Physics 2210 Fall 2015woolf/2210_Jui/oct21.pdf= 342 SCβ pix/s π€Μ . Totally Inelastic Collision...
Transcript of Physics 2210 Fall 2015woolf/2210_Jui/oct21.pdf= 342 SCβ pix/s π€Μ . Totally Inelastic Collision...
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 ππ, position ππ and velocity οΏ½βοΏ½π (both ππ and οΏ½βοΏ½π are functions of time) for π = 1,2,3,β― ,π. The net force on the πππ particle can be written as
οΏ½βοΏ½π = οΏ½οΏ½βοΏ½πππβ π
+ οΏ½βοΏ½π(ext)
where οΏ½βοΏ½ππ is the (internal) force exerted by another (πth) particle in the group,
and οΏ½βοΏ½π(ext)
is the net external force (vector sum of forces on the πth particle, not exerted by another particle in the group). We now sum over the group: Remember we generally are not allowed to do thisβ¦ but here we are treating the group as a single object!!! (Definition) The net force on the group is given by
οΏ½βοΏ½ = οΏ½οΏ½βοΏ½π
π
π=1
= οΏ½οΏ½οΏ½βοΏ½πππβ π
π
π=1
+ οΏ½οΏ½βοΏ½π(ext)
π
π=1
Now: By Newtonβs 3rd Law (units 10-13 are all about the consequences of N3L)
Each οΏ½βοΏ½ππ in the sum β β οΏ½βοΏ½πππβ πππ=1 is cancelled by an equal and opposite οΏ½βοΏ½ππ
Collective Motion (continued) i.e.
πππ βοΏ½οΏ½οΏ½βοΏ½πππβ π
π
π=1
β‘ 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.
οΏ½βοΏ½ = οΏ½οΏ½βοΏ½π
π
π=1
= οΏ½οΏ½βοΏ½π(ext)
π
π=1
Now By Newtonβs 2nd Law, the net force on the ππ‘π‘ particle is related to its acceleration by
οΏ½βοΏ½π β‘ποΏ½βοΏ½πππ
β‘π2ππππ2
=1ππ
οΏ½βοΏ½π
Note THIS DOES NOT MEAN
οΏ½βοΏ½π =1ππ
οΏ½βοΏ½π(ext)
Because the (ext) desgnation here means outside of the group, but for an individual particle you have to count the βexternal (to the particle) forceβ exerted by other particles in the group!!!
Newtonβs 2nd Law for the Collective Multiplying οΏ½βοΏ½π by ππ and summing over the group, interchanging the order of summation and differentiation (derivatives are βlinearβ)
οΏ½πππ2ππππ2
π
π=1
=π2
ππ2οΏ½ππππ
π
π=1
= οΏ½οΏ½βοΏ½π
π
π=1
= οΏ½οΏ½βοΏ½π(ext)
π
π=1
= οΏ½βοΏ½
Note that
οΏ½ππππ
π
π=1
= π β1ποΏ½ππππ
π
π=1
= ππ πΆπΆ
And so: π2
ππ2ππ πΆπΆ = π
π2π πΆπΆππ2
β‘ ππππΆπΆππ
β‘ ππ΄πΆπΆ = οΏ½βοΏ½
Which looks just like Newtonβs 2nd Law for a particle of mass π = β ππππ=1 ,
located at the center of mass of the group. We have implicitly defined the velocity and acceleration of the center-of-mass by
ππΆπΆ β‘ππ πΆπΆππ
, π΄πΆπΆ β‘πππΆπΆππ
β‘π2π πΆπΆππ2
Unit 11
οΏ½βοΏ½ β‘ ποΏ½βοΏ½ [ Units: kg β m/s ]
Unit 11
οΏ½βοΏ½ β‘ ποΏ½βοΏ½ [ Units: kg β 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
οΏ½βοΏ½ β‘ ποΏ½βοΏ½ [ Units: kg β m/s ]
Newtonβs Second Law: (time) rate of change of the momentum vector is equal to the net force (vector sum of all βexternalβ forces) on a body.
