Spring 2001
Physics I
Final Exam Solution Sheet
1. C
2. A
3. C
4. C
5. B
6. B
7. C
8. B
9. B
10. E
11. E
12. A
13. C
14. E
15. C
Part B-Short Answer
#1
Turn around point
+
velocity
time
0
+
acceleration
time
0
#2
Acceleration looks just like force.
1.5
1
Force
0.5
0
0
2
4
6
8
10
12
-0.5
-1
-1.5
Time
#3
While it is true that the acceleration due to gravity is dependent upon distance from the
center of the earth, and so does theoretically change when the height of the object
changes, the earth is so large compared to the additional height that we can throw an
object that this additional distance makes absolutely no detectable difference in the
acceleration due to gravity.
#4
B is correct.
The forces act on different objects so they don’t cancel out. The acceleration is the result
of the net force acting on THAT object.
Part C
#1Conservation of momentum
Using a point in time where the bullet has passed through the first block but not yet hit
the second block as our initial time:
Pi = MBVB,i + M1.2V1.2,i
Using a point in time where the bullet is embedded in the second block as our final time:
Pf = MBVB,f + M1.2V1.2,f + M1.8V1.8,f
So
MBVB,i + M1.2V1.2,i = MBVB,f + M1.2V1.2,f + M1.8V1.8,f
But V1.2,i = V1.2,f = 0.63 m/s
And VB,f = V1.8,f =1.4 m/s so
0.0035(VB,i) + M1.2(.63) =M1.2(.63) + (1.8 +0.0035)(1.4)
VB,i= 721.4 m/s
#2
First kinematics, then newton’s second law.
xf = xo+vot+1/2at2
0.8 = 0 + 0+1/2a(9)
so
a=0.178 m/s2
N
push
W
Rotating the coordinate system so that the x axis is along the incline and the y axis is
perpendicular to the incline, Newton’s second law along the x axis(with up being the
positive direction) gives:
Push-Wsin30o= ma
Push= (2)(.178)+(2)(9.8)(Sin30o)
Push=10.16 N
#3
first conservation of energy then conservation momentum
m gh  1 / 2m v2
9.8(2)  1 / 2v 2
v  6.36 m / s
pi  p f
( 2)(6.26)  ( 2)(1)  (3)(v f )
( 2)(6.26)  2
3
v f  4.84 m/s
vf 
#4
a) U=qV=-(1.6x10-19)(10)=1.6x10-18 J
b) U=qV=-(1.6x10-19)(10)=1.6x10-18 J
c)U=KE so
+1.6x10-18=1/2mv2
v = 1.87 x106 m/s
d) FB=qvxB=qvB=(1.6x10-19)(v)(5)
=1.50 x 10-12 N