physics 2210 fall 2015woolf/2210_jui/oct21.pdf= 342 scโ pix/s ๐คฬ . totally inelastic collision...
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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 ๐๐, position ๐๐ and velocity ๏ฟฝโ๏ฟฝ๐ (both ๐๐ and ๏ฟฝโ๏ฟฝ๐ are functions of time) for ๐ = 1,2,3,โฏ ,๐. The net force on the ๐๐ญ๐ญ particle can be written as
๏ฟฝโ๏ฟฝ๐ = ๏ฟฝ๏ฟฝโ๏ฟฝ๐๐๐โ ๐
+ ๏ฟฝโ๏ฟฝ๐(ext)
where ๏ฟฝโ๏ฟฝ๐๐ is the (internal) force exerted by another (๐th) particle in the group,
and ๏ฟฝโ๏ฟฝ๐(ext)
is the net external force (vector sum of forces on the ๐th particle, not exerted by another particle in the group). We now sum over the group: Remember we generally are not allowed to do thisโฆ but here we are treating the group as a single object!!! (Definition) The net force on the group is given by
๏ฟฝโ๏ฟฝ = ๏ฟฝ๏ฟฝโ๏ฟฝ๐
๐
๐=1
= ๏ฟฝ๏ฟฝ๏ฟฝโ๏ฟฝ๐๐๐โ ๐
๐
๐=1
+ ๏ฟฝ๏ฟฝโ๏ฟฝ๐(ext)
๐
๐=1
Now: By Newtonโs 3rd Law (units 10-13 are all about the consequences of N3L)
Each ๏ฟฝโ๏ฟฝ๐๐ in the sum โ โ ๏ฟฝโ๏ฟฝ๐๐๐โ ๐๐๐=1 is cancelled by an equal and opposite ๏ฟฝโ๏ฟฝ๐๐
Collective Motion (continued) i.e.
๐๐๐ โ๏ฟฝ๏ฟฝ๏ฟฝโ๏ฟฝ๐๐๐โ ๐
๐
๐=1
โก 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.
๏ฟฝโ๏ฟฝ = ๏ฟฝ๏ฟฝโ๏ฟฝ๐
๐
๐=1
= ๏ฟฝ๏ฟฝโ๏ฟฝ๐(ext)
๐
๐=1
Now By Newtonโs 2nd Law, the net force on the ๐๐ก๐ก particle is related to its acceleration by
๏ฟฝโ๏ฟฝ๐ โก๐๏ฟฝโ๏ฟฝ๐๐๐
โก๐2๐๐๐๐2
=1๐๐
๏ฟฝโ๏ฟฝ๐
Note THIS DOES NOT MEAN
๏ฟฝโ๏ฟฝ๐ =1๐๐
๏ฟฝโ๏ฟฝ๐(ext)
Because the (ext) desgnation here means outside of the group, but for an individual particle you have to count the โexternal (to the particle) forceโ exerted by other particles in the group!!!
Newtonโs 2nd Law for the Collective Multiplying ๏ฟฝโ๏ฟฝ๐ by ๐๐ and summing over the group, interchanging the order of summation and differentiation (derivatives are โlinearโ)
๏ฟฝ๐๐๐2๐๐๐๐2
๐
๐=1
=๐2
๐๐2๏ฟฝ๐๐๐๐
๐
๐=1
= ๏ฟฝ๏ฟฝโ๏ฟฝ๐
๐
๐=1
= ๏ฟฝ๏ฟฝโ๏ฟฝ๐(ext)
๐
๐=1
= ๏ฟฝโ๏ฟฝ
Note that
๏ฟฝ๐๐๐๐
๐
๐=1
= ๐ โ1๐๏ฟฝ๐๐๐๐
๐
๐=1
= ๐๐ ๐ถ๐ถ
And so: ๐2
๐๐2๐๐ ๐ถ๐ถ = ๐
๐2๐ ๐ถ๐ถ๐๐2
โก ๐๐๐๐ถ๐ถ๐๐
โก ๐๐ด๐ถ๐ถ = ๏ฟฝโ๏ฟฝ
Which looks just like Newtonโs 2nd Law for a particle of mass ๐ = โ ๐๐๐๐=1 ,
located at the center of mass of the group. We have implicitly defined the velocity and acceleration of the center-of-mass by
๐๐ถ๐ถ โก๐๐ ๐ถ๐ถ๐๐
, ๐ด๐ถ๐ถ โก๐๐๐ถ๐ถ๐๐
โก๐2๐ ๐ถ๐ถ๐๐2
Unit 11
๏ฟฝโ๏ฟฝ โก ๐๏ฟฝโ๏ฟฝ [ Units: kg โ m/s ]
Unit 11
๏ฟฝโ๏ฟฝ โก ๐๏ฟฝโ๏ฟฝ [ Units: kg โ 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
๏ฟฝโ๏ฟฝ โก ๐๏ฟฝโ๏ฟฝ [ Units: kg โ m/s ]
Newtonโs Second Law: (time) rate of change of the momentum vector is equal to the net force (vector sum of all โexternalโ forces) on a body.
