Transcript lecture17
Linear Momentum and Second Newton’s Law 2nd Newton’s Law for any object: Definition of acceleration: F ma We can write 2nd Newton’s Law as: dv F m dt Definition of momentum: Change in momentum: dp mdv p mv Second Newton’s Law in terms of momentum: dp F dt dv a dt or d(mv) F dt or Fdt dp Impulse Fdt dp p2 t2 J Fdt dp p 2 p1 p t1 p1 J Fev t 2 t1 Fev t p The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. Collisions and Impulse The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete. Example: In an egg-tossing contest, two people toss a raw egg back and forth. After each successful toss, each person takes a step back. Catching the egg without breaking it becomes harder and harder. Usually the trick is moving your hand down with the egg when you receive it. This works better because: A. It decreases the change in momentum B. It decreases the impulse C. It decreases the force on the egg • If the flying egg has speed v, the change in momentum is: Δp = 0 – mv = - mv (independent of how you catch it) • The impulse is just the same! J = Δp = - mv • By moving your hand with the egg, you are increasing the time interval over which this Δp must take place. So the average force on the egg Fev p / t decreases. By the way: Catching the egg is harder and harder because its speed becomes larger (and the required change in momentum, too), so exerting a small force becomes harder as well. System of particles. Conservation of momentum dP1 F12 F13 ... dt dP2 F21 F23 ... dt dP1 F31 F32 ... dt ................................ dPi dt Fi F1 F2 F3 F13 F23 F21 F12 F32 F31 Pi Ptot total momentum! Fi Fext total external force! F12 F21 F13 F31 F23 F32 dPtot Fext dt Fext 0 dPtot 0 Ptot const The total momentum of an isolated system of objects remains constant Example: Two guys of masses m1 =75kg and m2=90kg pull on both ends of a rope on an ice rink. After a couple of seconds, the thin one is moving at 0.2 m/s. What is the speed of the big one? No friction ptotal is conserved Fext = 0 No net vertical force p1,i p2,i p1,f p2,f 0 0 mv 1 1,f m2v2,f 75 kg m1 v1,f 0.2 m/s 0.17 m/s m2 90 kg v2,f Compare changes in linear momentum: m v (90 kg)(0.17 m/s - 0) 15 kg m/s p1 m1 v1,f v1,i (75 kg)(0.2 m/s - 0) 15 kg m/s p2 2 2,f v2,i The force produced a momentum transfer. Collisions (and explosions) A vA vB B •Momentum is conserved in all collisions. •Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic. •With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy. Completely inelastic collisions The objects stick together after collision, so there is only one final velocity ' ' ' ' A B A A B B A A B B v v v m v m v m v m v mAvA mBvB (mA mB )v m A v A mB v B v m A mB Example: A ball of mass, m1 = 2 kg has a horizontal velocity, of v1 = 7 m/s. The ball collides into a cart full of sand, as shown below. The cart has a mass, m2 = 8 kg and a horizontal velocity v2 = 1 m/s. The ball and the cart are moving towards each other. Find the velocity of the cart and the ball after the ball collides with the cart and gets stuck in the sand. m1 = 2 kg v1 = 7 m/s m2 = 8 kg v2 = 1 m/s v - ? v m1 1 v 2 m2 m1v1 m2 v2 (m1 m2 )v m1v1 m2 v2 v (m1 m2 ) 2kg 7m / s 8kg 1m / s v 0.6m / s (2kg 8kg) Example (Big block, small block): Consider the following two collisions between two blocks of masses m and M (> m). In both cases, one of the blocks is initially moving with speed v and the other is at rest. After the collision, they move together. The final speed of the two objects is larger when: A. The big block is initially at rest. B. The small block is initially at rest. C. The speed is the same in both cases. mvm Mv M m M v f A. vm v vM 0 v fA mv m M mvm Mv M vf m M B. vm 0 vM v v fB Mv v fA m M Example: An eagle of mass, mA = 6.0 kg moving with speed vA = 5.0 m/s is on collision course with a second eagle of mass mB = 4.0 kg moving at vB = 10.0 m/s in a direction perpendicular to the first. After they collide, they hold onto one another. At what speed they moving after the collision? mA=6.0 kg vA=5.0 m/s mB=4.0 kg vB=10.0 m/s v-? vB pB vA pA m m A m A 6.0 kg 4.0 kg 10.0 kg p A mA p A 6.0kg 5.0m / s 30.0kg m / s pB mB pB 4.0kg 10.0m / s 40.0kg m / s p v m p A2 pB2 m p pA pA p p A2 p B2 30.0kg m / s 2 40.0kg m / s 2 10kg 5m / s