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Problem 1: (Chuyển động của vật trên mặt phẳng nghiêng)
A sled is a quick push up the snow slop. The sled slides up and then comes back down; the whole
trip take time t. If the coeffient of sliding friction between the sled and snow is
µ
, find the time tu it
took the sled to reach the top point of its trajectory. The slope makes the angle
θ
with horizontal.
Solution:
Newton’s second law applied ti the sled gives accelerations upwards (positive direction up) and
downwards (positive direction dow) as follows:
( )
sin
u
a g cos
θ µ θ
= − +
and
( )
sin
d
a g cos
θ µ θ
= −
If u and v are, respectively, the initial velocity when moving upwards and final velocity when
moving downwards, then according to the (constant acceleration) kinematical formula involving
squares of velocites, (try to prove this formula!). If tu and td are, respectively, the time it takes to
move down, then according to the definition of acceleration and using the privious equation. Combing
the latter equation with the first two, we get (noting that
tan

• square: bình phương
• equation: phương trình
• slide: trượt
• prove: chứng minh
• involve (+ing): bao gồm
Problem 2: (Bài toán ném xiên)
Figure (hình) 1 shows a pirate ship, moored(bỏ neo) 560m from a fort defending the harbor(bến
cảng, cảng) entrance(đi vào) of an island. The harbor defense cannon, located at sea level, muzzle
velocity of 82m/s.
a) To what angle must the cannon be elavated (độ nâng) to hit the pirate ship?
b) What are times of flight for the elevation angle calculated above?
c) How far should be pirate ship be from the fort if it is to be beyond(quá, vượt xa hơn) range of
the cannon balls?
Solution: As known, the equation of the trajectory of the ball is:
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( ) ( )
2
2 2
0
tan 1
2
g
y X X
v cos
θ
θ
= −
Where
θ

( ) ( )
0
3X v cos t
θ
=
Soving Eq (3) for t give, for
0
27
θ
=
,
0
0 0
560
7,7
82. 27
X R
t s
v cos v cos cos
θ θ
= = = =
Repeating the calculation for
0
63
θ
=
yields t = 15s. It is reasonable that the time of flight for the
higher elevation angle should be larger.
c) We have known that the maximun range corresponds to an elevation angle
θ

• Cannon: pháo, đại bác
• Muzzle velocity: vận tốc đầu nòng (súng)
• Elevated angle: góc tầm (góc ngắm)
• (to) hit: trúng
• (the) commandant (of the fort): người chỉ huy (pháo đài)
• The time flight: thời gian bay
• Corresponds: tương ứng, đúng với, phù hợp với
Problem 3: (Bài toán tìm nhiệt dung riêng)
Two identical light metal containers are filled with equal amounts(lượng, số lượng) of water and
placed in a room with constant air temprature. A heavy ball is submerged(chìm, ngập dưới) into the
center of one of the containers on a thin nonconducting string. The mass of ball equals to the mass of
water, and the density of the ball is much greater than that of water.
Both containers are heated to the boiling point of water and are then allowed to cool. The container
with the ball in it takes k times longer to cool down to the room temprature than the container without
the ball. Find specific heat of the material of the ball C
b
in terms of k and the specific heat of water
C
w
.
Solution: The equation that relates the rate of heat flow to the temprature defferece is
( )
R
dQ
T T
dt
α
= −
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( )
Q mc T= ∆
. Therefore,
b w w b
w w
mC T mC T C C
k k
mC T C
∆ + ∆ +
= ⇒ =
∆
And, finaly,
( )
1
b W
C k C= −
New Words:
• identical: đồng chất, giống hệt nhau
• light: nhẹ
• metal: kim loại
• temprature: nhiệt độ; the current temprature: nhiệt độ hiện thời
• nonconducting string: dây không dẫn (nhiệt)
• density: mật độ, khối lượng riêng
• boiling point: điểm sôi
• specific heat: nhiệt dung riêng
• to relate sth to/with sth: liên hệ cái gì với cái gì
• difference: hiệu
• to depend on: phụ thuộc vào
• minimizing: làm giảm thiểu
• discrepancy: sự khác biệt

sinl
α
, and the
time required to cross with minimum lateral displacement is given by:
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( ) ( )
1
.k vcos t l because v kvcos
α α
= =
Combining these equations gives:
min
.sin .t k cos
α α
=
, or:
2
min
1
.
k
t t
k
−
=
New Words:
• speed: tốc độ
• current: dòng nước (the speed of the current : tốc độ dòng nước)
• lateral displacement: độ dịch chuyển theo dòng nước (nghĩa trong bài)

A
σ
∆
, where
2A Rt
π
∆ =
is the cross-sectional
area of the ballon material at its mid-plane. Here t is the thickness(độ dày) of th ballon material,
obtained(thu được) from
2
4
m
d
R t
π
=
. Combining(kết hợp) the equations givens:
9
16
PVd
m
σ
=
New Words:
• balloon: khí cầu
• pressure: áp suất (atmospheric pressure: áp suất khí quyển)
• material: vật liệu
• (to) burst: nổ
• density: mật độ hay khối lượng riêng

f
n
=
−
Here R is the radius of curvature of the concave surface, n is the index of refraction of lens, ans f
1
is focal length of the lens.
Also, for a concave surface with radius R, the focal length f
2
can be found as:
2
2
R
f =
The rays from the object are passing through the lens, reflecting from the convex surface, and
passing through the lens again to emerge from the flat side to form a virtual image:
( )
2 1 2 1 1 2
2 1
1 1 1 1 1 2 1 2 2
n
n
a b f f f f f R R R
−
− = + + = + = + =
When the hiker turns, the rays from the object passing through the lens also form a virtual image:
3 1
1 1 1 1n
a b f R
−

Problem 7: (Dao động cơ học)
When the system shown in the diagram is in equilibrium,
the right spring is stretched by X
1
. The coefficient of the
static friction the blocks is
s
µ
; there is no friction between
the bottom block and the supporting surface. The force
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