Physics · Simulation and virtual lab Free
Motion of a trolley: ticker-timer, tape chart and graphs
Let a trolley run down a sloping runway pulling a ticker-tape, cut the tape into 0.1 s strips and turn them into velocity–time graphs.
- You will learn to
- Measure distance, time and speed with a ticker-timer and a tape chart.
- Distinguish uniform motion from uniformly accelerated motion using the dots and graphs.
- Find the acceleration from the gradient of a velocity–time graph.
- You will measure
- Length of each 0.10 s strip (cm); then velocity (m/s) and acceleration (m/s²).
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: trolley, runway and wooden block, ticker-timer with 50 Hz supply, ticker-tape, scissors, ruler, graph paper.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To measure distance, speed and acceleration of a trolley with a ticker-timer, to tell uniform motion from uniformly accelerated motion, and to find the acceleration from a velocity–time graph.
What to do
- Set the slope to 3° (friction compensated) and a push of 0.5 m/s. Release the trolley. Look at the dots: are they equally spaced? Tap Record the tape.
- Set the slope to 6° and no push. Release, then record. How do the strips in the tape chart change?
- Repeat with 4° and 8°. Each run is one line on the velocity–time graph.
- Read the acceleration of each run from the box under the graph. How does it depend on the slope?
- Set the slope to 1° and a push of 0.8 m/s: the trolley slows down. What does the tape chart show now?
- Switch the graph to Distance–time and compare the shapes for uniform and accelerated motion.
Results
Questions
What time interval separates two dots made by a 50 Hz ticker-timer?
1 ÷ 50 = 0.02 s. Five dot spaces (one strip) take 5 × 0.02 = 0.10 s.
How can you tell from the tape that the trolley is accelerating?
The spaces between the dots get longer: the trolley travels further in each 0.02 s, so its speed is increasing.
Why does a tape chart look like a velocity–time graph?
Every strip lasts the same time (0.10 s), so the length of each strip is proportional to the average velocity in that interval; standing the strips side by side in order gives velocity against time.
A strip is 4.5 cm long. Find the average velocity during that strip.
v = 0.045 m ÷ 0.10 s = 0.45 m/s.
What does the gradient of a velocity–time graph give? And the area under it?
The gradient gives the acceleration; the area under the graph gives the distance travelled.
Why is the runway tilted slightly when we want constant velocity?
To compensate for friction: the component of the weight down the slope balances the friction, so there is no resultant force and the trolley keeps a constant velocity after a push.
For teachers: expected results and how the model works
Model: friction coefficient 0.05, so the runway is friction-compensated at 2.9°; acceleration a = g(sin θ − 0.05 cos θ): 4° → 0.20 m/s², 6° → 0.54 m/s², 8° → 0.88 m/s²; 3° gives almost constant velocity (0.02 m/s²). From rest on 6° the trolley covers the 1.44 m run in about 2.3 s (23 strips). Below 2.9° a pushed trolley decelerates and stops (1° with 0.8 m/s: −0.32 m/s²). The ticker-timer is 50 Hz; strips are 5 spaces = 0.10 s; the velocity of a strip is its average velocity, plotted at the middle of its time interval, so the v–t line gives the start velocity at t = 0. The runway angle is drawn 2.6 times steeper so that it can be seen.
Exam practice Exam set
National-exam style questions on this simulation (19 marks). Try each one on paper first, then open the marking scheme.
A ticker-timer makes 50 dots per second. A strip of 10 dot spaces is 15 cm long. Calculate (a) the time for the strip, (b) the average velocity. (3 marks)
Marking scheme: (a) 10 × 0.02 = 0.20 s (1). (b) v = 0.15 ÷ 0.20 (1) = 0.75 m/s (1).
The first 5-space strip of a tape is 2.0 cm long and the sixth strip is 7.0 cm long. Calculate the acceleration of the trolley. (4 marks)
Marking scheme: u = 0.020 ÷ 0.10 = 0.20 m/s (1); v = 0.070 ÷ 0.10 = 0.70 m/s (1); the strips are 0.50 s apart (5 × 0.10 s) (1); a = (0.70 − 0.20) ÷ 0.50 = 1.0 m/s² (1).
A car starts from rest and accelerates uniformly at 2.0 m/s² for 6.0 s. Calculate (a) its final velocity, (b) the distance it travels. (4 marks)
Marking scheme: (a) v = u + at = 0 + 2.0 × 6.0 (1) = 12 m/s (1). (b) s = ut + ½at² = ½ × 2.0 × 6.0² (1) = 36 m (1).
Sketch distance–time and velocity–time graphs for (a) a trolley moving at constant velocity and (b) a trolley with uniform acceleration from rest. (4 marks)
Marking scheme: (a) s–t: straight sloping line through the origin (1); v–t: horizontal line (1). (b) s–t: curve getting steeper (1); v–t: straight sloping line through the origin (1).
Distinguish between distance and displacement, and between speed and velocity. (4 marks)
Marking scheme: Distance is the length of the path (scalar) (1); displacement is the straight-line distance in a stated direction (vector) (1). Speed is distance ÷ time (scalar) (1); velocity is displacement ÷ time, with direction (vector) (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Motion of a trolley: ticker-timer, tape chart and graphs
Aim: To measure distance, speed and acceleration of a trolley with a ticker-timer, to tell uniform motion from uniformly accelerated motion, and to find the acceleration from a velocity–time graph.
Apparatus (real laboratory): trolley, runway and wooden block, ticker-timer with 50 Hz supply, ticker-tape, scissors, ruler, graph paper.
