Physics I – Exam 2 – Fall 2006
Answer Key
Part A – 1: 23, 2: A, 3: D, 4: A
4 pts each.
Subtract 2 points from #2 for each incorrect additional answer, minimum of 0 points.
5: A, 6: A, 7: C, 8: B, 9: C, 10: B, 11: A, 12: C
2 pts each.
For #11, 1 point partial credit if it agrees with #6 but is wrong. (Don't miss please.)
B-1 20 Points
The key to this problem is the elastic collision. The easiest method to solve the problem
is to use equations 45a to find V2i and 45b or 44x to find V2f.
m  m2
2 m2
v1,f  1
v1,i 
v 2,i   3.5  (0.5)( 4)  (1.5)( v 2,i )  v 2,i  1.0 m/s
m1  m 2
m1  m 2
2 m1
m  m1
v 2,f 
v1,i  2
v 2,i  (0.5)(4)  (0.5)(1.0)  1.5 m/s
m1  m 2
m1  m 2
Looking For
using conservation of momentum or equation 45.
using KE before = KE after or equation 45.
constant V2 before and after collision.
straight line V2 during collision..
correct value of V2i.
correct value of V2f.
VA (m/s)
4
3
2
1
t (msec)
0
-1
5
10
-2
-3
-4
VB (m/s)
2
1.5
1
0.5
t (msec)
0
-0.5
-1
-1.5
-2
5
10
B-2 20 Points
The key to this problem is conservation of mechanical energy and the Work – Kinetic
Energy Theorem. KE + PE = 0 and KE = W.
KE starts at 0 and goes up as much as PE goes
KE (J)
down. KE max = 4 J. Then back to zero.
4
Since the force is spring + gravity, the force curve is
3
a straight line. It begins positive (increasing KE)
2
and goes through 0 where KE is max. The area
1
under the first triangle is W = 4 J. That means
0
Fmax is 160 N. (Don't forget to convert cm!)
Looking For
using KE + PE = 0.
using KE = W = ∫Fdx or equiv.
KE is mirror of PE, begins at 0, and reaches a
maximum of 4 J.
F is straight line sloping down.
F = 0 at y = 5 cm.
F max = 160 N.
y (cm)
5
10
5
10
0
Net Force (N)
160
0
y (cm)
-160
Part C (28 Points)
This problem uses the idea of energy in rotational motion and calculation of rotational
inertia.
Itotal = Iwheel + 3Ichild + Iparent = 2125 + 3[(20)(2.5)2] + [(80)(2.5)2] = 3000 kg m2
There is net movement of 60 kg down 5 m. PE = (60)(9.8)(–5) = –2940 J.
Work done by friction = ∫d = (–339)() = –1065 J.
KE = –PE + Wnon-c = 2940–1065 = 1875 J.
KE initial = ½ I 2 = 1500 J. KE final = ½ I f2 = 1500+1875 = 3375 J. f = 1.5 rad/s.
Looking For
correct equation for adding rotational inertia of masses.
correct value of rotational inertia = 3000 kg m2.
PE = mgh, whether plus or minus, whether correct m or not.
correct value of = –2940 J including negative sign.
using ∫d to find work done by friction torque..
correct value = –1065 J including negative sign.
using KE + PE = Wnon-c or equivalent expression.
using KE = ½ I 2.
correct final answer f = 1.5 rad/s.