3. Conservation of Momentum If two isolated objects interact (collide or separate), then the total momentum of the system is conserved (constant). Conservation of Energy Energy is neither created nor destroyed.

4. Elastic Collisions Elastic collisions are ideal collisions where zero kinetic energy is converted into thermal energy (heat) …100% conserved. *Springs, magnets, and billiard balls can approximate elastic collisions but they only truly exist at the atomic level.

5. Elastic Collisions Elastic collisions are ideal collisions where zero kinetic energy is converted into thermal energy (heat) …100% conserved. Mathematically: p i =p f and KE i =KE f OR m 1 v i + m 2 v i = m 1 v + m 2 v AND ½ m 1 v i 2 + ½ m 2 v i 2 = ½ m 1 v 2 + ½ m 2 v 2

6. Inelastic Collisions Most collision are considered INELASTIC because some of the kinetic energy is converted into thermal energy (heat). KE NOT CONSERVED. *SPECIAL CASE: When two objects collide they may combine/stick together to become one object. This is a perfectly inelastic collision or completely inelastic. m 1 v i + m 2 v i = m T v (10 kg)(10m/s)+(10kg)(2m/s)= (20 kg)v f 100 Ns + 20 Ns=(20kg)v f 120 N/s /20 kg=V f 6 m/s=V f GIVEN: m 1 =m 2 =10 kg v 1 =10 m/s, v 2 =2 m/s V f =? KE i =½(10kg)(10m/s) 2 + ½(10kg)(2m/s) 2 ……500 J+20 J= 520 J KE f =½(20 kg)(6 m/s) 2 ……360 J……..520 J-360 J=160 J transferred

8. Conservation of Momentum If two isolated objects interact (collide or separate), then the total momentum of the system is conserved (constant). ELASTIC COLLISIONINELASTIC COLLISION SEPARATION OR EXPLOSION Types of Collisions

9. What is velocity of ball & pin Ex. Bowling ball= 7 kg, v=3m/sWhat is velocity of ball & pin AFTER collision ? (assume elastic) Bowling pin= 2 kgAFTER collision ? (assume elastic) Pi=Pf and KEi=KEf and V 1 + V 1 = V 2 + V 2 V 1 + V 1 = V 2 + V 2 3 m/s + V 1 = V 2 Two Unknowns… Two Equations. m 1 v i + m 2 v i = m 1 v + m 2 v (7kg)(3m/s)=(7kg)v 1 + (2kg) v 2 21 Ns=(7kg)v 1 + (2kg) v 2 Substitute first equation into second… 21 Ns=(7kg)v 1 + (2kg) (3 m/s + V 1 ) Solve for V 1 21 Ns=(7kg) v 1 + 6 Ns + (2 kg) v 1 OR 15 Ns= (9kg) v 1 OR v 1 =1.7 m/s Substitute back into first equation… 3 m/s + 1.7 m/s =v 2 or v 2 =4.7 m/s

10. A 0.05 kg bullet with velocity 150 m/s is shot into a 3 kg ballistic pendulum. Find how high the pendulum rises after the bullet gets stuck inside. First use conservation of momentum to find the final velocity of the bullet/pendulum. This is a perfectly inelastic collision. m 1 v 1i +m 2 v 2i = (m 1 + m 2 )vf (0.05kg)(150m/s)+(3kg)(0m/s) = (3.05kg)vf vf= 2.5 m/s Now that we know the ballistic pendulum with the bullet in it begins to swing with a speed of 2.5 m/s, we use conservation of energy to find how high it swings. ½ mv 2 = mgy ½ m (2.5m/s) 2 = mgy y= 0.32 m BALLISTIC PENDULUM (A Perfectly Inelastic Collision)

16. Δp = F Δt … Δp = F Δt