Physics 2210 Fall 2015

smartPhysics 10 Center-of-Mass

11 Conservation of Momentum 10/21/2015

Collective Motion and Center-of-Mass Take a group of particles, each with mass 𝑚𝑖, position 𝑟𝑖 and velocity �⃗�𝑖 (both 𝑟𝑖 and �⃗�𝑖 are functions of time) for 𝑖 = 1,2,3,⋯ ,𝑁. The net force on the 𝒊𝐭𝐭 particle can be written as

�⃗�𝑖 = ��⃗�𝑗𝑖𝑗≠𝑖

+ �⃗�𝑖(ext)

where �⃗�𝑗𝑖 is the (internal) force exerted by another (𝑗th) particle in the group,

and �⃗�𝑖(ext)

is the net external force (vector sum of forces on the 𝑖th particle, not exerted by another particle in the group). We now sum over the group: Remember we generally are not allowed to do this… but here we are treating the group as a single object!!! (Definition) The net force on the group is given by

�⃗� = ��⃗�𝑖

𝑁

𝑖=1

= ���⃗�𝑗𝑖𝑗≠𝑖

𝑁

𝑖=1

+ ��⃗�𝑖(ext)

𝑁

𝑖=1

Now: By Newton’s 3rd Law (units 10-13 are all about the consequences of N3L)

Each �⃗�𝑗𝑖 in the sum ∑ ∑ �⃗�𝑗𝑖𝑗≠𝑖𝑁𝑖=1 is cancelled by an equal and opposite �⃗�𝑖𝑗

Collective Motion (continued) i.e.

𝐍𝟑𝐋 →���⃗�𝑗𝑖𝑗≠𝑖

𝑁

𝑖=1

≡ 0

for any group of partcles. So the net force on the group is always equal to just the sum over the external forces on the individual particles in the group.

�⃗� = ��⃗�𝑖

𝑁

𝑖=1

= ��⃗�𝑖(ext)

𝑁

𝑖=1

Now By Newton’s 2nd Law, the net force on the 𝑖𝑡𝑡 particle is related to its acceleration by

�⃗�𝑖 ≡𝑑�⃗�𝑖𝑑𝑑

≡𝑑2𝑟𝑖𝑑𝑑2

=1𝑚𝑖

�⃗�𝑖

Note THIS DOES NOT MEAN

�⃗�𝑖 =1𝑚𝑖

�⃗�𝑖(ext)

Because the (ext) desgnation here means outside of the group, but for an individual particle you have to count the “external (to the particle) force” exerted by other particles in the group!!!

Newton’s 2nd Law for the Collective Multiplying �⃗�𝑖 by 𝑚𝑖 and summing over the group, interchanging the order of summation and differentiation (derivatives are “linear”)

�𝑚𝑖𝑑2𝑟𝑖𝑑𝑑2

𝑁

𝑖=1

=𝑑2

𝑑𝑑2�𝑚𝑖𝑟𝑖

𝑁

𝑖=1

= ��⃗�𝑖

𝑁

𝑖=1

= ��⃗�𝑖(ext)

𝑁

𝑖=1

= �⃗�

Note that

�𝑚𝑖𝑟𝑖

𝑁

𝑖=1

= 𝑀 ∙1𝑀�𝑚𝑖𝑟𝑖

𝑁

𝑖=1

= 𝑀𝑅𝐶𝐶

And so: 𝑑2

𝑑𝑑2𝑀𝑅𝐶𝐶 = 𝑀

𝑑2𝑅𝐶𝐶𝑑𝑑2

≡ 𝑀𝑑𝑉𝐶𝐶𝑑𝑑

≡ 𝑀𝐴𝐶𝐶 = �⃗�

Which looks just like Newton’s 2nd Law for a particle of mass 𝑀 = ∑ 𝑚𝑖𝑁𝑖=1 ,

located at the center of mass of the group. We have implicitly defined the velocity and acceleration of the center-of-mass by

𝑉𝐶𝐶 ≡𝑑𝑅𝐶𝐶𝑑𝑑

, 𝐴𝐶𝐶 ≡𝑑𝑉𝐶𝐶𝑑𝑑

≡𝑑2𝑅𝐶𝐶𝑑𝑑2

Unit 11

�⃗� ≡ 𝑚�⃗� [ Units: kg ∙ m/s ]

Unit 11

�⃗� ≡ 𝑚�⃗� [ Units: kg ∙ m/s ]

Momentum Newton did not actually formulate his laws in terms of acceleration. Instead he used a quantity called momentum Definition of momentum: velocity multiplied by mass

�⃗� ≡ 𝑚�⃗� [ Units: kg ∙ m/s ]

Newton’s Second Law: (time) rate of change of the momentum vector is equal to the net force (vector sum of all “external” forces) on a body.

