Download Answers to Data Analysis Assessment Task for Unit 3

Motion data analysis Task.
Two students examine the motion of an oscillating spring. Firstly, examining how much is stretches
when a mass of 150g is hung from it. The estimated the extension at 11.8 cm
Figure 1
Figure 2
Tension
Force due to gravity
1. On Figure B, draw and label the forces acting. Forces should be equal in size but opposite in
direction (2 marks)
2. Determine an estimate for the spring constant for this spring.
(2 marks)
F = kx,
F = mg = 0.150* 9.8 = 1.47 N
k = F/x = 1.47/0.118 = 12.5 N/m
The students lifted the mass so that the spring was not
intension (not extended) and then released it. They then used
Tracker to examine a recording of the motion of the spring.
They marked the video by hand.
Figure 3
3. Indicate the forces and their relative sizes of the for the four diagrams below. Position A
represents the starting position for the experiment and position C is when the spring is at its
longest extension (4 marks).
Figure 4
4. What is the net force acting at position A?
(1 mark)
F = mg = 0.150* 9.8 = 1.47 N
A graph of the displacement vs time for the first three oscillations is shown below. Distances in cm.
Figure 5
5. Describe how the shape of this graph relates to the motion of the spring. (3 marks)
 Negative values as it is dropping from start point
 Each trough then crest in the wave shape indicated an oscillation
 Turning points indicate zero velocity, where the spring stops and starts changing direction
 Fastest speed is when slope is steepest -> mid-way between max and min heights
 After the first drop the masses rise to a similar point but not quite to the starting position
 The oscillations are reaching very similar heights each time
6. Indicate on the graph where the maximum speeds occur in the first oscillation.
(1 mark)
7. Determine the maximum speed of the mass in first fall. (2 marks)
Gradient between two point, (.30, -10), (0.33, -13.25)
=(-13.25 - -10)/( 0.33 - 0.30) = - 108.3 cm/s = - 1.08 m/s ~ -1.1 m/s (2 s.f.)
8. What is the magnitude of the mass’ momentum when it reaches this maximum speed?
(1 mark)
P = mv = 0.150kg *1.08 m/s = 0.1625 kg m/s ~ 0.16 kg m/s
9. What was the change in momentum the for the mass when it reaches the same position on its
way up?
(3 marks)
ΔP = Pf - Pi ,
need final velocity, gradient from graph = (-9.5 – -11.5)/(0.73 – 0.70)
= 66.7 cm/s = 0.667 m/s ~ 0.67 m/s
= 0.150(0.667 - - 1.08)
= 0.26 kg m/s (up)
10. Estimate the average force the object experiences during this change in momentum?
(3 marks)
Use I = ΔP => Ft = ΔP => F = ΔP/t
time from graph = 0.72 – 0.32 = 0.40s
F = 0.26 / 0.40 = 0.65 N
11. What is the total energy of this system?
(2 marks)
Take total energy as the grav. Potential from drop height to bottom of spring stretch. Approx. 21 cm
Egp = mgh = 0.15 * 9.8 * 0.21 = 0.3087 J ~ 0.31 J
12. How much elastic potential does the spring hold at the bottom of the first oscillation?
Eep at bottom = energy at top = 0.31 J
13. Use this value to determine check you spring constant estimate. How do they compare?
Eep= ½ k x2 => k = 2 Eep/ x2
x = 21 cm
= 2 * 0.31/(0.21)2 = 14 N/m
This value is a little bit larger compared to 13 N/m
14. Using the data above and your knowledge of conservation of energy calculate a prediction for
the maximum kinetic energy and thus a predicted maximum speed.
(4 marks)
Max kinetic energy will occur at the mid-point of the extension. Taking into account the energy loss
after the initial drop the mid-point is when x = 21/2 = 10.5 cm or 0.105 m from max height of
oscillation. (1)
Take k as an average of the two values (14 + 13)/2 = 13.5 N/m
At this point Etotal =
Epg
+
Eep
+ Ek
0.3087 J = 0.15*9.8*0.105 + ½*13.5*(0.105)2 + Ek
Ek = 0.3087 – 0.2287 = 0.0799 J
Ek = ½mv2= 0.0799 J
v = sqrt(2*0.0799)/0.15 = 1.03 m/s
15. Comment on how your calculated maximum speed and the measured maximum speed
compare and give explanation for the discrepancy .
(2 marks)
Calculated value 1.02 m/s and gradient of first fall 1.08 m/s compare quite well.
There is particular difficulty in reading the graph accurately and this finding the correct
gradient and thus velocity.
The graph generated by the students came from video marked by hand. This may mean the that marks
on the video and thus the corresponding displacement time graph is inaccurate.
Graph can only accurately be read to 2 s.f on the time axis so spring const and all other calcs to only 2
s.f.
16. How much energy appears to be ‘lost’ after three oscillations? (2 marks)
After initial drop the spring never restore to that high but consistently reaches ~ -1.7 cm
This suggests this energy may not be ‘lost’ to the system by heat/sound etc.
Egp = mgh = 0.15 * 9.8 * 0.017 = 0.0250 J ~ 0.025J
17. How does this loss correlate with the conservation of energy?
(2 marks)
Spring does not restore energy completely, loss due to heat, sound deformation (2)
Drop of spring my not have been perfect; the spring may not have been completely slack, of
may have been held too high so there was a minimal free fall before the spring started to
stretch. The calculations assume perfect drop. (2)
18. What could these students do to improve the experiment?
(3 marks)