the collision in terms of the initial velocity of m 2 if m 1 is originally at rest? m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f m 2 v 2i =m 1 v 1f +m 2 v 2f v 2i =v 1f +v 2f (v 1i -v 2i )=(v 2f -v 1f ) -v 2i = v 2f -v 1f v 2f =0 v 1f =v 2i

11. PHY 231 11 elastic collision of unequal masses Given m 2 =3m 1. What is the velocity of m 1 and m 2 after the collision in terms of the initial velocity of the moving bullet if a) m 1 is originally at rest b) m 2 is originally at rest A) m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f 3m 1 v 2i =m 1 v 1f +m 2 v 2f 3v 2i =v 1f +3v 2f (v 1i -v 2i )=(v 2f -v 1f ) -v 2i = v 2f -v 1f B) m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f m 1 v 1i =m 1 v 1f +m 2 v 2f v 1i =v 1f +3v 2f (v 1i -v 2i )=(v 2f -v 1f ) v 1i = v 2f -v 1f v 2f =v 2i /2 v 1f =3v 2i /2 v 2f =v 1i /2 v 1f =-v 1i /2