query 182

course phy201

018. `query 18

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Question: `qQuery intro problem sets

Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical

displacement.

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Your solution:

The intial horizontal velocity is the vave for that position. We use the time for the vertical velocity and the vertical displacement

to find the horizontal range of a projectile.

confidence rating #$&*

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Given Solution:

`a** We treat the vertical and horizontal quantities independently.

We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So

we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion.

We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial

vertical velocity. We divide this into the vertical displacement to find the elapsed time.

We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells

us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve

the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **

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Self-critique (if necessary):

ok, I understand this.

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Self-critique rating #$&*3

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Question: `qQuery class notes #17

Why do we expect that in an isolated collision of two objects the momentum change of each object must be equal and opposite to that of

the other?

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Your solution:

momentum change must be equal and opposite to that of the other because once they collide the momentum of the greater force is

transfered to to that of the lesser force.

confidence rating #$&*

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Given Solution:

`a Briefly, the force exerted on each object on the other is equal and opposite to the force exerted on it by the other, by Newton's

Third Law.

By assumption the collision is isolated (i.e., this is a closed system); the two objects interact only with one another. So the net

force on each object is the force exerted on it by the other.

So the impulse F_net `dt on one object is equal and opposite the impulse experienced by the other.

By the impulse-momentum theorem, F_net `dt = `d ( m v). The impulse on each object is equal to its change in momentum.

Since the impulses are equal and opposite, the momentum changes are equal and opposite.

**COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change.

Momentum has nothing directly to do with energy.

Two colliding objects exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2

`dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 =

`dp1 +(-`dp1) = 0. **

STUDENT QUESTION

Are impulses the same as momentum changes?

INSTRUCTOR RESPONSE

impulse is F * `dt

momentum is m v, and as long as mass is constant momentum change will be m `dv

by the impulse-momentum theorem impulse is equal to change in momentum (subject, of course, to the conditions of the theorem)

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Self-critique (if necessary):

ok

so impulse or fnet * 'dt ON the object is equal and opposite to the force exerted BY the other

so the net force on each object is the force exerted ON it BY the other.

Impulse is equal to change in momentum.

since impulses are equal and opposite momentum changes are equal and opposite.

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Self-critique rating #$&*2

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Question: `qWhat are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before

and after collision, are both moving along a common straight line? How are these quantities related?

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Your solution:

m

v0

fnet

vf

not sure

confidence rating #$&*

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Given Solution:

`a** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the

after-collision velocities v1' and v2'.

Total momentum before collision is m1 v1 + m2 v2.

Total momentum after collision is m1 v1' + m2 v2'.

Conservation of momentum, which follows from the impulse-momentum theorem, gives us

m1 v1 + m2 v2 = m1 v1' + m2 v2'. **

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Self-critique (if necessary):

ok, so

before collison is m1 v1 + m2 v2

after collison is m1 v1' + m2 v2'

so the momentum is the mass * velocity of the first object + the mass * the velocity of the second object before collision which equal

the mass * the change in veloictyin first object + the mass * the change in velocity for the second object after collision

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Self-critique rating #$&*2

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Question: `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much

thermal energy is produced in the collision?

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Your solution:

m= 7650kg

v0= 95km/hr= 95km * 1000/3600= 26.39m/s

vf=0

mom= m*v

7650kg * 26.39m/s= 201883.2kg m/s

confidence rating #$&*

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Given Solution:

`aThere is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy

converted to thermal energy.

The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4

m/s)^2 = 2,650,000 Joules, so that their total KE is 2 * 2,650,000 J = 5,300,000 J.

This KE is practically all converted to thermal energy.

`1Query* gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the

track)?

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Your solution:

Ok so to find th thermal energy we need to find KE

kE= .5m v^2

KE= .5(7650kg) (26.39m/s)^2

KE= 3825 * 696.43m^2/s^2

= 2,665,872J

since their are 2 cars the KE = 2 * 2665872J= 5,331,744J

confidence rating #$&*

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Given Solution:

`a**GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION:

Until just now I did not think I could work that one, because I did not know the mass, but I retried it.

Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0.

PE is all gravitational so that `dPE = (y2 - y1).

The only other force acting in the direction of motion is friction.

Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and

.5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m)

It looks like the M's cancel so I don't need to know mass.

.5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so

v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.

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Self-critique (if necessary):

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&#This looks good. Let me know if you have any questions. &#