# acceleration using light gates Watch

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Hi!

If I wanted to measure the acceleration of a falling object using light gates, how many would I need to use? Because I can't see how you would calculate it using two light gates...from my understanding (which may be completely wrong), the first light gate would start timing when the object breaks the light beam and when the second light beam is broken, you would get the time taken for the object to fall through the distance between the two light gates. But the object would have an initial speed 'u' when it breaks the first light beam so how can you calculate acceleration using only one time?

s = ut + 0.5at^2 ...and you don't have both 'u' or 'a' so how do you calculate 'a'?

thanks so much

If I wanted to measure the acceleration of a falling object using light gates, how many would I need to use? Because I can't see how you would calculate it using two light gates...from my understanding (which may be completely wrong), the first light gate would start timing when the object breaks the light beam and when the second light beam is broken, you would get the time taken for the object to fall through the distance between the two light gates. But the object would have an initial speed 'u' when it breaks the first light beam so how can you calculate acceleration using only one time?

s = ut + 0.5at^2 ...and you don't have both 'u' or 'a' so how do you calculate 'a'?

thanks so much

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#2

If you knew the length of the object, the time between the gates being tripped and the time each gate was tripped for then you could calculate it that way I guess.... (Get velocities at each gate, then use the time between both of them being triggered)

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(Original post by

If you knew the length of the object, the time between the gates being tripped and the time each gate was tripped for then you could calculate it that way I guess.... (Get velocities at each gate, then use the time between both of them being triggered)

**NoHands**)If you knew the length of the object, the time between the gates being tripped and the time each gate was tripped for then you could calculate it that way I guess.... (Get velocities at each gate, then use the time between both of them being triggered)

**while**it was tripping the light gates too, right?

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#4

(Original post by

but the velocities you would get would be average velocities because it would be accelerating

**alaskadish**)but the velocities you would get would be average velocities because it would be accelerating

**while**it was tripping the light gates too, right?
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#5

(Original post by

Hi!

If I wanted to measure the acceleration of a falling object using light gates, how many would I need to use? Because I can't see how you would calculate it using two light gates...from my understanding (which may be completely wrong), the first light gate would start timing when the object breaks the light beam and when the second light beam is broken, you would get the time taken for the object to fall through the distance between the two light gates. But the object would have an initial speed 'u' when it breaks the first light beam so how can you calculate acceleration using only one time?

s = ut + 0.5at^2 ...and you don't have both 'u' or 'a' so how do you calculate 'a'?

thanks so much

**alaskadish**)Hi!

If I wanted to measure the acceleration of a falling object using light gates, how many would I need to use? Because I can't see how you would calculate it using two light gates...from my understanding (which may be completely wrong), the first light gate would start timing when the object breaks the light beam and when the second light beam is broken, you would get the time taken for the object to fall through the distance between the two light gates. But the object would have an initial speed 'u' when it breaks the first light beam so how can you calculate acceleration using only one time?

s = ut + 0.5at^2 ...and you don't have both 'u' or 'a' so how do you calculate 'a'?

thanks so much

You need the distance between the gates, say S1 between the 1st two and S2 between the 2nd and 3rd.

Time between 1st two is t1 and t2; and between 2nd and 3rd is t2 and t3

Those are your measurements.

Here is a graph of what's happening. It's a velocity-time graph showing the times and distances travelled (area under). It also shows the velocities at the start and end of each section. V1 V2 and V3

You know that the distance travelled in a section is average velocity times time, and that average velocities are (V1 + V2) / 2 in S1 and (V2 + V3) / 2 in S2

The times are t2 - t1 and t3 - t2

Average velocity = distance / time

So

**average velocity in first section = S1 / (t2 - t1)**

**Average velocity in 2nd section = S2 / (t3 - t2)**

These are all things you have measured in the experiment.

The average velocity in the 1st section is marked as U on the graph. It is the velocity at exactly half the time interval between t1 and t2. (For uniform acceleration)

The average velocity in the 2nd section is marked as V on the graph and occurs at exactly half the time interval there.

These are marked with the dotted lines.

So to find the acceleration you need to apply V = U + gT

g = (V - U) / T [acceleration is change in velocity divided by time taken]

T is the time interval between those two halfway points. It's marked on the graph. Can you work out what it is?

If so, you can now find the acceleration because you can calculate V and U from your measurements as explained above in bold type.