𝑑�⃗�𝑑𝑑

= �⃗�

These are vector relations: Definition of momentum:

𝑝𝑥 ≡ 𝑚𝑣𝑥 ≡ 𝑚𝑑𝑑𝑑𝑑

, 𝑝𝑦 ≡ 𝑚𝑣𝑦 ≡ 𝑚𝑑𝑦𝑑𝑑

, 𝑝𝑧 ≡ 𝑚𝑣𝑧 ≡ 𝑚𝑑𝑧𝑑𝑑

Newton’s Second Law: 𝑑𝑝𝑥𝑑𝑑

= 𝐹𝑥 ,𝑑𝑝𝑦𝑑𝑑

= 𝐹𝑦,𝑑𝑝𝑧𝑑𝑑

= 𝐹𝑧

Total Momentum of a System of Particles Take a group of particles, each with mass 𝑚𝑖, position 𝑟𝑖 and velocity �⃗�𝑖 (both 𝑟𝑖 and �⃗�𝑖 are functions of time) for 𝑖 = 1,2,3,⋯ ,𝑁. Definition of Total Momentum: vector sum of individual momenta

𝑃 ≡��⃗�𝑖

𝑁

𝑖=1

≡�𝑚𝑖

𝑁

𝑖=1

�⃗�𝑖 ≡�𝑚𝑖𝑑𝑟𝑖𝑑𝑑

𝑁

𝑖=1

= 𝑀𝑉𝐶𝐶

𝑃𝑥 ≡�𝑝𝑖𝑥

𝑁

𝑖=1

≡�𝑚𝑖

𝑁

𝑖=1

𝑣𝑖𝑥 ≡�𝑚𝑖𝑑𝑑𝑖𝑑𝑑

𝑁

𝑖=1

, 𝑃𝑦 ≡�𝑝𝑖𝑦

𝑁

𝑖=1

≡�𝑚𝑖

𝑁

𝑖=1

𝑣𝑖𝑦 ≡�𝑚𝑖𝑑𝑦𝑖𝑑𝑑

𝑁

𝑖=1

, ⋯

Differentiating the total momentum w.r.t. time:

𝑑𝑑𝑑𝑃 = �

𝑑𝑑𝑑�⃗�𝑖

𝑁

𝑖=1

= ��⃗�𝑖

𝑁

𝑖=1

= ��⃗�𝑖(ext)

𝑁

𝑖=1

= �⃗�

Which is the alternate form of Newton’s Second Law for the group. This leads to the Law of Conservation of Momentum: If the net (external) force on the group of particles is zero, the total momentum is conserved (this is true for each component independetly)

𝐹𝑥 = 0 →𝑑𝑑𝑑𝑃𝑥 ≡

𝑑𝑑𝑑�𝑝𝑖𝑥

𝑁

𝑖=1

= 0, 𝐹𝑦 = 0 →𝑑𝑑𝑑𝑃𝑦 ≡

𝑑𝑑𝑑�𝑝𝑖𝑦

𝑁

𝑖=1

= 0, ⋯

Poll 10-21-01

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. If you throw a ball off the cart towards the left, will the cart be put into motion (neglect friction between cart and ground)?

A. Yes, and it moves to the right. B. Yes, and it moves to the left. C. No, it remains in place

Poll 10-21-02

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical surface that is firmly attached to the cart. If the ball bounces straight back as shown in the picture, will the cart be put into motion after the ball bounces back from the surface?

A. Yes, and it moves to the right. B. Yes, and it moves to the left. C. No, it remains in place

Poll 10-21-03

Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck.

Which box ends up moving faster?

