Chapter 09 Center of Mass and Linear Momentum

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Chapter 09 Center of Mass and Linear Momentum • Center of mass: The center of mass of a body or a system of bodies is the point that moves as if all of the mass are concentrated there and all external forces are applied there. • Note that HRW uses “com” but I will use “c.m.” because “c.m.” is more standard notation for the “center of mass”. System of Particles

Consider two masses m1 at x = x1 and m2 at x2. m x + m x x ≡ 1 1 2 2 c.m. + m1 m2

• Center of mass for a system of n particles: n n 1 n = 1 = 1 = yc.m. mi yi z m z xc.m. mi xi ∑ c.m. ∑ i i ∑ M i=1 M = M i=1 i 1

• M is just the total mass n M = m of the system ∑ i i=1

v = ˆ + ˆ + ˆ • Using vectors, we have: rc.m. x c.m.i yc.m. j zc.m.k

1 n • Therefore: v = v rc.m. ∑ mi ri M i=1 • For a solid body, we can treat it as a continuous distribution of matter dm 1 1 1 = y = y dm = xc.m. x dm c.m. ∫ zc.m. z dm M ∫ M M ∫ • If the object has uniform density, dm = (M/V)dV 1 1 x = x dV y = y dV = 1 c.m. ∫ c.m. zc.m. z dV V V ∫ V ∫ • If an object has a point, a line or a plane of symmetry, the center of mass of such an object then lies at that point, on that line or in that plane. • Sample 9-2: the figure shows a uniform metal plate P of radius 2R from which a disk of radius R has been stamped out (removed). Using the x-y coordinate system shown, locate the center of mass of the plate.

Notice that the object is symmetric

about the x-axis, so yc.m. = 0. We just need to calculate xc.m. = 0. Technique: Use symmetry as much as possible. C

The center of mass of the large disk must be at the center of the disk by symmetry. The same holds true for the small disk.

The hole is as if the smaller disk S had “negative” mass to counteract the solid larger mass disk. Superimpose the two disks. The overlap of the “positive” mass C with the “negative” mass will S result mathematically in a hole.

The center of mass will be somewhere over here.

center of mass (c.m. or com) of Plate C: xC = 0 − center of mass (c.m. or com) of Disk S: xS = R m x + m x 0 + m (−R)  − m  x = C C S S = S = S R c.m. + +  +  mC mS mC mS  mC mS  Superimpose the two disks. The overlap of the “positive” mass C with the “negative” mass will S result mathematically in a hole.

Now we need mC and mS. Both disks have the same uniform mass density ρ (but with different “signs”). Thus, = ρ = ρ()()π 2 = ρ π 2 mC CVC (2R) 4 R Likewise, = ρ = −ρ()()π 2 = −ρ π 2 mS SVS (R) R Superimpose the two disks. The overlap of the “positive” mass C with the “negative” mass will S result mathematically in a hole.

= ρ = ρ()()π 2 = ρ π 2 mC CVC (2R) 4 R = ρ = −ρ()()π 2 = −ρ π 2 mS SVS (R) R m x + m x 0 + m (−R)  − m  x = C C S S = S = S R c.m. + +  +  mC mS mC mS  mC mS   − ρ(πR 2 )  1 =  R = R  2 2   ρ(4πR ) − ρ(πR )  3 • Newton’s 2nd law for a system of particles

We know n r = 1 r n rc.m. mi ri r = Σ r ∑ M rc.m. mi ri M i=1 i=1 take derivative with respect to time r = r + r + + r M vc.m. m1v1 m2v2 ... mn vn take derivative with respect to time again r = r + r + + r M a c.m. m1a1 m2a 2 ... mna n r = r + r + + r = r Ma c.m. F1 F2 ... Fn Fnet Newton’s second law for a system of particles = = = Fnet,x Ma c.m.,x Fnet,y Ma c.m.,y Fnet,z Ma c.m.,z v = v ⇒ Fnet Ma c.m.

Fnet is the net force of all external forces that act on the system. M is the total mass of the system. ac.m. is the acceleration of the center of the mass Linear Momentum • The linear momentum of a particle is a vector defined as pv = mvv

• Newton’s second law in terms of momentum dpv d dv dm v = (mv) = m + v = mav = F dt dt dt dt net

Most of the time the mass v doesn’t change, so this term v = dP is zero. Exceptions are Fnet dt rockets (Monday) The figure gives the linear momentum versus time for a particle moving along an axis. A force directed along the axis acts on the particle. (a) Rank the four regions indicated according to the magnitude of the force, greatest first

(b) In which region is the particle slowing? • Linear momentum of a system of particles

n n v = v = v = v P ∑ p i ∑ m i v i M v c.m . i=1 i=1 v = v P Mvc.m.

