Mechanics Exercise Class Ⅰ

英文数字、算式表达法 = 　 is equal to / equals > 　 is larger than

< is less than >> is much greater than

x

a = b + c 　　 a is equal to b plus ca = b - c 　　 a is equal to b minus ca = b x c 　　 a is equal to b times c / a equals to b multiplied by ca = b/c 　　 a is equal to b divided by c / a equals to b over c

   (square) root x / the square root of x  

   cube root (of) x ;  fourth root (of) x ;   nth root (of) x  3 x 4 x n x

英文数字、算式表达法

0

x2 　　 x square / x to the second power / x to the power twox3 　　 x cube / x to the third power / x to the power threex7 the seventh power of x （ x to the seventh power ） lognX 　 log x to the base n

f( x )    fx / f of x / the function f of x  

    the limit as x approaches zero  

   the integral from zero to infinity  

0xlim

Brief Review

netF ma

212

D C Av

2vaR

Newton’s Second Law Drag Force

Centripetal Acceleration

Displacement r

Velocitydtrdv

Acceleration 2

2

dtrd

dtvda

Projectile Motion jvivv ˆsinˆcos

2

00

2

0 cos2tan

vgxxy (trajectory)

1. A model rocket fired vertically from the ground ascends with a constant vertical acceleration of the 4.00m/s2for 6.00 s. Its fuel is then exhausted ,so it continues upward as a free-fall particle and then falls back down. (a) What is the maximum altitude reached? (b) What is the total time elapsed from the takeoff until the rocket strikes the ground? Solution:(a) One key idea is since the fuel is exhausted and before the rocket strikes the ground, its acceleration is g of magnitude. And when the fuel is exhausted ,the velocity is

0

y

2 21 1 1

1 1 4 6 722 2

s a t m

1 1 1 4 6 24v a t m / s (upward)

then we can get the position of the rocket

2 2 22 1

20 24 29 39

9 8v vs . mg .

1 2 72 29 4 101 39s s s . . m

Second key idea is the velocity equals zero, when the rocket is at maximum altitude. So

So the maximum altitude is

(b) Since the fuel is exhausted and before the rocket returns, the time interval is

2 12

2

0 24 2 459 8

v vt . sa .

23

10 101 39 9 82

. ( . )t

Then the rocket from the maximum altitude falls back to the ground. From the Eq. 2-50, we can obtain

1 2 3 6 2 45 4 55 13 00t t t t . . . s

3 4 55t . s

Work out this equation, yielding

So the total time elapsed from the takeoff until the rocket strikes the ground is

2. A block of mass m1 on a frictionless inclined plane of angle is connected by a cord over a massless , frictionless pulley to a second block of mass m2 hanging vertically. What (a) the magnitude of the acceleration of each block and (b) The direction of the acceleration of the hanging block? (c) what is the tension of the cord? Solution:Choose m2 to be a system and draw it’s free-body diagram

m2

m2g

T m1

m2

With Newton’s second law applied to the m2 system, we can obtain

2 2 2m g T m a (1)

(2)

The acceleration components a1 and a2, have the same value since the string does not stretch, thus

1 1 2T m g sin m a

m1g

NTChoose m1 to be a system and draw it’s

free-body diagram and we can writeParallel direction 1 1 1T m g sin m a

1m g cos N Perpendicular direction

2 2T m ( g a )

Now we can solve equation (1) and (2) simultaneously for T and a2. First solve equation (1) for T

(3)

2 2 1 2 2m ( g a ) m g sin m a Then substitute for T into equation (2):

2 12

1 2

m m sina gm m

Solving for a2, we find

Substituting for a2 for Eq. (3), we find that tension in the rope is of magnitude

1 2

1 2

1mm ( sin )T gm m

Discussion : If the direction of the acceleration of the hanging block is vertically downward. On the other hand if the direction is vertically upward.

0sin12 mm

0sin12 mm

3 A truck moves along a straight road with a constant speed of 30m/s, and a projectile is launched from it. What is the (a) magnitude and (b) direction of the projectile’s initial velocity relative to the truck if it can fall back to the launching point on the truck after the truck moves 60m. The drag force is negligible.

0y yv v gt , 0x xv v

0xx v t 20

12yy v t gt

Solution: Choose the truck as reference. Set the launching point as origin, and construct coordinates O-xy as shown.Set the launching time as starting point of time.

sin,cos 0000 vvvv yx

200 2

1sin,cos gttvytvx

gaa yx ,0

so

The loading time:

st 23060

And at the loading time 0,0 yx

2cos0 0 v

20 2

212sin0 gv

(1)

(2)

Eq (1) yields

Eq (2) yields

2,0cos

smv /8.90

So, if the projectile is thrown vertically relative to the truck, it will fall back to the launching point after the truck moves 60m.

Solution： Isolating the bodies ,m,m 21

2m0cosNgmR 2

12amsinN

1m sinamgmcosN 211

cosamamsinN 2111

2

12

212 sinmm

sing)mm(a

2

12

21

sinmmcosgmmN

For

For

Solving the equations for a2 and N:

4 An inclined plane of mass m2 can slide on a smooth horizontal surface, with the inclined angle of α. An athlete of mass m1 slides on the inclined plane frictionlessly, what is (a) the relative acceleration of the athlete with respect to the inclined plane and (b) the pressure force on the plane from the athlete?

5 A parachute athlete dives with an initial speed of v0, and the drag force is proportional to the square of the speed as: av2. The total mass of the parachute and the athlete is m. What is the function of v=v(t)?Hint: this can be utilized in the integration

)v1(21

)v1(21

v11

2

Solution:

,mdt

avmgdv ,

dtdvmavmg

2

2

,mdt

)mg/av1(mgdv

2

,gdtmg/av1

dv2

Transferring to

the above equation can be written as

,gdtv1

dv22

,gdtv12

dvv12

dv

）（）（

,gdtv12

v1dv12

v1d

）（）（

）（）（

Cgt2)v1v1ln(

C)

v1v1ln(

0

0

)v1v1ln(gt2)

v1v1ln(

0

0

Quadrature

Then

)1v1v1e(

)1v1v1e(

1v

0

0gt2

0

0gt2

set Cmga

mga ,2