Math  /  Data & Statistics

QuestionEXPERIMENT 19: UNIFORM VELOCITY AlM: To investigate the uniform velocity of a moving trolley. APPARATUS Ticker timer Ticker tape Power source Trolley Trolley track METHOD 1 Altach a long strip of paper tape to the trolley and pass the tape through the ticker fimer. 2 Connect the ticker timer to the power source
3. Raise the one end of the runway sufficiently so that the trolley moves down the runway at constant speed. (This adjustment which compensates for friction is most important for the success of this investigation) 4 Cut off the beginning and end portions of the tape where the motion was not uniform, and keep only the portion of the tape where the dots are evenly spaced.
5. Measure the length of the tape and count the number of spaces Record the results in Table 1.
6. Now mark the tape in lengths of 5 spaces ( 0,10 s0,10 \mathrm{~s} if the frequency of the timer is 50 Hz ) each Measure the displacement (from the first chosen dot) for successive time intervals, ie for t=5t=5 time intervals, displacement = the total length of tape for 26 spaces For each 5 space interval., calculate the average velocity during that interval by dividing the length of the interval spaces), in meters, by 0,10 s0,10 \mathrm{~s} (the time for 5 spaces). Record all results in Table
5. RESULTS 1 Copy Table 1 into your practical book. Record the length of the tape as well as the number of spaces as determined in step 5 of the method. The frequency of the ticker timer depends on the power source used. If connected to 220 VAC , the frequency is 50 Hz \begin{tabular}{|c|c|} \hline \multicolumn{2}{|l|}{ TABLE 1} \\ \hline Length of tape (x)m(x) \mathrm{m} & \\ \hline No. of spaces ( π\pi ) & \\ \hline Frequency of timer (1) & 50 Hz \\ \hline Period of timer (T) & \\ \hline Total time (n×T)(n \times T) & \\ \hline Average velocity (πf)\left(\frac{\pi}{f}\right) & \\ \hline \end{tabular} 2 Copy Table 2 into your practical book and then completa the open cells. \begin{tabular}{|c|c|c|c|c|} \hline \multicolumn{5}{|l|}{1 TABLE 2} \\ \hline \begin{tabular}{l} Time I (s) \end{tabular} & \begin{tabular}{l} \begin{tabular}{c} 2 \\ Δt(s)\Delta t(s) \end{tabular} \end{tabular} & \begin{tabular}{l} Displacement -x \\ (m) \end{tabular} & \begin{tabular}{l} Δx(m)\Delta x(m) \end{tabular} & \begin{tabular}{l} v=Δx5Δt(ms)v=\frac{\Delta x^{5}}{\Delta t}\left(m s^{\prime}\right) \end{tabular} \\ \hline & \%: : 2 & 2=2= & & \\ \hline & & & & \\ \hline & & & & \\ \hline & & a & & \\ \hline & & & & \\ \hline & & & & \\ \hline & & & & \\ \hline \end{tabular}

CONCLUSION AND QUESTIONS 1 What is the main reason for failure when performing this experiment? How can you thy to compensate for that? \qquad \qquad
2. What is the meaning of the phrase frequency of the timen \qquad 3 What is the meaning of the period of the timen? \qquad 4 Is it necessary to use the whole length of the tape for your calculations? Explain \qquad
5. Show by a calculation that the time for one interval is 0,1 s \qquad
6. Plot a graph of displacement vs. time on a graph paper. What is the shape of the graph? \qquad

Studdy Solution
Answer the concluding questions.
1. **What is the main reason for failure when performing this experiment? How can you try to compensate for that?**
The main reason for failure could be friction not being adequately compensated for. Ensure the runway is raised correctly to counteract friction.
2. **What is the meaning of the phrase frequency of the timer?**
The frequency of the timer is the number of ticks (or dots) it makes per second, which in this case is 50 Hz.
3. **What is the meaning of the period of the timer?**
The period of the timer is the time interval between two successive ticks, which is the reciprocal of the frequency.
4. **Is it necessary to use the whole length of the tape for your calculations? Explain.**
No, it is not necessary to use the whole length of the tape. Only the portion where the motion was uniform should be used to ensure accurate calculations.
5. **Show by a calculation that the time for one interval is 0.1 s.**
Given the frequency f=50f = 50 Hz, the time for 5 spaces (5 ticks) is:
Time for one interval=5×150 s=0.1 s \text{Time for one interval} = 5 \times \frac{1}{50} \text{ s} = 0.1 \text{ s}
6. **Plot a graph of displacement vs. time on graph paper. What is the shape of the graph?**
The graph of displacement vs. time for uniform velocity should be a straight line, indicating constant velocity.

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