• Newton’s 2nd law for a system of particles v dP dv v = M c.m. = Mav = F dt dt c.m. net v v dP F = net dt Daily Quiz, February 16, 2005

A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule then dissociates with the hydrogen atom having momentum mpv along the +x axis. What happens to the helium atom? Daily Quiz, February 16, 2005

A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule

then dissociates with the hydrogen atom having momentum mpv along the +x axis. What happens to the helium (mHe = 4mp) atom? 1) stays at x=0 2) goes along +x at speed v 3) goes along −x at speed v 4) goes along −x at speed v/4 5) none of the above Daily Quiz, February 16, 2005

A helium atom and a hydrogen atom can bind to form the metastable molecule HeH (lifetime of about 1µs). Consider one such molecule at rest in the lab frame at the origin. This molecule

then dissociates with the hydrogen atom having momentum mpv along the +x axis. What happens to the helium (mHe = 4mp) atom? pinitial = 0 means pfinal = 0 4) goes along −x at speed v/4 mpv+ mHevHe = 0 = mpv + 4mpvHe = 0 − => vHe = mpv/4 Collisions take time!

Even something that seems instantaneous to us takes a finite amount of time to happen. Impulse and Change in Momentum v v = dp v = v v − v = ∆v = tf v Fnet ⇒ dp Fdt ⇒ pf pi p Fdt dt ∫ti

Call this change in momentum the “Impulse” and give it the symbol J.

v − v = ∆v = v = tf v pf pi p J Fdt ∫t i Actual force function versus time.

Average force versus collision time. Impulse and Change in Momentum Call this change in momentum the “Impulse” and give it the symbol J. v − v = ∆v = v = tf v pf pi p J Fdt ∫t i The instantaneous force is hard (or very difficult) to know in a real collision, but we can use the average force Favg. Thus,

∆ J = Favg t. Daily Quiz, February 18, 2005

Foam Table An egg was dropped on the table broke, but the egg dropped on the foam pad didn’t break. Why didn’t this egg break? Daily Quiz, February 18, 2005

Foam Table An egg was dropped on the table broke, but the egg dropped on the foam pad didn’t break. Why didn’t this egg break? 1) Foam egg’s speed was less. 2) Foam egg’s change in momentum was less. 3) Foam egg’s collision time was greater. 4) What do you mean, it did break! 0) none of the above Daily Quiz, February 18, 2005

Foam Table An egg was dropped on the table broke, but the egg dropped on the foam pad didn’t break. Why didn’t this egg break? 1) Foam egg’s speed was less. The eggs were dropped from same height, so their speeds were the same. mgh = 1/2mv2 Daily Quiz, February 18, 2005

Foam Table An egg was dropped on the table broke, but the egg dropped on the foam pad didn’t break. Why didn’t this egg break? 2) Foam egg’s change in momentum was less. The table egg’s momentum stopped at the table, but the foam egg bounced up making its change in momentum greater than the table egg! Daily Quiz, February 18, 2005

Foam Table An egg was dropped on the table broke, but the egg dropped on the foam pad didn’t break. Why didn’t this egg break? 3) Foam egg’s collision time was greater. ∆ Since J constant, if t is large then Favg will be small. ∆ J = Favg t. Conservation of Linear Momentum

• For a system of particles, if it is both isolated (the net external force acting on the system is zero) and closed ( no particles leave or enter the system )…. r r dP If Σ F = 0 then = 0 dt r r r Therefore P = constant or Pi = Pf

then the total linear momentum of the system cannot change. Law of conservation of linear momentum • Conservation of linear momentum along a specific direction: Σ If Fx = 0 Then Pi, x = Pf, x Σ If Fy = 0 Then Pi, y = Pf, y

If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change. Linear Momentum • The linear momentum of a particle is a vector defined as pv = mvv

• Newton’s second law in terms of momentum dpv d dv dm v = (mv) = m + v = mav = F dt dt dt dt net

Most of the time the mass v doesn’t change, so this term v = dP is zero. Exceptions are Fnet dt rockets (Tuesday) Conservation of Linear Momentum

If the component of the net external force on a closed system is zero along an axis, then the component of the linear momentum of the system along that axis cannot change. Collisions

In absence of external forces,

Linear momentum is conserved. Mechanical energy may or may not be conserved. Elastic collisions: Mechanical energy is conserved. Inelastic collisions: Mechanical energy is NOT conserved.

