18 Open Query

course Phy 201

6/30, 10:30 PM

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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:

You can use the vertical velocity and range and acceleration of 9.8m/s/s to find the change in time

You would use the equation

Vf^2=v0^2+2a(‘dt)

Once you have ‘dt you can find the horizontal displacement by using vf of x and the time to find displacement. Acceleration would be constant in this case. So you would take vf(‘dt) to find displacement.

Confidence rating: 3

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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

Self-critique rating:ok

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Question: `qQuery class notes #17

Why do we expect that in a 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:

Because the forces are going to be equal and opposite which will result in an equal and opposite momentum change. I am not sure of the details.

Confidence rating:

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Given Solution:

`a**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 object 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. **

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Self-critique (if necessary): Are impulses the same as momentum changes?

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)

Self-critique rating: ok

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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:

For each object we use the mass, initial velocities and the velocities after the collision

You can find the momentum changes by taking Mass(v) and mass(velocity after the collision)

Momentum will always be equal and opposite so (mass)(v1)+mass(v2)=mass(vf1)+mass(vf2)

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'. **

`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:

We need to find the KE that is lost during the collision, this would be 7640kg(v)^2(.5)

95km/hour*1000m/1km*1 hour/3600sec= 26.4m/s

7640kg(26.4)^2(.5)= 26623387 Joules.

Would you double this since there are two trains?

Confidence rating: 2

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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 = 265,000 Joules, so that their total KE is 2 * 265,000 J = 530,000 J.

This KE is practially 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:

1.7m/s

PE=m(9.8m/s/s)(28m)= 274.4m

Frition= (1/5)(45m)

Initial KE = M(1.7)^2(1/2)

KF final= m(v^2)(.5)

So we add these so they equal 0

9-274.4+1.445+KEfinal=0 I get a negative number which you can not take a square root of. I made it positive and this is the answer I got 22.97m/s

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):

the frictional force is 1/5 m g, which is multiplied by the 45 m

My answer is slightly off. I think this has to do with the friction. How was 88m^2/s^2 calculated?

.2 * g * 45m = 88 m^2 / s^2, with g = 9.8 m/s^2.

Self-critique rating:

3

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&#Good responses. Let me know if you have questions. &#