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(Original post by

Here's one way of doing it. You need 3 gates.

You need the distance between the gates, say S1 between the 1st two and S2 between the 2nd and 3rd.

Time between 1st two is t1 and between 2nd and 3rd is t2

Those are your measurements.

Here is a graph of what's happening. It's a velocity-time graph showing the times and distances travelled (area under). It also shows the velocities at the start and end of each section. V1 V2 and V3

You know that the distance travelled in a section is average velocity times time, and that average velocities are (V1 + V2) / 2 in S1 and (V2 + V3) / 2 in S2

The times are t2 - t1 and t3 - t2

Average velocity = distance / time

So

These are all things you have measured in the experiment.

The average velocity in the 1st section is marked as U on the graph. It is the velocity at exactly half the time interval between t1 and t2. (For uniform acceleration)

The average velocity in the 2nd section is marked as V on the graph and occurs at exactly half the time interval there.

These are marked with the dotted lines.

So to find the acceleration you need to apply V = U + gT

g = (V - U) / T [acceleration is change in velocity divided by time taken]

T is the time interval between those two halfway points. It's marked on the graph. Can you work out what it is?

If so, you can now find the acceleration because you can calculate V and U from your measurements as explained above in bold type.

**Stonebridge**)Here's one way of doing it. You need 3 gates.

You need the distance between the gates, say S1 between the 1st two and S2 between the 2nd and 3rd.

Time between 1st two is t1 and between 2nd and 3rd is t2

Those are your measurements.

Here is a graph of what's happening. It's a velocity-time graph showing the times and distances travelled (area under). It also shows the velocities at the start and end of each section. V1 V2 and V3

You know that the distance travelled in a section is average velocity times time, and that average velocities are (V1 + V2) / 2 in S1 and (V2 + V3) / 2 in S2

The times are t2 - t1 and t3 - t2

Average velocity = distance / time

So

**average velocity in first section = S1 / (t2 - t1)****Average velocity in 2nd section = S2 / (t3 - t2)**These are all things you have measured in the experiment.

The average velocity in the 1st section is marked as U on the graph. It is the velocity at exactly half the time interval between t1 and t2. (For uniform acceleration)

The average velocity in the 2nd section is marked as V on the graph and occurs at exactly half the time interval there.

These are marked with the dotted lines.

So to find the acceleration you need to apply V = U + gT

g = (V - U) / T [acceleration is change in velocity divided by time taken]

T is the time interval between those two halfway points. It's marked on the graph. Can you work out what it is?

If so, you can now find the acceleration because you can calculate V and U from your measurements as explained above in bold type.

thanks so much again

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#7

(Original post by

thank you so much for all that I was wondering how to do it using two light gates because everywhere I've read it mentions two light gates and never three (including my textbook which shows a picture of an electromagnet and two light gates and says 'acceleration can be calculated using time intervals and distances') so you can't do it with two right?

thanks so much again

**alaskadish**)thank you so much for all that I was wondering how to do it using two light gates because everywhere I've read it mentions two light gates and never three (including my textbook which shows a picture of an electromagnet and two light gates and says 'acceleration can be calculated using time intervals and distances') so you can't do it with two right?

thanks so much again

**You can do it with 2 gates if**

a) the object starts from rest at the 1st gate. Then you know that u=0 and you measure s and t. [s = ut + Â½gtÂ² would give the value of g]

You need 3 values to calculate the 4th.

If the object was not at rest when it passed the 1st gate (of two) then you only have 2 known values, s and t; with u, v and g unknown.

or

b) you use 2 gates and measure the distance between the point the object starts to fall and the 1st gate. This gives you S1 and you know that u=0 in the 1st section. It's more or less the same as my method. Possibly a bit easier mathematically.

In the graph v1 and t1 would be zero at the origin. You would need to be able to start the timer when the object starts to fall.

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#8

**alaskadish**)

but the velocities you would get would be average velocities because it would be accelerating

**while**it was tripping the light gates too, right?

Stomebridge's method is much better, but not every school has enough light gates to give each group three!

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(Original post by

a) the object starts from rest at the 1st gate. Then you know that u=0 and you measure s and t. [s = ut + Â½gtÂ² would give the value of g]

You need 3 values to calculate the 4th.

If the object was not at rest when it passed the 1st gate (of two) then you only have 2 known values, s and t; with u, v and g unknown.

or

b) you use 2 gates and measure the distance between the point the object starts to fall and the 1st gate. This gives you S1 and you know that u=0 in the 1st section. It's more or less the same as my method. Possibly a bit easier mathematically.

