its (a) Problem 2- Elastic collision between a bullet and a rod (a) A bullet with mass m and initial speed v along the horizontal direction collides elastically against a long uniform rod of length L and mass M>> m, pivoted at its upper end. The collision occurs at a distance L/2 from the rod's upper end, with the rod initially at rest with its axis along the vertical direction. Assume that L=1m, v= 100m/s, m = 1g and M = 1kg. Estimate the maximum angle, between the vertical axis and the rod axis, reached by the rod. (Hint: use cos 8max 10max/2). (b) Estimate the time it takes for the rod, from the time. the initial collision with the bullet, to reach the position of maximum angle. (c) Suppose that we take the rod to a distant planet, whose mass is twice that of the Earth and whose radius is also twice that of the Earth. Estimate the period of small oscillations of the rod on this distant planet? (Hint: approximate g = 10 m/s² on Earth; in parts (b) and (c) one significant digit is enough)
its (a) Problem 2- Elastic collision between a bullet and a rod (a) A bullet with mass m and initial speed v along the horizontal direction collides elastically against a long uniform rod of length L and mass M>> m, pivoted at its upper end. The collision occurs at a distance L/2 from the rod's upper end, with the rod initially at rest with its axis along the vertical direction. Assume that L=1m, v= 100m/s, m = 1g and M = 1kg. Estimate the maximum angle, between the vertical axis and the rod axis, reached by the rod. (Hint: use cos 8max 10max/2). (b) Estimate the time it takes for the rod, from the time. the initial collision with the bullet, to reach the position of maximum angle. (c) Suppose that we take the rod to a distant planet, whose mass is twice that of the Earth and whose radius is also twice that of the Earth. Estimate the period of small oscillations of the rod on this distant planet? (Hint: approximate g = 10 m/s² on Earth; in parts (b) and (c) one significant digit is enough)
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I attached the question and the answer
What I don’t understand in part a is the part after we find angular velocity how was the semplification made to find the maximum angle just the simplification
And in part b I don’t understand where pi/4 comes from
Transcribed Image Text:(Hint: in points (a) and (b), approximate g = 10 m/s²)
Problem 2- Elastic collision between a bullet and a rod
(a) A bullet with mass m and initial speed v along the horizontal direction collides elastically against a long uniform rod of length L and mass M>> m, pivoted
at its upper end. The collision occurs at a distance L/2 from the rod's upper end, with the rod initially at rest with its axis along the vertical direction. Assume
that L=1m, v = 100m/s, m = 1g and M = 1kg. Estimate the maximum angle, between the vertical axis and the rod axis, reached by the rod. (Hint: use.
cos max 1-
/2).
Omax/
(b) Estimate the time it takes for the rod, from the time of the initial collision with the bullet, to reach the position of maximum angle.
(c) Suppose that we take the rod to a distant planet, whose mass is twice that of the Earth and whose radius is also twice that of the Earth. Estimate the
period of small oscillations of the rod on this distant planet? (Hint: approximate g = 10 m/s² on Earth; in parts (b) and (c) one significant digit is enough)
Transcribed Image Text:1 m/v²-v₁² ) = uhrad,
but n
is oo
=
M,V
Problem 2- Elastic collision, bullet and rod
Conseration of argular momentum gives:
mv v 1/2 = -mv = 2 + [win
where
I=
is the rod's moment of
ML²
inertia, and win its angular relocity
immediately after the collision. Then,
Win =
mvL
I
The numerical value of Win is
20 Ano
= 3 = (1-(05@max)
3m V
ML
des tant planet,
where h = L/2. The time, AT, it takes
to reach maximum angle is T/4. Thus:
IV 33,14. (15) 0.6
DT =
Tt O
(c) The acceleration of gravity on the
Eerth is
GME
g=
and on the
r2
W₁₁ = 0.35!
I
Mah
GMP
·湖·5米·
=
The period of oscillation of the rod
on the distant planet is
T= 2πTV
=
Problem 3- Satellites
2x
1/2
27 (1)
2s
MIM U..
with W₁ = Mg (1-cosman) and a
we obtain
1-cosmex =
1-cos@man =
st=_3
1m
max
6x 10
With reference to the figure, using the work- (b) The period of small oscillations of
energy principle in the form Wg = 1 / 1 (1) - W²23) the rod is
I
MgL
Substituting numbers, and usy gs107/11,
we obtain
10 m/s
OM/
st inflate
Lw²
W₁ =
3g
9x105²3x103
Finally, using cos@mex ≈ 1-Omer,
oblan
man
T= 2 = 21
n
0.25
= 0,08 rad
(b) The mechanical energy of a satellite on
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thank you for the explanation
2 things that still arent clear why are we doing initial minus final instead of final minus initial and you didnt answer my question regarding why the period is divided by 4 how do we deduce that
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