What to do (with the simulation on www.donatstem.co.rw)
- Set the slope to 3° (friction compensated) and a push of 0.5 m/s. Release the trolley. Look at the dots: are they equally spaced? Tap Record the tape.
- Set the slope to 6° and no push. Release, then record. How do the strips in the tape chart change?
- Repeat with 4° and 8°. Each run is one line on the velocity–time graph.
- Read the acceleration of each run from the box under the graph. How does it depend on the slope?
- Set the slope to 1° and a push of 0.8 m/s: the trolley slows down. What does the tape chart show now?
- Switch the graph to Distance–time and compare the shapes for uniform and accelerated motion.
Results
| # | Run | Slope (°) | Strip | t (s) | Length (cm) | v (m/s) |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
| 9 | | | | | | |
| 10 | | | | | | |
| 11 | | | | | | |
| 12 | | | | | | |
Graph
Stick the strips (or plot their velocities) side by side to make a velocity–time graph for each run. Draw the best straight line and find its gradient (the acceleration).
Questions
- What time interval separates two dots made by a 50 Hz ticker-timer? [2]
- How can you tell from the tape that the trolley is accelerating? [2]
- Why does a tape chart look like a velocity–time graph? [2]
- A strip is 4.5 cm long. Find the average velocity during that strip. [2]
- What does the gradient of a velocity–time graph give? And the area under it? [2]
- Why is the runway tilted slightly when we want constant velocity? [2]
- A ticker-timer makes 50 dots per second. A strip of 10 dot spaces is 15 cm long. Calculate (a) the time for the strip, (b) the average velocity. [3]
- The first 5-space strip of a tape is 2.0 cm long and the sixth strip is 7.0 cm long. Calculate the acceleration of the trolley. [4]
- A car starts from rest and accelerates uniformly at 2.0 m/s² for 6.0 s. Calculate (a) its final velocity, (b) the distance it travels. [4]
- Sketch distance–time and velocity–time graphs for (a) a trolley moving at constant velocity and (b) a trolley with uniform acceleration from rest. [4]
- Distinguish between distance and displacement, and between speed and velocity. [4]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Motion of a trolley: ticker-timer, tape chart and graphs
Expected results
Model: friction coefficient 0.05, so the runway is friction-compensated at 2.9°; acceleration a = g(sin θ − 0.05 cos θ): 4° → 0.20 m/s², 6° → 0.54 m/s², 8° → 0.88 m/s²; 3° gives almost constant velocity (0.02 m/s²). From rest on 6° the trolley covers the 1.44 m run in about 2.3 s (23 strips). Below 2.9° a pushed trolley decelerates and stops (1° with 0.8 m/s: −0.32 m/s²). The ticker-timer is 50 Hz; strips are 5 spaces = 0.10 s; the velocity of a strip is its average velocity, plotted at the middle of its time interval, so the v–t line gives the start velocity at t = 0. The runway angle is drawn 2.6 times steeper so that it can be seen.
Answers
- What time interval separates two dots made by a 50 Hz ticker-timer? [2]
1 ÷ 50 = 0.02 s. Five dot spaces (one strip) take 5 × 0.02 = 0.10 s.
- How can you tell from the tape that the trolley is accelerating? [2]
The spaces between the dots get longer: the trolley travels further in each 0.02 s, so its speed is increasing.
- Why does a tape chart look like a velocity–time graph? [2]
Every strip lasts the same time (0.10 s), so the length of each strip is proportional to the average velocity in that interval; standing the strips side by side in order gives velocity against time.
- A strip is 4.5 cm long. Find the average velocity during that strip. [2]
v = 0.045 m ÷ 0.10 s = 0.45 m/s.
- What does the gradient of a velocity–time graph give? And the area under it? [2]
The gradient gives the acceleration; the area under the graph gives the distance travelled.
- Why is the runway tilted slightly when we want constant velocity? [2]
To compensate for friction: the component of the weight down the slope balances the friction, so there is no resultant force and the trolley keeps a constant velocity after a push.
- A ticker-timer makes 50 dots per second. A strip of 10 dot spaces is 15 cm long. Calculate (a) the time for the strip, (b) the average velocity. [3]
(a) 10 × 0.02 = 0.20 s (1). (b) v = 0.15 ÷ 0.20 (1) = 0.75 m/s (1).
- The first 5-space strip of a tape is 2.0 cm long and the sixth strip is 7.0 cm long. Calculate the acceleration of the trolley. [4]
u = 0.020 ÷ 0.10 = 0.20 m/s (1); v = 0.070 ÷ 0.10 = 0.70 m/s (1); the strips are 0.50 s apart (5 × 0.10 s) (1); a = (0.70 − 0.20) ÷ 0.50 = 1.0 m/s² (1).
- A car starts from rest and accelerates uniformly at 2.0 m/s² for 6.0 s. Calculate (a) its final velocity, (b) the distance it travels. [4]
(a) v = u + at = 0 + 2.0 × 6.0 (1) = 12 m/s (1). (b) s = ut + ½at² = ½ × 2.0 × 6.0² (1) = 36 m (1).
- Sketch distance–time and velocity–time graphs for (a) a trolley moving at constant velocity and (b) a trolley with uniform acceleration from rest. [4]
(a) s–t: straight sloping line through the origin (1); v–t: horizontal line (1). (b) s–t: curve getting steeper (1); v–t: straight sloping line through the origin (1).
- Distinguish between distance and displacement, and between speed and velocity. [4]
Distance is the length of the path (scalar) (1); displacement is the straight-line distance in a stated direction (vector) (1). Speed is distance ÷ time (scalar) (1); velocity is displacement ÷ time, with direction (vector) (1).