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(phys221)[2009](f)midterm~ma_yxf^_10542.pdf
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Solution to Midterm Problems

Question 1. (prepared by Xiaoguang Ma)
(a) By conservation of energy, we have

(1)
where h=2R.


Fig 1. Geometry of the problem
From Fig 1, we have
(2)
From Newtons 2nd law, we have
(3)
where N is the constraint force exerted by the track, and N should be finite value to make sure that the block contact with the track, which requires that

(4)

With (1), (2), (3), (4), one can easily determine when the block loses contact with the track

,
and


(b) At position Q, from conservation of energy, we have

(5)


Fig 2 The acceleration a of the block at position Q
So the centripetal acceleration is

Then the total acceleration

The direction is

(c) The trajectory of the block is a parabola


Fig 3. The trajectory of the leaving block (dashed parabola)
(d) In this problem, energy is conserved throughout the entire motion. Thus the
speed of block just before it hits the ground is given by

(6)
Where h=2R is the initial height in Fig 1.
Then the final speed is
.
Since theres no force on the block in the horizontal direction after the block leaves the track, the x-component of velocity is a constant.


So the direction of the velocity(see in Fig 3) is


Question 2. (prepared by Yongky Utama)
a.

soit is not conservative



therefore it is conservative.

is a central force therefore it is conservative. b. Its impossible to define a potential foras it is not conservative. For, if we set the origin to be the reference point, because of path-independence we can choose a path consisting of 3 straight lines,


(
then



As
is a central force, if reference point is chosen to be at the origin:

c. is not conservative so it cant be central.

One can show that is not a central by counter argument. For example, at the force which is not parallel to the direction of . So its not a central force.




is obviously central (i.e. magnitude only depending on ,

direction only towards ) .


Question 3. (prepared by Alan Fung)
(a) By the formulae sheet,
, while the maximum value of
is given by
, where
, and
. Given that ,

,
and ,




and


Therefore,
and

(b) From (a),
.
After manipulating the equation, we have
Therefore, the latitude = .


(c)
If
the projectile will never come back. Thus
The speed of escape is the same for
. In fact, the answer is independent of the launch angle
.
Question 4 (prepared by Binping Jin)
(a)


The equation of orbit of a particle moving under a central force is


Which can be re-written as
-----------------(*)
For
,
is real and Eq. (*) becomes



This is the simplest type