A. Box 1 B. Box 2 C. Same

Example 11-1 (1/3) A 12 kg block is at rest on a level floor. A 419 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty? (%i1) /* Completely inelastic collision initially Let putty be m1=0.419kg , block m2=12 kg

Total momentum: */

P: m1*v1 + m2*v2;

(%o1) m2 v2 + m1 v1

(%i2) Pi: P, v1=v0, v2=0;

(%o2) m1 v0

(%i3) Pf: P, v1=vf, v2=vf;

(%o3) m2 vf + m1 vf

(%i4) soln1: solve(Pi=Pf, vf);

m1 v0

(%o4) [vf = -------]

m2 + m1

(%i5) vf: rhs(soln1[1]);

m1 v0

(%o5) -------

m2 + m1

... continued

Example 11-1 (2/3) A 12 kg block is at rest on a level floor. A 419 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty? (%i1) (%i6) /* second part: friction force does work Wf on block+putty */ M = m1 + m2;

(%o6) M = m2 + m1

(%i7) KEi: 0.5*M*vf^2;

2 2

0.5 m1 v0 M

(%o7) -------------

2

(m2 + m1)

(%i8) KEf: 0;

(%o8) 0

(%i9) /* normal force is equal to weight in this problem */

N: M*g;

(%o9) g M

(%i10) /* friction force is in the -x direction */

Ff: -mu_k*N;

(%o10) - g mu_k M

... continued

Example 11-1 (3/3) A 12 kg block is at rest on a level floor. A 419 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty? (%i11) /* work done by friction force */ Wf: Ff*Dx;

(%o11) - Dx g mu_k M

(%i12) /* KEi + Wf = KEf by work-energy theorem: solve for v0 */

soln2: solve(KEi+Wf=KEf, v0); (sqrt(2) m2 + sqrt(2) m1) sqrt(Dx g mu_k)

(%o12) [v0 = - -----------------------------------------,

m1

(sqrt(2) m2 + sqrt(2) m1) sqrt(Dx g mu_k)

v0 = -----------------------------------------]

m1

(%i13) /* take positive root */

v0: rhs(soln2[2]);

(sqrt(2) m2 + sqrt(2) m1) sqrt(Dx g mu_k)

(%o13) -----------------------------------------

m1

(%i14) v0, m1=0.419, m2=12.0, Dx=0.15, g=9.81, mu_k=0.40, numer;

(%o14) 32.15864420812297

Answer: v0 = 32.2 m/s

Conservation of Momentum in a Collision Collision Experiment: A cart of mass 𝑚1 is traveling at speed 𝑣𝑖 in the +𝑑 direction towards a second cart of mass 𝑚2, which is at rest. They collide and stick together. What is their (common) speed 𝑣𝑓 after the collision?

http://www.physics.utah.edu/~jui/2210_s2015/collision01/inelastic_cars_x264.avi

From http://physics.wfu.edu/demolabs/demos/avimov/bychptr/chptr3_energy.htm

Case 1: 𝑚2 = 𝑚1

= 1.0 SC This is called a “totally inelastic collision”

Conservation of Momentum in a Collision Theory: Total momentum is conserved in the 𝑑 direction because no external forces with non-zero 𝑑-components act on the group (they interact but the internal forces must cancel because of N3L)

𝑃𝑖𝑥 = 𝑚1𝑣𝑖 𝑃𝑓𝑥 = 𝑚1 + 𝑚2 𝑣𝑓

Setting 𝑃𝑖𝑥 = 𝑃𝑓𝑥 we get 𝑚1𝑣𝑖 = 𝑚1 + 𝑚2 𝑣𝑓

So the final speed of the conjoined carts is given by

𝑣𝑓 =𝑚1

𝑚1 + 𝑚2𝑣𝑖

Or: the ratio 𝑣𝑓/𝑣𝑖 is given by 𝑣𝑓𝑣𝑖

=𝑚1

𝑚1 + 𝑚2

Predictions for three cases (1) 𝑚1 = 1.0 SC, 𝑚2 = 1.0 SC: 𝑣𝑓/𝑣𝑖 = 1/2 = 0.500 (2) 𝑚1 = 2.0 SC, 𝑚2 = 1.0 SC: 𝑣𝑓/𝑣𝑖 = 2/3 = 0.667 (3) 𝑚1 = 1.0 SC, 𝑚2 = 2.0 SC: 𝑣𝑓/𝑣𝑖 = 1/3 = 0.333

