南京大学 大学物理Solutions to the Sample Midterm Exam Problems

And the equation of motion for the wedge is
.
(4)
&& = N sin α . MX (5) &=& & tan α . Eliminating & & and & & from Eq.(4), we have The last equation of Eq. (4) gives & y x y x && = ( N cosα − 3mg ) cot α . − N sin α − 3mX (6)
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Sum /100 April 23, 2006 Sunday
University Physics Sample Midterm Examination School of Intensive Instruction in Sciences and Arts, Nanjing University
T ′ = T,
The ratio then is
U'=
U . 2
T' = −1 . U'
(c) The total energy is
E′ = T ′ +U ′ = 0 .
Then, the orbit is a parabola and the earth will escape the solar system. (4) (20pts) A slider of mass 2m is initially at 2m rest at the bottom edge of a wedge of mass M v0 α M and angle α . The wedge has a frictionless m surface and rests on a frictionless table. At t = 0 , a bullet of mass m and velocity v0 traveling parallel to the upper surface of the wedge hits and sticks in the slider. (a) What is the maximum height reached by the slider? (b) What is the wedge's velocity at the time the slider reaches its maximum height? (c) How far has the wedge moved at that time? Solution: (a) Immediately after the bullet hits and sticks in the slider, we have mv0 = (2m + m)v . Thus, the velocity of the slider (with the ห้องสมุดไป่ตู้ullet) is v = v0 3 . Assume the maximum height of the slider is hmax , then the conservations of energy gives
G = 6.67 × 10 −11 m 3 / s 2 ⋅ kg g n = 9.806m/s 2
O
a
x
&= m& x
m xg . l
By integrating both sides with respect to x , and noting that
& &dx = x
we have
& dx dx & & dx & = vdv dx = dx =x dt dt g l
1 2 1 m 2 cos 2 α 2 mv0 = v0 + 3mghmax . 6 2 M + 3m
Thus, we obtain
hmax
(b) From Eq. (2), we have
2 ⎛ 1 m cos 2 α ⎞ v0 ⎟ ⎜ − = ⎟. 6g ⎜ ⎝ 3 M + 3m ⎠
(3)
V =
1 1 (3m)v 2 = (M + 3m)V 2 + 3mghmax . 2 2
(1) (2)
And the conservation of the momentum gives
Eliminating V in Eq.(1) with the use of Eq.(2), we have
3
3mv cos α = (M + 3m)V .
F ( x) = −
(b) The equilibrium separation is determined by
dU 12a 6b = 13 − 7 . dx x x
dU dx
=−
x = xm 16
12a 6b + 7 = 0. x 13 xm m
⎛ 2a ⎞ xm = ⎜ ⎟ . Thus, we have ⎝ b ⎠ (c) As x approaches ∞ , U = 0 . Then, the minimum energy needed to break the molecule is a b b2 . E d = U (∞) − U ( x m ) = − 12 + 6 = x m x m 4a
∫
or
v
0
vdv =
∫
x
a
xdx
1 2 1g 2 v = ( x − a 2 ) or v = 2 2 l
At the instant it slides off the table, x = l and
g( x 2 − a 2 ) . l
v=
(b) From the result above, we have
∫
or
l
a
=
t=
l l ln( x + x 2 − a 2 ) = a g
l ⎛ l + l 2 − a2 ln ⎜ g ⎜ a ⎝
⎞ ⎟. ⎟ ⎠
(2) (20pts) The potential energy function for the force between two atoms in a diatomic molecule can be express as
[
]
(8)
(5) (20pts) A uniform rod of A d θ mass m and length l is C held flat on a horizontal table and at right angle to an edge A of the table. The center of mass C of the rod is beyond the edge of a distance d . The rod is released at rest and starts to rotate about the edge A and eventually slides off the table. The coefficient of static friction between the rod and the edge of the table is µ . (a) Find the rotational inertias of the rod, I C about the center of mass C and I A about point A . (b) Calculate the angular velocity of the rod as the function of rotation angle θ before sliding occurs. (c) Determine the angle θ 0 when sliding of the rod begins. Solution: The rotational inertia about C is
3mv cos α mv 0 cos α = . M + 3m M + 3m
(c) The equation of motion for the slider ( in the wedge frame) can be written as
&& = 3m& & x − N sin α − 3mX & N cos α − 3mg = 3m& y dy = dx tan α
(3) (20pts) Assume that the Earth’s orbit around the Sun is circular, and the Sun’s mass is much larger than the Earth’s mass. (a) Compute the ratio of the kinetic and potential energies of the Earth. (b) The Sun undergoes a supernova explosion and its mass suddenly reduces to half of its original value. Assume that the energy released in the explosion has no effect on the Earth. Compute the ratio of the kinetic and potential energies of the Earth at the instant after the explosion. (c) What is the new total energy of the Earth? What is the shape of the orbit now? Will the earth escape the solar system? Solution: (a) For circular motion, we have
By using Eq.(5) to eliminate N in Eq. (6), we finally have
&& = X
Thus,
3mg cot α M (1 + cot 2 α ) + 3m
(const.)
(7)
t=
V && X
and
S=
v 2 m M (1 + cot 2 α ) + 3m sin 2α 1 && 2 1 V 2 . Xt = = 0 && 12 g 2 2 X ( M + 3m) 2
U ( x) =
a b − 6 12 x x
where a and b are positive constants and x is the distance between two atoms. (a) Find the force between two atoms as the function of x . (b) What is the equilibrium separation between two atoms? (c) Find the minimum energy needed to break the molecule apart. Solution: (a) The force between two atoms is
Physical Constants Gravitational constant Standard gravity acceleration Select five out of following six problems. (1) (20pts) As shown in the accompanying figure, a uniform chain of total length l a hangs with a piece of length a (0 < a < l ) over the edge of a plane table. The initial velocity of the chain is zero and there are no friction between the surface of the table and the chain. (a) Find the velocity of the chain at the instant it slides off the table. (b) Calculate the time in which the chain slides from the table. Solution: (a) The equation of motion is
1 1 me v 2 = − U . 2 2
T 1 =− . U 2
If the mass of the sun suddenly goes to 1 2 of its original value, the kinetic energy remains unchanged but the potential energy is halved:
T=
and
1 me v 2 2
2
U =−
GM S me R
By equating the gravitational force to the centripetal force, we have
GM S me v2 = me R R2
or
GM S me = me v 2 . R
So
T=
The ratio is
9(l 2 − a 2 ) . l
1
dx = dt
It can be rewritten as
g x2 − a2 . l
=
dx x −a dx x2 − a2
2 2
g dt . l g t dt , l ∫0
Taking integration of both sides of above equation, we obtain