๐๏ฟฝโ๏ฟฝ๐๐
= ๏ฟฝโ๏ฟฝ
These are vector relations: Definition of momentum:
๐๐ฅ โก ๐๐ฃ๐ฅ โก ๐๐๐๐๐
, ๐๐ฆ โก ๐๐ฃ๐ฆ โก ๐๐๐ฆ๐๐
, ๐๐ง โก ๐๐ฃ๐ง โก ๐๐๐ง๐๐
Newtonโs Second Law: ๐๐๐ฅ๐๐
= ๐น๐ฅ ,๐๐๐ฆ๐๐
= ๐น๐ฆ,๐๐๐ง๐๐
= ๐น๐ง
Total Momentum of a System of Particles Take a group of particles, each with mass ๐๐, position ๐๐ and velocity ๏ฟฝโ๏ฟฝ๐ (both ๐๐ and ๏ฟฝโ๏ฟฝ๐ are functions of time) for ๐ = 1,2,3,โฏ ,๐. Definition of Total Momentum: vector sum of individual momenta
๐ โก๏ฟฝ๏ฟฝโ๏ฟฝ๐
๐
๐=1
โก๏ฟฝ๐๐
๐
๐=1
๏ฟฝโ๏ฟฝ๐ โก๏ฟฝ๐๐๐๐๐๐๐
๐
๐=1
= ๐๐๐ถ๐ถ
๐๐ฅ โก๏ฟฝ๐๐๐ฅ
๐
๐=1
โก๏ฟฝ๐๐
๐
๐=1
๐ฃ๐๐ฅ โก๏ฟฝ๐๐๐๐๐๐๐
๐
๐=1
, ๐๐ฆ โก๏ฟฝ๐๐๐ฆ
๐
๐=1
โก๏ฟฝ๐๐
๐
๐=1
๐ฃ๐๐ฆ โก๏ฟฝ๐๐๐๐ฆ๐๐๐
๐
๐=1
, โฏ
Differentiating the total momentum w.r.t. time:
๐๐๐๐ = ๏ฟฝ
๐๐๐๏ฟฝโ๏ฟฝ๐
๐
๐=1
= ๏ฟฝ๏ฟฝโ๏ฟฝ๐
๐
๐=1
= ๏ฟฝ๏ฟฝโ๏ฟฝ๐(ext)
๐
๐=1
= ๏ฟฝโ๏ฟฝ
Which is the alternate form of Newtonโs 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)
๐น๐ฅ = 0 โ๐๐๐๐๐ฅ โก
๐๐๐๏ฟฝ๐๐๐ฅ
๐
๐=1
= 0, ๐น๐ฆ = 0 โ๐๐๐๐๐ฆ โก
๐๐๐๏ฟฝ๐๐๐ฆ
๐
๐=1
= 0, โฏ
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 ๐1 is traveling at speed ๐ฃ๐ in the +๐ direction towards a second cart of mass ๐2, which is at rest. They collide and stick together. What is their (common) speed ๐ฃ๐ 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: ๐2 = ๐1
= 1.0 SC This is called a โtotally inelastic collisionโ
Conservation of Momentum in a Collision Theory: Total momentum is conserved in the ๐ direction because no external forces with non-zero ๐-components act on the group (they interact but the internal forces must cancel because of N3L)
๐๐๐ฅ = ๐1๐ฃ๐ ๐๐๐ฅ = ๐1 + ๐2 ๐ฃ๐
Setting ๐๐๐ฅ = ๐๐๐ฅ we get ๐1๐ฃ๐ = ๐1 + ๐2 ๐ฃ๐
So the final speed of the conjoined carts is given by
๐ฃ๐ =๐1
๐1 + ๐2๐ฃ๐
Or: the ratio ๐ฃ๐/๐ฃ๐ is given by ๐ฃ๐๐ฃ๐
=๐1