Case 1: m1 = m2

Totally Inelastic Collision: m2 = m1

This window dump missed the first digitized point

We have 𝑚1 = 1.0 SC 𝑚2 = 1.0 SC Before the collision: �⃗�1𝑖 = (1.0 SCU)*(342 pix/s) 𝚤̂ = 342 SC⋅pix/s 𝚤̂ �⃗�2𝑖 = (1.0 SCU)*(0.0 pix/s) SC⋅pix/s 𝚤̂ = 0 𝚤̂ 𝑃𝑖 = �⃗�1𝑖 + �⃗�2𝑖 = 342 SC⋅pix/s 𝚤̂

Totally Inelastic Collision : m2 = m1

After the collision:

𝑃𝑓 = (1.0 + 1.0) SCU * (168 pix/s) 𝚤̂ = 336 SC⋅pix/s 𝚤̂ Predicted 𝑣𝑓/𝑣𝑖 = 1/2 = 0.500 Measured: 𝑣𝑓/𝑣𝑖 = 168/342 = 0.491.

Within a few % of 𝑷𝒊 ‼!

Within a few % of 𝐩𝐩𝐩𝐩𝐩𝐩𝐭𝐩𝐩𝐩 !!!

t (s) x (pix) 0.00 100 0.10 141 0.20 172 0.30 212 0.40 243 0.50 282 0.60 321 0.70 351 0.80 386 0.90 402 1.00 419 1.10 437 1.20 452 1.30 469 1.40 484 1.50 500 1.60 516 1.70 529 1.80 543

Unit 12

Unit 12

Conservation of Momentum in Elastic(?) Collisions

http://www.physics.utah.edu/~jui/2210_s2015/collision02/elastic_cars_bs_x264.avi

From http://physics.wfu.edu/demolabs/demos/avimov/bychptr/chptr3_energy.htm

𝑚1 = 2.0 SC 𝑚2 = 1.0 SC

Collision Experiment: A cart of mass 𝑚1 is traveling with velocity +𝑣𝑖 (in the +𝑑 direction) towards a second cart of mass 𝑚2, which is at rest. They collide elastically. What are their velocities after the collision?

(Almost) Elastic Collision Case 1: m2 = 0.5 m1 Digitized 𝑑1 and 𝑑2 every 3 frames (0.10 s) Since 𝑚2 = 2 (SC) and 0.5 𝑚1 = 1.0 (SC) then

𝑋𝐶𝐶 =𝑚1𝑑1 + 𝑚2𝑑2𝑚1 + 𝑚2

=2𝑑1 + 𝑑2

3

Measurement of V1, V2: (from slope near time of collision) V1i = Vi = 340 pix/s V1f=133 pix/s V2f = 421 pix/s

t (s) x1 (pix) x2 (pix) xcm (pix) 0 110 406 212.07 0.1 148 406 236.97 0.2 176 406 255.31 0.3 211 406 278.24 0.4 240 406 297.24 0.5 276 406 320.83 0.6 313 406 345.07 0.7 341 406 363.41 0.8 375 428 393.28 0.9 390 473 418.62 1 403 519 443.00 1.1 415 553 462.59 1.2 424 597 483.66 1.3 435 630 502.24 1.4 444 673 522.97

0.00s

0.10s

0.20s

0.40s

0.30s

0.50s

0.60s

0.70s

0.80s

0.90s

1.00s

1.10s

1.20s

1.30s

1.40s

The collision as seen in the center of mass frame

Comparison of Prediction and Measurement • From (1) Conservation of Momentum and (2) Conservation of Energy

𝑣1𝑓 =𝑚1 −𝑚2

𝑚1 + 𝑚2𝑣𝑖 , 𝑣2𝑓 =

2𝑚1

𝑚1 + 𝑚2𝑣𝑖

• The experimental Results gave us V1i = Vi = 340 pix/s V1f=133 pix/s V2f = 421 pix/s 1. V1f/Vi: predicted:

𝑣1𝑓𝑣𝑖

=𝑚1 −𝑚2

𝑚1 + 𝑚2=

2 − 12 + 1

= 0.333

Measured: V1f/Vi = 133/340 = 0.391 2. V1f/Vi: predicted:

𝑣2𝑓𝑣𝑖

=2𝑚1

𝑚1 + 𝑚2=

42 + 1

= 1.333

Measured: V2f/Vi = 421/340 = 1.238

Conclusion: the Collision was NOT completely elastic