Lecture 7 Chapter 9 Systems of Particles Wednesday Warm-up problem Puzzle Question: Cite two possible reasons why it appears that some basket ball players and dancers have a greater hang time. Physlets

Topics Center of mass Newton’s 2nd law for a system of particles Linear momentum, 2nd law in terms of P Conservation of P

Center of mass The center of a body or a system of bodies is the point that moves as though all of the mass were concentrated there and all external forces were applied there.

Center of Mass Why is it important? For any rigid body the motion of the body is given by the motion of the cm and the motion of the body around the cm. As an example find the center of mass of the following system x m M. x cm d m x cm = M (d-x cm ) x cm = M/(m + M) d

Problem dimensions Find x cm and y cm x cm M = m 1 x 1 + m 2 x 2 + m 3 x 3 y cm M = m 1 y 1 + m 2 y 2 + m 3 y 3 x cm 15 = 3*0 + 4*2 + 8*1 x cm = 16/15 = 1.1 m y cm 15 = 3*0 +4*1 + 8*2 y cm = 20/15 = 1.33 m.

If the 8 kg mass increases, how does the cm change? x cm =(m 1 x 1 + m 2 x 2 + m 3 x 3 )/(m 1 +m 2 +m 3 ) When m 3 gets very large, suppose we neglect the other masses. x cm ~(m 3 x 3 )/(m 3 ) ~ x 3 The center of mass moves towards the large object as it should

Center of Mass for a system of particles x cm = (m 1 x 1 + m 2 x 2 + m 3 x 3 )/M

Newton’s Second Law for a System of particles: F net = Ma cm take d/dt on both sides take d/dt again Identify ma as the force on each particle

Linear Momentum form of Newton’s 2nd Law Important because it is a vector quantity that is conserved in interactions. Now take derivative d/dt of

Law of Conservation of Linear Momentum If F net = 0 on a closed system where no mass enters or leaves the system, then dP/dt = 0 or P = constant. P i = P f for a closed isolated system Also each component of the momentum P x,P y,P z is also constant since F x, F y, and F z all = 0 If one component of the net force is not 0, then that component of momentum is not a constant. For example, consider the motion of a horizontally fired projectile. The y component of P changes while the horizontal component is fixed after the bullet is fired.

Air track examples Two carts connected by a spring. Set them into oscillation by pulling them apart and releasing them from rest. Note cm does not move. Now repeat with spring between two carts. Analogous to exploding mass. Again observers in two different inertial reference frames will measure different values of momentum, but both will agree that momentum is conserved. Note if the net force vanishes in one inertial frame it will vanish in all inertial frames

F(t) -F(t) Change in momentum of ball on left or right.

Andy Rodick has been clocked at serving a tennis ball up to 149 mph(70 m/s). The time that the ball is in contact with the racquet is about 4 ms. The mass of a tennis ball as is about 300 grams. F avg = J/  t =  p/  t  p =70( 0.3) - 0 = 21 kg-m/s F avg = 21/0.004 = 5000 Newtons What is the acceleration of the ball? a = F avg /m = 5000N/0.3 kg = 16,667 m/s 2 What is the average force exerted by the racquet on the ball? What distance the racquet go through while the ball is still in contact? V 2 f - V 2 i =2ax x = V 2 f /2a = (70) 2 /2(16667) = 0.15 m

hihi Initial -v f Before bounce vfvf After bounce BOUNCING BALL E = PE = mgh E = KE = 1/2mv 2 hfhf

Measuring velocities and heights of balls bouncing from a infinitely massive hard floor Type of Ball Coefficient of Restitution (C.O.R.) Rebound Energy/ Collision Energy (R.E./C.E.) Superball Racquet ball Golf ball Tennis ball Steel ball bearing Baseball Foam rubber ball Unhappy ball Beanbag R.E./C.E.= H i /H f Almost elastic collision Almost inelastic collision C.O.R. = V f /V i

Conservation of Momentum In a closed isolated system containing a collision, the linear momentum of each colliding body may change but the total linear momentum P of the system can not change, whether the collision is elastic or inelastic. F net = 0, dP/dt = 0, Hence, P = constant. Each x,y, and z component of the momentum is a constant.

One Dimension Collision m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Total momentum before = Total momentum after 1/2 m 1 v 1i 2 = 1/2 m 1 v 1f 2 + 1/2 m 2 v 2f 2 Kinetic energy is conserved too.

Balls bouncing off massive floors, we have m 2 >>m 1 m1m1 m2m2

Lecture 8 Chapter 10 Collisions Thursday

Colliding pool balls The excutive toy m 2 = m 1

Two moving colliding objects: Problem 45 m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 1/2 m 1 v 1i 2 + 1/2 m 2 v 2i 2 = 1/2 m 1 v 1f 2 + 1/2 m 2 v 2f 2 -v v v 1f v m1m1 m2m2 Just before each hits the floor Just after the big ball bounced Just after the little ball bounced off the big ball

-v v For m 2 =3m 1, v 1f = 8/4V =2V superball has twice as much speed. How high does it go? 4 times higher V 1f =2V V 2f =0

-v v How high does it go? V 1f =3V V 2f = -V 9 times higher For maximum height consider m 2 >> m 1 V 1f =3V V 2f = -V