Physics I – Exam 1 – Fall 2005
Answer Key
Part A-1 – 1: 12, 2: D, 3: F, 4: B, 5: B
4 pts each.
Partial credit +4 points if 4 and 5 are wrong but the same (+2 each).
Part A-2 – 6: E, 7: D, 8: 0, 9: G, 10: B, 11: H
2 pts each.
B 24 Points
This problem relies on the impulse-momentum theorem (eq. 12). The X part has constant
forces in the two time intervals, and so the graph of Px is two straight line segments with
slopes of 2 and 1 kg m/s2 respectively. The graph of Py is more complicated since the
two parts are parabolas. Students should know this in analogy with the relationship
between velocity and displacement, or alternatively they could figure it out by calculating
the values at intermediate points.
Features Looking For
use impulse-momentum theorem
Px = 0 at t = 0.
Px at 1 sec = +2 more than Px at 0 sec.
Px at 2 sec = +1 more than Px at 1 sec.
straight line Px from 0 to 1 sec.
straight line Px from 1 to 2 sec.
Py = 0 at t = 0.
Py at 1 sec = +1 more than Py at 0 sec.
Py at 2 sec = +1 more than Py at 1 sec.
correct parabola for Py from 0 to 1 sec. (slope = 0 at t = 0, slope = 1 at t = 1)
correct symmetry of curve around t =1 sec. (any shape with proper symmetry.)
–1 point if numbers are otherwise correct but no units or wrong units (only deduct once)
Fx (N)
Fy (N)
2
2
1
Area = 2 N s
1
Area = 1 N s
0
t (sec)
1.0
0
2.0
px kg m/s
Area = 1 N s
Area = 1 N s
t (sec)
1.0
2.0
1.0
2.0
py kg m/s
3
3
2
2
1
1
t (sec)
0
1.0
2.0
t (sec)
0
Part C
Must show work to receive credit.
C-1 20 points
This is a straightforward projectile motion problem.
First find the time to hit the wall using motion in the X (horizontal) direction.
(Keeping a few extra digits in intermediate calculations.)
VX , 0  V0 cos()  2.598 m / s
x  1.50  (2.598) t  t  0.57735 s
Find initial vertical velocity:
Vy , 0  V0 sin( )  1.500 m / s
Plug into the Y equation for final vertical location:
y f  y 0  Vy,0 t  12 a y t 2  1.00  1.500  0.57735  4.9  (0.57735) 2  0.23 m
C-2 24 points
Use Conservation of Momentum separately in the X and Y directions.
Since both pucks have the same mass, divide the momentum equations (44x,44y) by
mass and deal only with velocity components.
V1, x  V1 cos(30)  23 V1
V1, y  V1 sin( 30)  12 V1
V2, y  V2 sin( 60)   23 V2
V2,x  V1 cos(60)  12 V2
Momentum equations with common factor of m divided out:
10.0  V1, x  V2, x  23 V1  12 V2
0  V1, y  V2, y  12 V1  23 V2
Solve the Y equation and get V1  3 V2 , put that into X equation and get V2 = 5.0 m/s.
V1  3 V2  3 (5.0)  8.7 m / s.