But, Linear momentum is always conserved. Conservation of Linear Momentum

v = v p i p f ⇒ v + v = v + v p 1i p 2 i p 1f p 2 f

v + v = v + v m 1 v 1i m 2 v 2 i m 1 v 1f m 2 v 2 f Center of Mass motion is constant Center of Mass Motion

v = v + v = v = + v P p 1i p 2 i M v c .m . (m 1 m 2 ) v c .m . v P ⇒ vv = c .m . + m 1 m 2 Inelastic Collisions

Perfectly inelastic collision: The two masses stick together v + v = v + v = ()+ v m 1 v 1i m 2 v 2 i m 1 v 1f m 2 v 2 f m 1 m 2 V f

0 v m V = 1 vv (= vv ) f + 1i c .m . m 1 m 2 Inelastic Collisions

Was the mechanical energy:

conserved (Ei = Ef); lost (Ei > Ef); or gained (Ei < Ef); in the collision? Inelastic Collisions

How much mechanical energy was lost in the collision? 2 2 1 v 2 1  m  2 1  m  2 ()m + m V = ()m + m  1  v =  1  v ⇒ 1 2 f 1 2  +  1i  +  1i 2 2  m 1 m 2  2  m 1 m 2  1 1  m 2  1  m m  E = E − E = m v 2 −  1  v 2 =  1 2  v 2 lost i f 1 1i  +  1i  +  1i 2 2  m 1 m 2  2  m 1 m 2 

1  m m  E =  1 2  v 2 lost  +  1i 2  m 1 m 2  Elastic Collisions

Perfectly elastic collision: Mechanical energy is conserved v + v = v + v ()= v m 1 v 1i m 2 v 2 i m 1 v 1f m 2 v 2 f M v c .m .

1 1 1 1 E = m v 2 + m v 2 = m v 2 + m v 2 = E i 2 1 1i 2 2 2 i 2 1 1f 2 2 2 f f Elastic Collisions

Perfectly elastic collision: Mechanical energy is conserved v + v = v + v ()= v m 1 v 1i m 2 v 2 i m 1 v 1f m 2 v 2 f M v c .m . m − m 2 m v = 1 2 v + 2 v 1f + 1i + 2 i m 1 m 2 m 1 m 2 2 m m − m v = 1 v + 2 1 v 2 f + 1i + 2 i m 1 m 2 m 1 m 2 What about two (or more) dimensions?

Simply break the momenta and velocities into their x-, y-, and z-components. • Inelastic collisions in two dimensions r = r Pi Pf

Pi x = Pf x Pi y = Pf y

For the case shown here:

= + m1v1i m1v1 f cosθ1 m2v2 f cosθ2 = + 0 m1v1 f sin θ1 m2v2 f sin θ2 Conservation of Linear Momentum

then the total linear momentum of the system cannot change. Law of conservation of linear momentum Linear Momentum is conserved v = v Pf Pi Let the mass M change (M + dM), which in turn makes the velocity change. Rocket exhausts −dM in time dt at a velocity U relative to our inertial reference frame. Mv = −dM U + (M + dM)(v + dv) Velocity of rocket Velocity of rocket Velocity of exhaust   =   +         relative to ref. frame relative to exhaust  relative to ref. frame + = + (v dv) vrel U U A S U

vrel v + dv Linear Momentum is conserved v = v Pf Pi = + − U (v dv) vrel Substitute and divide by dt Mv = −dM U + (M + dM)(v + dv) dM dv − v = M dt rel dt U A S U

vrel v + dv Linear Momentum is conserved v = v Pf Pi = + − U (v dv) vrel Substitute and divide by dt Mv = −dM U + (M + dM)(v + dv) = − dM vf = − Mf dM dv vrel ⇒ dv vrel M ∫vi ∫Mi M  M  − =  i  ⇒ vf vi vrel ln   Mf  A S U

vrel v + dv Problem 09-05

Find the center of mass of the ammonia molecule.

