Making sense of data - Stopping distances

Analysis:

The following is the data given.

Using simple geometry, I can work out the corresponding height of a particular distance up slope.                                              860mm

                                        172mm                   d

e.g. When d = 800mm

 =

              h = 160mm

Kinetic energy = mv2 (where m is the mass in kg, v is the velocity in ms-1)

Potential energy = mgh (where m is the mass in kg again, g is the gravitation force in

                    and h is the height in m)

Kinetic energy = Potential energy (with the assumption that all energy has conserved during the whole experiment, and that the potential energy has fully transferred into kinetic energy)

mv2 = mgh

   v12 = 2gh  (in this case, v is the velocity of the ball at the bottom of the

slope noted by v1)

e.g. when d = 800mm, h = 160mm

         v12 = 2 x 10N/kg x 160mm

            = 2 x 10N/kg x (160 x 10-3)m

            = 3.136ms-1 

v2 = u2 +2as (where v is the final speed and u is the initial speed both in ms-1,

           a is the acceleration and s is the distance travelled in meter)

From the case of the speed of the small ball bearing from the bottom of the slope to the point where it stopped, the initial speed (u) or (v1 as used before) was the speed at the bottom of the slope, and the final speed (I will use v2) was the speed at the point where it stopped. Obviously when a substance stops, its speed is 0ms-1, so in this case v2 = 0. Besides, the acceleration of the ball was negative as the speed of the ball is decreasing until it stopped.

So when substitute everything that I mentioned above

0 = u2 - 2as (where a now stands for deceleration)

-u2 = -2as

u2 = 2as

3.136ms-1 = 2a x 1098.3mm

3.136ms-1 = 2a x (1098.3 x 10-3)m

     a = 1.428ms-2 

so the deceleration of the carpet to the ball is 1.428. for the calculation shown above, there is an assumption that the acceleration of the ball bearing on the rough carpet is exactly the same all the way through.

When comparing u2 = 2as with the equation of a straight line y = mx + c, I can see that u2 = 2as is a straight line equation where the y-axis is u2 instead of y, the x-axis is s instead of x, the gradient m = 2a and the y interception is at 0.

And now I can use the gradient to find the rate of deceleration from the carpet to the ball.

Gradient = 2a =

                 2a =

                 2a = 3

                  a = 1.5

So this, 1.5 ms-2, is the average deceleration the ball had from the carpet. Although it is a little bit out from 1.428 ms-2 , the deceleration of a particular case i.e. 800 mm up slope. it is acceptable because the average deceleration is calculated under the assumption of a made above.

There are several possible errors which could make this experiment not as accurate as it should be, and they are

1. the meter ruler --- the limit of accuracy was = 0.5mm, this is due to the distance apart from each unit of the meter ruler is 1mm and as the data of the stopping distance given, all decimal places have been rounded up
2. the air resistance --- although the air resistance that acts on the ball when it decelerated on the carpet would be small, it could still have an effect on the rolling ball. However, I have ignored it throughout all my calculation and analysis, so this could have made it inaccurate.
3. assumption --- during the analysis, I have made two assumption, which is the energy wasn’t lost, all potential energy was converted into kinetic energy, and the deceleration in each case is constant until the ball stopped. This again has affected the result as things are not likely to be as assumed in real life.
4. friction --- throughout the analysis, I haven’t mentioned about friction at all. As friction is proportional to force, so without taking this into account definitely lead to inaccuracy.

END