๐1 + ๐2
Predictions for three cases (1) ๐1 = 1.0 SC, ๐2 = 1.0 SC: ๐ฃ๐/๐ฃ๐ = 1/2 = 0.500 (2) ๐1 = 2.0 SC, ๐2 = 1.0 SC: ๐ฃ๐/๐ฃ๐ = 2/3 = 0.667 (3) ๐1 = 1.0 SC, ๐2 = 2.0 SC: ๐ฃ๐/๐ฃ๐ = 1/3 = 0.333
Case 1: m1 = m2
Totally Inelastic Collision: m2 = m1
This window dump missed the first digitized point
We have ๐1 = 1.0 SC ๐2 = 1.0 SC Before the collision: ๏ฟฝโ๏ฟฝ1๐ = (1.0 SCU)*(342 pix/s) ๐คฬ = 342 SCโ pix/s ๐คฬ ๏ฟฝโ๏ฟฝ2๐ = (1.0 SCU)*(0.0 pix/s) SCโ pix/s ๐คฬ = 0 ๐คฬ ๐๐ = ๏ฟฝโ๏ฟฝ1๐ + ๏ฟฝโ๏ฟฝ2๐ = 342 SCโ pix/s ๐คฬ
Totally Inelastic Collision : m2 = m1
After the collision:
๐๐ = (1.0 + 1.0) SCU * (168 pix/s) ๐คฬ = 336 SCโ pix/s ๐คฬ Predicted ๐ฃ๐/๐ฃ๐ = 1/2 = 0.500 Measured: ๐ฃ๐/๐ฃ๐ = 168/342 = 0.491.
Within a few % of ๐ท๐ โผ!
Within a few % of ๐ฉ๐ฉ๐ฉ๐ฉ๐ฉ๐ฉ๐ญ๐ฉ๐ฉ๐ฉ !!!
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
๐1 = 2.0 SC ๐2 = 1.0 SC
Collision Experiment: A cart of mass ๐1 is traveling with velocity +๐ฃ๐ (in the +๐ direction) towards a second cart of mass ๐2, which is at rest. They collide elastically. What are their velocities after the collision?
(Almost) Elastic Collision Case 1: m2 = 0.5 m1 Digitized ๐1 and ๐2 every 3 frames (0.10 s) Since ๐2 = 2 (SC) and 0.5 ๐1 = 1.0 (SC) then
๐๐ถ๐ถ =๐1๐1 + ๐2๐2๐1 + ๐2
=2๐1 + ๐2
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 โข From (1) Conservation of Momentum and (2) Conservation of Energy
๐ฃ1๐ =๐1 โ๐2
๐1 + ๐2๐ฃ๐ , ๐ฃ2๐ =
2๐1
๐1 + ๐2๐ฃ๐
โข The experimental Results gave us V1i = Vi = 340 pix/s V1f=133 pix/s V2f = 421 pix/s 1. V1f/Vi: predicted:
๐ฃ1๐๐ฃ๐
=๐1 โ๐2
๐1 + ๐2=
2 โ 12 + 1
= 0.333
Measured: V1f/Vi = 133/340 = 0.391 2. V1f/Vi: predicted:
๐ฃ2๐๐ฃ๐
=2๐1
๐1 + ๐2=
42 + 1
= 1.333
Measured: V2f/Vi = 421/340 = 1.238
Conclusion: the Collision was NOT completely elastic