In the graph v1 and t1 would be zero at the origin.

**Stonebridge**)**You can do it with 2 gates if**a) the object starts from rest at the 1st gate. Then you know that u=0 and you measure s and t. [s = ut + Â½gtÂ² would give the value of g]

You need 3 values to calculate the 4th.

If the object was not at rest when it passed the 1st gate (of two) then you only have 2 known values, s and t; with u, v and g unknown.

or

b) you use 2 gates and measure the distance between the point the object starts to fall and the 1st gate. This gives you S1 and you know that u=0 in the 1st section. It's more or less the same as my method. Possibly a bit easier mathematically.

In the graph v1 and t1 would be zero at the origin.

**You would need to be able to start the timer when the object starts to fall.**
(Original post by

This is quite right, but you can still get a pretty good value if you ignore that effect. The amount of error introduced will depend upon the length of the object as it passes through the light gates, amongst other things. Doing it this way still helps you to get the hang of the theory, gives you a broadly acceptable value, and is good for starting a discussion about the shortcomings of the experiment (the main one of which you have just identified!).

Stomebridge's method is much better, but not every school has enough light gates to give each group three!

**Pangol**)This is quite right, but you can still get a pretty good value if you ignore that effect. The amount of error introduced will depend upon the length of the object as it passes through the light gates, amongst other things. Doing it this way still helps you to get the hang of the theory, gives you a broadly acceptable value, and is good for starting a discussion about the shortcomings of the experiment (the main one of which you have just identified!).

Stomebridge's method is much better, but not every school has enough light gates to give each group three!

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#10

Without seeing the equipment you have it's difficult to say.

One way uses a ball bearing held up by an electromagnet. Switching off the power to the electromagnet starts the timer and the ball begins to drop.

You then need the time as it passes a point on its way down. That would actually need only one "gate" but would need the timer to be triggered by the switch on the magnet and still log 2 times.

The 3 gate method works fine if you have a timer that can log the 3 times from the 3 gates. You can drop the ball from anywhere. It doesn't even need to have been at rest at the start.

With 2 gates (or one and a switch) you need to be able to log the time when the object

I don't know how your timer/equipment is set up so it's difficult to be more precise.

One way uses a ball bearing held up by an electromagnet. Switching off the power to the electromagnet starts the timer and the ball begins to drop.

You then need the time as it passes a point on its way down. That would actually need only one "gate" but would need the timer to be triggered by the switch on the magnet and still log 2 times.

The 3 gate method works fine if you have a timer that can log the 3 times from the 3 gates. You can drop the ball from anywhere. It doesn't even need to have been at rest at the start.

With 2 gates (or one and a switch) you need to be able to log the time when the object

**starts**to fall**from rest**.I don't know how your timer/equipment is set up so it's difficult to be more precise.

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(Original post by

Without seeing the equipment you have it's difficult to say.

The 3 gate method works fine if you have a timer that can log the 3 times from the 3 gates. You can drop the ball from anywhere. It doesn't even need to have been at rest at the start.

With 2 gates (or one and a switch) you need to be able to log the time when the object

I don't know how your timer/equipment is set up so it's difficult to be more precise.

**Stonebridge**)Without seeing the equipment you have it's difficult to say.

**One way uses a ball bearing held up by an electromagnet. Switching off the power to the electromagnet starts the timer and the ball begins to drop.**

You then need the time as it passes a point on its way down. That would actually need only one "gate" but would need the timer to be triggered by the switch on the magnet and still log 2 times.You then need the time as it passes a point on its way down. That would actually need only one "gate" but would need the timer to be triggered by the switch on the magnet and still log 2 times.

The 3 gate method works fine if you have a timer that can log the 3 times from the 3 gates. You can drop the ball from anywhere. It doesn't even need to have been at rest at the start.

With 2 gates (or one and a switch) you need to be able to log the time when the object

**starts**to fall**from rest**.I don't know how your timer/equipment is set up so it's difficult to be more precise.

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#12

I have used two light gates and a card on the trolley with two slots , one at each end . with all the dimensions known, then it calculates the velocities at each end and the acceleration . This was using Easysense equipment and was about 10 years ago. Not sure what is available now because I am still trying to get hold of old kit for cost reasons!

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