Lab112 Newton Second Law Sample Assignment
Lab 112: Newton’s Second Law
1. Objective:
- To study newton’s second law of motion through experiment
- Compare the experimental data with calculations using one-dimensional equation, and therefore verify Newton’s second law.
2. Theoretical background
- ∑ F=ma, Newton’s second law states that the net force on an object is the product of its mass and acceleration.
- Vf2-Vo2=2a(x-x0), equation of motion involves constant acceleration. This formula is used to calculated the unknown final velocity.
- Vf=V0+at, this equation is used to calculate the unknown time.
3. Procedures
- Adjust the air track to be horizontal so that the glider doesn’t move.
- Place the 2 photogates and measure each of position on the air track.
- Weigh the glider on a scale and record value.
- Position glider on the air track and attach 40 grams of weight on glider using a cord.
- Set up loggerPro.
- Perform Part I of experiment by releasing the glider from its initial position and record the data of glider passing through 2 photogates using LoggerPro.
- Proceed to add 2 weights, and 4 weights on glider, respectively, and collect data on each.
- With data of Part I obtained. Part II is performed with same procedures but an inclined angle.
- Measure the inclined angle of air track and adjust photogates to be at same position of the air track.
- Repeat steps from Part I.
4. Results
a. Experimental Data
Mg=178.83g=0.17883kg, Mg is the mass of glider
Mh=40g=0.04kg, Mh is the mass of a hanging weight, M1=50.28g, or 0.05028kg
X0=90m, L=54.5m, X0 is the initial position of glider, and L is distance between 2 photogates.
ϴ=5o, angle of inclined air track.
|
Total Glider mass (kg) |
Hanging mass Mh(kg) |
Acceleration (m/s2) |
Time to travel distance L (s) |
Velocity at gate 1 V1 (m/s) |
Velocity at gate 2 V2 (m/s) |
|
Mg |
0.04 |
1.79 |
0.457 |
0.915 |
1.753 |
|
Mg+2M1 |
0.04 |
1.279 |
0.549 |
0.768 |
1.486 |
|
Mg+4M1 |
0.04 |
0.953 |
0.633 |
0.667 |
1.284 |
|
Mg ϴ=5o |
0.04 |
1.141 |
0.569 |
0.732 |
1.396 |
|
Mg+2M1 ϴ=5o |
0.04 |
0.522 |
0.845 |
0.486 |
0.938 |
|
Mg+4M1 ϴ=5o |
0.04 |
0.149 |
1.608 |
0.242 |
0.486 |
b. Calculation
(1)Finding acceleration
a=F/M= Mh(g)/(Mg+Mh). a=(40*9.8)/(178.83+40)=1.79m/s2, we can the proceed to find values of acceleration by adding more mass to the glider.
(2)Finding acceleration at inclined angle
a=F/M=[Mgsin(5o)-Mh]*g / (Mg+Mh), a=[178.83*0.087-40]*(-9.8) / (178.83+40)=1.093m/s2.
(3)Finding velocity
V2=2ax, the v0 is zero and x is displacement from glider to gate 1, x is 0.196m for V1, and 0.741m for V2
V=sqrt[2*1.79*0.196]=0.838m/s
(4)Finding time
T=(V2-V1)/a, T=(1.629-0.838)/1.79=0.442s
|
Total Glider mass (kg) |
Hanging mass M(kg) |
Acceleration (m/s2) |
Time to travel distance L (s) |
Velocity at gate 1 V1 (m/s) |
Velocity at gate 2 V2 (m/s) |
|
Mg |
0.04 |
1.79 |
0.442 |
0.838 |
1.629 |
|
Mg+2M1 |
0.04 |
1.227 |
0.534 |
0.694 |
1.347 |
|
Mg+4M1 |
0.04 |
0.933 |
0.612 |
0.605 |
1.176 |
|
Mg ϴ=5o |
0.04 |
1.093 |
0.565 |
0.656 |
1.272 |
|
Mg+2M1 ϴ=5o |
0.04 |
0.48 |
0.853 |
0.434 |
0.844 |
|
Mg+4M1 ϴ=5o |
0.04 |
0.161 |
1.475 |
0.251 |
0.488 |
c. Error analysis
|
Mg (g) |
Pulse(s) |
V1(m/s) |
V2(m/s) |
a (m/s2) |
Pulse(s) |
V1(m/s) |
V2 (m/s) |
a(m/s2) |
|
178.83 |
.0150 |
.0770 |
.1237 |
.0020 |
.0043 |
.0773 |
.1240 |
.0477 |
|
279.39 |
.0153 |
.0744 |
.1373 |
.0517 |
.0083 |
.0521 |
.0944 |
.0418 |
|
379.95 |
.0210 |
.0620 |
.1078 |
.0196 |
.1329 |
.0090 |
.0020 |
.0117 |
|
178.83 |
3.39% |
9.19% |
7.59% |
.11% |
.76% |
11.81% |
9.75% |
4.36% |
|
279.39 |
2.87% |
10.73% |
10.18% |
4.21% |
.97% |
12.02% |
11.19% |
8.70% |
|
379.95 |
3.43% |
10.25% |
9.17% |
2.10% |
9.01% |
3.57% |
.41% |
7.26% |
5. . Discussion:
- Linear density of the cord may have affected the experiment in a very small subtle way. Because during the experiment, the weight of the cord was not taken into accounts. And therefore the calculated acceleration might differ from computer produced acceleration since cord’s mass is not part of total mass.
- If the flag is not parallel to the direction of movement, the calculated velocity would become too large because the wrongly aligned flag has a larger width than the parallel one (titled flag has a larger overall area) and therefore resulting larger velocity
- Increase in ϴ would result in the increase in calculated acceleration because of formula [Mgsin(ϴ)-Mh]*g / (Mg+Mh), and decrease in ϴ, logically results in decrease in acceleration because numerator is less than the original ϴ.
The experiment consists of 2 parts, with second part of experiment an inclined angle. The measured data allows us to find the acceleration of the glider with various masses using newton’s second law, F=ma. We can then proceed to calculate other missing values, velocity and time, from the experiment using equations of motion. There are errors produced as referred to table I and table II. If calculation were done correctly, then % difference could very well tell us something had gone wrong during the experiment. Factors such negligence of flag and cord’s mass, inaccurate measurements could result in error.
6. Conclusion
The newton’s second law of motion establishes the relationship between mass and acceleration, as their product gives the net force acting on an object. Acceleration is therefore inversely proportion to the mass of the object. In this experiment, newton’s law is well demonstrated as increase in mass leads to decrease in acceleration, or vise versa.
