Checking Newton Essay

Published: 2020-01-28 09:51:14
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Category: Newton

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In this experiment I will make measurements on a two pulley system and use them to check Newtons 2nd law:The rate of momentum of a body Is directly proportional to the external, Resultant force acting upon it. The change in momentum takes place in the Direction of that force. The apparatus is set up as shown in the diagram below: The masses m1 and m2 are initially set at 400 grams each. The ruler is set vertically. I measured and recorded the distance (S).

There must be the same friction at the pulleys; in order to comply with this took added 10 grams masses to m1 until the two masses jus begin to move. I recorded the extra masses m0 added, and the total mass M of m1 and m2. I then held m1 by hand, added one 10g mass, and released it. My partner at that exact moment started the stopwatch. When the m1 hit the soft material the stopwatch was stopped. I recorded the time taken in the table below and repeated this a further one more time so I could finally take an average reading.

Extra mass added to m1 Time 1 s Time 2 s Mean Time Timei 1/(m Kgs-1 0. 020 3. 0 4. 18 3. 59 12. 8 50 0. 040 1. 60 2. 14 1. 87 3. 49 25 0. 060 1. 62 1. 45 1. 53 2. 34 16. 6 0. 080 1. 19 1. 30 1. 24 1. 53 12. 5 0. 100 1. 19 1. 0 1. 09 1. 18 10 On then ext page is the graph that I plotted with t on t on the y- axis against 1/(m on the x- axis.

Theory: The string is assumed inextensible so then the tension is the same at all points. For the downward motion of the mass m2 I applied F=ma.

M1g T = m1a¦ 1 For the upward motion of m2 T m1g = m1a¦ 2 Adding 1 and 2 cancels out T M2g m1g = m2a+m1a (m2 m1)g = (m2+m1) a There for a = m2-m1/m2+m1 g¦ 3 Now s=ut + at U=0, and There for s=1/2at A=2s/ti??¦ 4 Substituting 4 in 3 for a 2s/t = (m/M g ( 2Ms= t. g. (m This can be written asti = 2Ms/g . 1/(mm, s and g are constants. So a graph of ti?? against 1/(m should result in a straight line through the origin.

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