I will investigate the change of velocity and acceleration of a laterally moving object attached by a string and pulley to a dropped object, when the mass of the dropped object is changed.

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Investigation on converting Gravitational Potential energy into horizontal and vertical Kinetic Energy

Investigation on converting Gravitational Potential energy into horizontal and vertical Kinetic Energy

 - Leszek Swirski


P – Planning

Aim

I will investigate the change of velocity and acceleration of a laterally moving object attached by a string and pulley to a dropped object, when the mass of the dropped object is changed.

Method

Apparatus

  • Trolley
  • Piece of Card
  • Runway
  • Weights
  • 2 Light Gates
  • Weights
  • Pulley
  • Computer
  • LogIt 9000
  • Insight Timing
  • Appropriate cables

Diagram

Set up procedure

This is how I will set up the experiment:

  1. Set up the apparatus as above, with Light gate A at 45 cm from where the centre of the piece of card (not trolley) will start to move from, and Light gate B at 45 cm further.
  2. Make sure that there is enough space between the pulley and light gate B for the piece of card to go through the light gate. If not, move the starting point of the trolley back, and the light gates accordingly.
  3. Plug Light gate A and Light gate B into the LogIt, and plug that into the computer.
  4. Start up Insight Timing on the computer

Then I am ready to start

Fair Test

I will make this a fair test by limiting the key factors:

  • Weight of trolley – I will use the same trolley every time to ensure the same weight
  • Distance travelled – I will keep this constant by releasing the weights from the same height and releasing the trolley form the same place every time. I will also not move the light gates
  • Friction/Air Resistance – Unfortunately, these cannot be avoided, however they will be minimal due to the equipment used, and will be relatively constant as I will not change apparatus.

Safety

There is little safety to be considered. No harmful substances are being used, neither are flames or solvents. I will need to take precautions when increasing the dropping mass, and make sure that all the weights are securely fixed. The main point would be to stop/catch the accelerated trolley before it falls off, especially at higher speeds.

Execution Procedure

This is how I will execute the procedure

  1. The starting point should be the centre of the piece of card. I will have 10g as the dropping weight
  2. I will start recording on Insight Timing, and let the weight drop, making the trolley and, more importantly, the piece of card move.
  3. I will make sure someone catches the trolley before it falls off (see safety)
  4. I will stop insight timing and move the trolley back to the start position
  5. I will then add another 10g to the weight and repeat from 2. until I have measure the velocity at a mass of 100g.

I will repeat this procedure two more times, so that I have 3 repeats

Theory

There are three points to be considered in the theory. The first is how the mass should affect the velocity. The second is how the height should affect the velocity. The third is how the mass will affect the acceleration.

First let me discuss the first point. This test is based on converting gravitational potential energy into kinetic energy.

Gravitational potential energy (g.p.e.) is mgh, and kinetic energy (k.e.) is ½mv2.

If we consider the mass of the trolley to be m1, and the dropping mass to be m2, this gives us g.p.e. as m2gh, as only the dropping weight has g.p.e. The letter ‘g’ is the gravitational constant, and is approximated as 9.81. The letter ‘h’ is the height. Equally, k.e. is ½(m1+ m2)v2, as both the dropping mass and the trolley will move.

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By Law of Conservation of Energy, energy can neither be created nor can it be destroyed, however it can be converted from one form to another. Therefore, if the g.p.e. is not ‘wasted’ in any other way (e.g. friction converting it into heat energy), it must all be converted into k.e.

This means that g.p.e. = k.e., therefore:

m2gh = ½(m1+ m2)v2

If I didn’t have to consider m1, this could be re-written as:

v2 = 2m2gh

       m2

This would mean that the m2s will cancel out, proving that the mass of an object has no effect on ...

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