Physics I – Exam 2 – Fall 2004
Answer Key
Part A –
1: D.
2: ABCD. Partial credit 1 point for each.
3: B.
4. E. Partial credit 2 points for F.
5. C.
6: C. Partial credit 2 points for D.
B-1 16 Points
U = m g h = 7840 at highest point.
K = ½ m v2 = 1000 at highest point.
K+U = 8840 everywhere.
Partial Credit
+4 points for K+U = constant.
+2 points for U curve same shape as h curve.
+2 points for Umax = 7840.
+2 points for Umax–Umin = 11760.
+4 points for K curve is mirror shape of U.
+2 points for K = 1000 at lowest point.
h (m)
10.0
5.0
0
d (km)
2.0
4.0
6.0
8.0
10.0
-5.0
U (J)
7840
3920
d (km)
0
-3920
2.0
4.0
6.0
8.0
10.0
12760
K (J)
8840
4920
1000
0
B-2 24 Points
p = F t (for constant F and p = 0 at t = 0).
l = r x p : x component of r is irrelevant
because p is in the x direction.
l is into the page or –Z direction.
Partial Credit
+2 points p = 0 at t = 0.
+2 points p is straight line.
+2 points p = 2 kg m/s at t = 10.
+2 points correct units of l (kg m2 /s).
+2 all components of l = 0 at t = 0.
+2 points lx = 0 for all times.
+2 points ly = 0 for all times.
+4 points lz is the only non-zero curve.
The next points apply to the non-zero l curve
as long as there is only one such non-zero curve.
+2 points l curve is a straight line.
+2 points l = + or – 0.5 p from p curve.
+2 points if non-zero component of l is negative.
d (km)
2.0
4.0
6.0
8.0
10.0
2.0
4.0
6.0
8.0
10.0
px (kg m/s)
2.0
1.0
0
t (s)
lz (kg m^2/s)
t (s)
0
2.0
-1
4.0
6.0
8.0
10.0
Part C
Must show work to receive credit.
C-1 16 Points
There are two basic approaches, both starting with conservation of momentum.
For simplicity, we will use units of mass such that the mass of H is 1 and unit of velocity
is 1,000,000 m/s. It is OK to do this as long as we are consistent.
Method 1:
Use Conservation of Momentum: (1)(5)  (4)(0)  (1)(3)  (4)( v 2 )  v 2  2
Check KE before and after:
Before: K  12 (1)(5) 2  12.5 After: K  12 (1)( 3) 2  12 (4)( 2) 2  4.5  8  12.5
Since kinetic energy is the same before and after, the collision is elastic:
Method 2:
The collision is evidently one-dimensional and momentum is conserved. Check what we
predict with the one-dimensional elastic collision equation for v1 final:
m  m2
1 4
Predicted: v1f  1
v1i 
(5)  3
m1  m 2
1 4
Since this is what we actually have, the collision must have been elastic because if it were
inelastic, the final velocity of mass 1 would be different from what was predicted.
C-2 20 points
Use Conservation of Energy. Only the unbalanced mass on the end counts for potential
energy because the rod itself is balanced (equal sides go up and down.)
Initial: K+U = 0+mgh = (4)(9.8)(0.25) = 9.8 J. (I defined U = 0 at bottom position.)
Final: K+U = ½ I 2+0. Then   2(mgh ) / I . Need to find I:
I  I rod  I mass  121 (3)(.5) 2  (4)(.25) 2  0.3125 kg m2.
  2(9.8) / 0.3125  7.9 rad/s.