# CSULA- PHYSICS 211 - DOC

Document Sample

```					CSULA- PHYSICS 211                            Summer’12                             Solution: Material to Study                                  Chapter 4

OBJECTIVE QUESTIONS

OQ-1 (7e- Q1)
The car's acceleration must have an inward component and a forward component: answer (e). Another argument: Draw
a final velocity vector of two units west. Add to it a vector of one unit south. This represents subtracting the initial
velocity from the final velocity, on the way to finding the acceleration. The direction of the resultant is that of vector (e).

OQ-4 (7e-None)
(b) acceleration = 1 g , and (c) the horizontal component of the velocity vix  vi cos
6
OQ-6 (7e-Q9)
The projectile on the moon is in flight for a time interval six times larger, with the same range of vertical speeds and with
the same constant horizontal speed as on Earth. Then its maximum altitude is (d) six times larger.
v i2 sin2  i
Using the equation for maximum height:                         h                      you can see that if g in the denominator changes to                                    1
g , then
2g                                                                                                   6
h goes 6 times greater.

CONCEPTUAL QUESTIONS
CQ-5 (7e-Q7)
The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of gravity. Its horizontal
component of acceleration is zero.

CQ-6 (7e-Q13)                                                                                                                                            v            v
The skater starts at the center of the eight, goes clockwise around the left circle and then                                                         a       a v              a        v
counterclockwise around the right circle.                                                                                                   v
a                                 a
a       a
v        a
CQ-7 (7e-Q10)                                                                                                                                    v                                v
(a) NO. Its velocity is constant in magnitude and direction.
(b) YES. The particle is continuously changing the direction of its velocity vector.

PROBLEMS

P2 (7e-P3)
The sun projects onto the ground the x component of her velocity: 5.00 m s cos     2.50 m s
60.0
P4 (7e-P4)
dx  d 
(a)     From      x  5.00sin  t       , the x component of velocity is:                  vx          5.00sin  t   5.00 cos t
dt  dt 
dvx
and ax           +5.00 2 sin  t .
dt
Similarly, vy   d   4.00  5.00cos t   0  5.00 sin  t & ay   d   5.00 sin  t   5.00 2 cos t .
                                                      
 dt                                                                  dt 
At t  0 ,       v  5.00 cos0i  5.00 sin 0ˆ 
ˆ              j           5.00 iˆ  0ˆj  m s    &                       ˆ
a  5.00 2 sin 0i  5.00 2 cos0ˆ 
j       0iˆ  5.00 ˆj 
2
m s2     .

(b)            ˆ
r  x i  yˆ 
j       4.00 m  ˆ   5.00 m    sin  t i  cos t ˆ  ,
j                          ˆ            j                v       5.00 m     cos t i  sin  t ˆ 

ˆ           j

&

a                           ˆ
 5.00 m   2 sin  t i  cos t ˆ 

j


(c)     The object moves in a circle of radius 5.00 m centered at 0, 4.00 m  .
P6 (7e-P8)
a  3.00ˆ m s2 ; v i  5.00i m s ; ri  0i  0ˆ
j                  ˆ             ˆ    j

(a)                             1        
rf  ri  v i t  a t 2   5.00t i  3.00t 2 ˆ  m
2        
ˆ 1
2
j


& v f  vi  at             5.00iˆ  3.00tˆj    m s

(b)                                           i
1
2
j       i       j 
t  2.00 s , rf  5.00  2.00 ˆ   3.00 2.002 ˆ  10.0ˆ  6.00ˆ m                                      so          x f  10.0 m   &   y f  6.00 m

ˆ                        ˆ       j   
v f  5.00i  3.00  2.00 ˆ  5.00i  6.00ˆ m s
j                                     
v f  v f  vxf  vyf 
2     2
 5.002   6.002  7.81 m s

P10 (7e-P10)
The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking the origin at the
point where the mug leaves the bar, the coordinates of the mug at any time are
1                                    1            1
x f  vxi t  ax t 2  vxi t  0 and y f  vyi t  ay t 2  0  gt 2
2                                                            2          2

When the mug reaches the floor, y f   h so:                                             1
 h   gt 2    which gives the time of impact as:                      t
2h
2                                                                      g

(a) Since x f  d when the mug reaches the floor, x f  vxi t becomes                                                  d  vxi
2h    giving the initial velocity as:     vxi  d
g
g                                                  2h

(b) Just before impact, the x component of velocity is still: vxf  vxi while the y component is:                                                          vyf  vyi  ay t  0  g
2h
g
Then the direction of motion just before impact is below the horizontal at an angle of:
 vyf            g 2h/ g            2h 
  tan 1       tan 1            tan 1  
 vxf            d g/ 2h            d 
                        
P11 (7e-P13)
so: 3vi sin i  i 
vi2 sin 2  i                vi2  sin 2 i  ;
2    2
v 2 sin 2 i                                 2 sin 2 i tan  i
h                    ;   R                             3h  R                                                             or                
2g                             g                                         2g                     g                        3 sin 2 i   2

Thus: i  tan 1    53.1
4
 
 3

P21 (7e-P21)
The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from
1
y f  yi  vyi t  ay t 2
2
1
2

40.0 m  0  0  9.80 m s2 t 2                
t  2.86 s
The extra time 3.00 s  2.86 s  0.143 s is the time required for the sound she hears to travel straight back to the player. It
covers distance  343 m s 0.143 s  49.0 m  x2   40.0 m 2 , where x represents the horizontal distance the rock travels.
x  28.3 m  vxi t  0t 2
28.3 m
 vxi                9.91 m s
2.86 s

P27 (7e-P25)
v2  20.0 m s
2

ac                   377 m s2                             The mass is unnecessary information.
r     1.06 m

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 1 posted: 10/4/2012 language: Latin pages: 2