ποΏ½βοΏ½ππ
= οΏ½βοΏ½
These are vector relations: Definition of momentum:
ππ₯ β‘ ππ£π₯ β‘ πππππ
, ππ¦ β‘ ππ£π¦ β‘ πππ¦ππ
, ππ§ β‘ ππ£π§ β‘ πππ§ππ
Newtonβs Second Law: πππ₯ππ
= πΉπ₯ ,πππ¦ππ
= πΉπ¦,πππ§ππ
= πΉπ§
Total Momentum of a System of Particles Take a group of particles, each with mass ππ, position ππ and velocity οΏ½βοΏ½π (both ππ and οΏ½βοΏ½π are functions of time) for π = 1,2,3,β― ,π. Definition of Total Momentum: vector sum of individual momenta
π β‘οΏ½οΏ½βοΏ½π
π
π=1
β‘οΏ½ππ
π
π=1
οΏ½βοΏ½π β‘οΏ½πππππππ
π
π=1
= πππΆπΆ
ππ₯ β‘οΏ½πππ₯
π
π=1
β‘οΏ½ππ
π
π=1
π£ππ₯ β‘οΏ½πππππππ
π
π=1
, ππ¦ β‘οΏ½πππ¦
π
π=1
β‘οΏ½ππ
π
π=1
π£ππ¦ β‘οΏ½ππππ¦πππ
π
π=1
, β―
Differentiating the total momentum w.r.t. time:
ππππ = οΏ½
ππποΏ½βοΏ½π
π
π=1
= οΏ½οΏ½βοΏ½π
π
π=1
= οΏ½οΏ½βοΏ½π(ext)
π
π=1
= οΏ½βοΏ½
Which is the alternate form of Newtonβs 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)
πΉπ₯ = 0 βπππππ₯ β‘
ππποΏ½πππ₯
π
π=1
= 0, πΉπ¦ = 0 βπππππ¦ β‘
ππποΏ½πππ¦
π
π=1
= 0, β―
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 π1 is traveling at speed π£π in the +π direction towards a second cart of mass π2, which is at rest. They collide and stick together. What is their (common) speed π£π 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: π2 = π1
= 1.0 SC This is called a βtotally inelastic collisionβ
Conservation of Momentum in a Collision Theory: Total momentum is conserved in the π direction because no external forces with non-zero π-components act on the group (they interact but the internal forces must cancel because of N3L)
πππ₯ = π1π£π πππ₯ = π1 + π2 π£π
Setting πππ₯ = πππ₯ we get π1π£π = π1 + π2 π£π
So the final speed of the conjoined carts is given by
π£π =π1
π1 + π2π£π
Or: the ratio π£π/π£π is given by π£ππ£π
=π1
π1 + π2
Predictions for three cases (1) π1 = 1.0 SC, π2 = 1.0 SC: π£π/π£π = 1/2 = 0.500 (2) π1 = 2.0 SC, π2 = 1.0 SC: π£π/π£π = 2/3 = 0.667 (3) π1 = 1.0 SC, π2 = 2.0 SC: π£π/π£π = 1/3 = 0.333
Case 1: m1 = m2
Totally Inelastic Collision: m2 = m1
This window dump missed the first digitized point
We have π1 = 1.0 SC π2 = 1.0 SC Before the collision: οΏ½βοΏ½1π = (1.0 SCU)*(342 pix/s) π€Μ = 342 SCβ pix/s π€Μ οΏ½βοΏ½2π = (1.0 SCU)*(0.0 pix/s) SCβ pix/s π€Μ = 0 π€Μ ππ = οΏ½βοΏ½1π + οΏ½βοΏ½2π = 342 SCβ pix/s π€Μ
Totally Inelastic Collision : m2 = m1
After the collision:
ππ = (1.0 + 1.0) SCU * (168 pix/s) π€Μ = 336 SCβ pix/s π€Μ Predicted π£π/π£π = 1/2 = 0.500 Measured: π£π/π£π = 168/342 = 0.491.
Within a few % of π·π βΌ!
Within a few % of π©π©π©π©π©π©ππ©π©π© !!!
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
π1 = 2.0 SC π2 = 1.0 SC
Collision Experiment: A cart of mass π1 is traveling with velocity +π£π (in the +π direction) towards a second cart of mass π2, which is at rest. They collide elastically. What are their velocities after the collision?
(Almost) Elastic Collision Case 1: m2 = 0.5 m1 Digitized π1 and π2 every 3 frames (0.10 s) Since π2 = 2 (SC) and 0.5 π1 = 1.0 (SC) then
ππΆπΆ =π1π1 + π2π2π1 + π2
=2π1 + π2
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 β’ From (1) Conservation of Momentum and (2) Conservation of Energy
π£1π =π1 βπ2
π1 + π2π£π , π£2π =
2π1
π1 + π2π£π
β’ The experimental Results gave us V1i = Vi = 340 pix/s V1f=133 pix/s V2f = 421 pix/s 1. V1f/Vi: predicted:
π£1ππ£π
=π1 βπ2
π1 + π2=
2 β 12 + 1
= 0.333
Measured: V1f/Vi = 133/340 = 0.391 2. V1f/Vi: predicted:
π£2ππ£π
=2π1
π1 + π2=
42 + 1
= 1.333
Measured: V2f/Vi = 421/340 = 1.238
Conclusion: the Collision was NOT completely elastic