Mass ratio: N/H = 13.9 H to triangle center: d = 9.40x10-11m N to hydrogen: L = 10.14x10-11m Problem 09-05 Find the center of mass of the ammonia molecule. Problem 09-05 Find the center of mass of the ammonia molecule. Problem 09-54 Find the distance the spring is compressed. m1=2.0kg, m2=1.0kg. Problem 09-54 Find the distance the spring is compressed. m1=2.0kg, m2=1.0kg. Problem 09-56 Find the final velocity of A and is the collision elastic?

Mass A: mA=1.6kg, vAi=5.5 m/s Mass B: mB=2.4kg, vBi=2.5 m/s, vBf=4.9 m/s Problem 09-56 Find the final velocity of A and is the collision elastic?

Mass A: mA=1.6kg, vAi=5.5 m/s Mass B: mB=2.4kg, vBi=2.5 m/s, vBf=4.9 m/s Problem 09-60 Find the final speeds of the ball and block.

Mass 1 (ball): m1=0.5kg, h=0.70m Mass 2 (block): m2=2.5kg, v2i=0.0 m/s Problem 09-60 Find the final speeds of the ball and block.

Mass 1 (ball): m1=0.5kg, h=0.70m Mass 2 (block): m2=2.5kg, v2i=0.0 m/s Problem 09-63 Find the mass m to stop M and the final height of m.

Mass 1 (baseball): m=?, hinitial =1.8m Mass 2 (basketball): M=0.63kg, vMf=0.0 m/s

Elastic collisions: Mechanical energy is conserved. 1 Basketball: Mgh = Mv2 ⇒ v = 2gh 2 rebounds upward with same speed -- only reversed direction Problem 09-63 Find the mass m to stop M and the final height of m.

Mass 1 (baseball): m=?, hinitial =1.8m Mass 2 (basketball): M=0.63kg, vMf=0.0 m/s

1 Baseball: mgh = mv2 ⇒ v = 2gh 2

M − m 2m M − m 2m v = v + v = 2gh − 2gh Mf M + m Mi M + m mi M + m M + m M − 3m = 2gh = 0 M + m ⇒ m = M / 3 = 0.21kg Problem 09-63 Find the mass m to stop M and final height of m.

Mass 1 (ball): m1=0.5kg, h=0.70m Mass 2 (block): m2=2.5kg, v2i=0.0 m/s Problem 09-97 Find the final speeds of the sleds. Mass 1(sleds): M=22.7kg,

Mass 2 (cat): m=3.63kg, vi=3.05 m/s Problem 09-97 Problem 09-97 Problem 09-130

Ball 1 vo=10.0m/s at contact point of balls 2 and 3. All three balls have mass m. Find the final velocities of all three balls. Problem 09-130

Ball 1 vo=10.0m/s at contact point of balls 2 and 3. All three balls have mass m. Find the final velocities of all three balls.

Pi = mvo 2 ball1 balls 2 & 3 P = mV + 2mv cosθ 1 f

θ = 30o since all three balls are identical Problem 09-130

Ball 1 vo=10.0m/s at contact point of balls 2 and 3. All three balls have mass m. Find the final velocities of all three balls. Problem 09-130

Ball 1 vo=10.0m/s at contact point of balls 2 and 3. All three balls have mass m. Find the final velocities of all three balls. A Quiz

p e− x

CERN

Consider a proton (mass mp and kinetic energy Ep) colliding head on with an electron (mass me) initially at rest. What is the maximum kinetic energy (Ee) that can be delivered to the electron? A Quiz

y Consider a proton (mass mp = 1836me and CERN kinetic energy Ep) colliding head on with an electron (mass m ) initially at rest. What is p e e− x the maximum kinetic energy (Ee) that can be delivered to the electron? 1) Ee = Ep 2) Ee < Ep 3) Ee > Ep 4) not enough information A Quiz = + Conservation of Momentum mpVi mpVf mp vf

1 2 = 1 2 + 1 2 Conservation of Energy mpVi mpVf mevf 2 2 2 1 1 E ≡ m V2 and E = m v2 p 2 p i e 2 e f 4m /m E = e p E < E e ()+ 2 p p 1 me/mp y Consider a proton (mass mp = 1836me and CERN kinetic energy Ep) colliding head on with an electron (mass m ) initially at rest. What is p e e− x the maximum kinetic energy (Ee) that can be delivered to the electron? 1) Ee = Ep 2) Ee < Ep 3) Ee > Ep 4) not enough information