Solution to Quiz 011-2
covers chapter 3 and 4


Question 1:
Which of the following statement is wrong?

A.     .               Correct Answer

B.     .

C.    .

D.     .

E.     .


Question 2:

  has a magnitude of 3 m and it is along the positive y axis.   

 has a magnitude of 6 m and it is 600 from the positive y axis.

Find

 

A.                has a magnitude of 6 m and it is - 1500 from the positive x axis.     Correct Answer

B.                 has a magnitude of 3 m and it is    1500 from the positive x axis.

C.                = 0.

D.                has a magnitude of 6 m and it is  300 from the positive x axis.

E.                 has a magnitude of 9 m and it is  1500 from the positive x axis.

Solution:

cx = bx - 2 ax

cx = - 6 sin 600  - 2 (0) = - 3 sqrt(3) m

cy = by - 2 ay

cx =  6 cos 600  - 2 (3) = -3 m

c = sqrt((- 6 sqrt(3))2+(-3)2) = 6 m


Question 3:
and  find the angle between them.

A.     1450          Correct Answer  

B.     -1450

C.     350

D.     -950

E.     3250

Solution:


Question 4:           
The distance between Riyadh and Dammam is 350 km.  When there is no wind, an airplane travels this distance in 30 minutes.  Suppose a wind of 50 km/h is blowing in the same direction as the plane, how long does the plane need to travel this distance?

A.    28 minutes        Correct Answer

B.    25 minutes

C.    32 minutes

D.    20 minutes

E.. 18 minutes

Solution:

speed of plane relative to wind = VPW = 350 km / 30 min = 700 km/h

speed of wind relative to ground = VWG= 50 km/h

speed of plane relative to ground = VPG= VPW + VWG= 700 + 50 = 750 km/h

time to travel = distance / velocity = 350 / 750 =0 .467 h (60 m /1 h)= 28 min


Question 4:    
A ball is thrown from the top of a building landed on the ground 30.0 m away from the building.  The velocity of ball, when it strikes the ground, has a magnitude of 60.0 m/s and makes an angle of 45.00 with the ground as shown in the figure.  What is the height of the building?

A.               27.6 m           Correct Answer

B.                30.0 m

C.               60.0 m

D.               33.9 m

E.       75.3 m

 

 

 

 

 

 

 

 

 

 

Solution:

From the figure, we can find vx and vy:

velocity along the horizontal direction =  vx = 60 cos 450 m

velocity along the vertical direction at time t = vy = - 60 sin 450 m

 

 

Since velocity along the horizontal direction is constant ( ax = 0),  we can use
x-x0 = vx0  t + ax t2/2 = vx0  t to determine the time of the projectile.

time of the projectile motion = t = (x-x0)/vx = 30/(60 cos 450) s

 

We can find the initial velocity along the vertical direction from vy = vy0 - g t

so vy0 = vy + gt

 

now, to find the height, we can use

y-y0 = vy0 t - g t2/2

         = (vy + gt) t - g t2/2

         = vy t + g t2/2

         = (- 60 sin 450 )(30/(60 cos 450)) + (9.8)(30/(60 cos 450)) 2/2

         = -30 + 9.8/4

        = -27.6 m

 

So height is 27.6 m