ch 9 linear momentum and collisions 9.1 linear momentum and its conservation linear – traveling...
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![Page 1: Ch 9 Linear Momentum and Collisions 9.1 Linear Momentum and Its Conservation Linear – traveling along a path p = mv single particle dp/dt = d(mv)/dt =](https://reader036.vdocuments.mx/reader036/viewer/2022062421/56649d6a5503460f94a47bff/html5/thumbnails/1.jpg)
Ch 9 Linear Momentum and Collisions9.1 Linear Momentum and Its Conservation
Linear – traveling along a path p = mv single particle dp/dt = d(mv)/dt = mdv/dt = ma
Fnet =dp/dt If Fnet =0, p=0
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CT1: Two carts of identical inertial mass are put back-to-back on a track. Cart A has a spring loaded piston; cart B is entirely passive. When the piston is released, it pushes against cart B, and
A. A is put in motion but B remains at rest.B. both carts are set into motion, with A gaining more speed than B.C. both carts gain equal speed but in opposite directions.D. both carts are set into motion, with B gaining more speed than A.E. B is put in motion but A remains at rest.
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CT2: Two carts are put back-to-back on a track. Cart A has twice the mass of cart B. Cart A has a spring loaded piston; cart B is entirely passive. When the piston is released, it pushes against cart B, and
A. A is put in motion but B remains at rest.B. both carts are set into motion, with A gaining more speed than B.C. both carts gain equal speed but in opposite directions.D. both carts are set into motion, with B gaining more speed than A.E. B is put in motion but A remains at rest.
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Internal Forces as Third Law Pairs
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External forces and Momentum Conservation
System of Particles (i = 1 to N)
Fneti = dpi/dt ptot = pi
Fneti = Fext + Fint = dpi/dt = dptot/dt
Internal forces occur in third law pairs
so their sum adds to zero.
Fext = dptot/dt
If Fext = 0, then ptot = 0 and
(ptot is conserved)
![Page 6: Ch 9 Linear Momentum and Collisions 9.1 Linear Momentum and Its Conservation Linear – traveling along a path p = mv single particle dp/dt = d(mv)/dt =](https://reader036.vdocuments.mx/reader036/viewer/2022062421/56649d6a5503460f94a47bff/html5/thumbnails/6.jpg)
ptot = pi
Fext = dptot/dt
If Fext = 0, then ptot = 0
P9.4 (p.260)
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The Average Force During a Collision
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Ch9 Linear Momentum and Collisions9.2 Impulse and Momentum I = Impulse (a vector) I = Area under F vs. t curve
I = Favt = (ptot/t)t = ptot
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P9.9 (p.261)
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PHYS201F07 Exam2Average 62 Lo 9.5 Hi 100
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PHYS201F07 TotalAverage 68 Lo 23 Hi 98
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CT3: A ball is dropped onto the floor below. Which of the following statements is true while the ball falls (neglect air resistance)? Consider the system as just the ball.
A. Momentum is conserved for the ball, but energy is not.
B. Energy is conserved for the ball, but momentum is not.
C. Both energy and momentum are conserved for the ball.
D. Neither energy nor momentum is conserved for the ball.
![Page 13: Ch 9 Linear Momentum and Collisions 9.1 Linear Momentum and Its Conservation Linear – traveling along a path p = mv single particle dp/dt = d(mv)/dt =](https://reader036.vdocuments.mx/reader036/viewer/2022062421/56649d6a5503460f94a47bff/html5/thumbnails/13.jpg)
A. abc
B. acb
C. bca
D. bac
E. cba
F. cab
Rank greatest to least
Concept Question 4
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CT6: Two carts are put back-to-back on a track. Cart A has a spring-loaded piston; cart B, which has twice the inertial mass of cart A, is entirely passive. When the piston is released, it pushes against cart B, and the carts move apart. How do the magnitudes of the final momenta and kinetic energies compare?
A. pA > pB; kA > kB
B. pA > pB; kA = kB
C. pA > pB; kA < kB
D. pA = pB; kA > kB
E. pA = pB; kA = kB
F. pA = pB; kA < kB
G. pA < pB; kA > kB
H. pA < pB; kA = kB
I. pA < pB; kA < kB
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CT5: Two identical looking balls are dropped from the same height (demonstration). Which of the following statements is true throughout the whole process (neglect air resistance)?
A. Momentum is conserved for each of the balls, but energy is not.
B. Energy is conserved for each of the balls, but momentum is not.
C. Neither energy nor momentum is conserved for either ball.
D. Energy and momentum are conserved for both balls. E. Momentum is conserved for one ball, but not the
other. Energy isn’t conserved for either ball. F. Momentum is conserved for one ball, but not the
other. Energy is conserved for both balls. G. Energy is conserved for one ball, but not the other.
Momentum is conserved for both balls. H. Energy is conserved for one ball, but not the other.
Momentum isn’t conserved for either ball.
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Ch 9 Linear Momentum and Collisions9.3 Collisions in One Dimension
A. Inelastic CollisionsMomentum is conserved, but energy isn’t. Totally inelastic collision when objects stick together. P9.15 (p.261) P9.16 (p.261)
B. Elastic CollisionsMomentum and energy are both conserved. P9.19 (p.262)
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CT7: A golf ball is fired at a bowling ball initially at rest and sticks to it. Compared to the bowling ball, the golf ball after the collision has
A. more momentum but less kinetic energy.B. more momentum and more kinetic energy.C. less momentum and less kinetic energy.D. less momentum but more kinetic energy.E. none of the above
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CT8: A golf ball is fired at a bowling ball initially at rest and bounces back elastically. Compared to the bowling ball, the golf ball after the collision has
A. more momentum but less kinetic energy.B. more momentum and more kinetic energy.C. less momentum and less kinetic energy.D. less momentum but more kinetic energy.E. none of the above
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Ch 9 Linear Momentum and Collisions9.4 Collisions in Two Dimensions
A. Inelastic CollisionsMomentum is conserved, but energy isn’t. Totally inelastic collision when objects stick together. P9.28 (p.262)B. Elastic CollisionsMomentum and energy are both conserved.
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The Center of Mass of Two Objects
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Ch 9 Linear Momentum and Collisions9.4 Collisions in Two Dimensions
For several masses
xcm = mixi/M where M = mi
ycm = miyi/M zcm = mizi/M
rcm = miri/M
The mi/M are the weighting factors.
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Ch 9 Linear Momentum and Collisions9.4 Collisions in Two Dimensions
vcm = drcm/dt = (midri/dt)/M = mivi/M
Mvcm = mivi = ptot
acm = dvcm/dt = (midvi/dt)/M = miai/M
Macm = Fext + Fint = Fext
Fext = Macm = dptot/dt
If Fext = 0, then ptotal = 0 and ptotal is
conserved.
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Consider a system of two particles in the xy plane: m1 = 2.00 kg is at
the location i + 2j m and has a velocity of 3i +0.5j m/s; m2 = 3.00 kg
is at -4i - 3j m and has velocity 3i – 2j m/s. (a) Plot these particles on
a grid or graph paper. Draw their position vectors and show their
velocities. (b) Find the position of the center of mass of the system
and mark it on the grid. (c) Determine the velocity of the center of
mass and also show it on the diagram. (d) What is the total linear
momentum of the system?
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P9.57 (p.266)
Find v0