Checking Newton’s 2nd law of motion

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

Time²

s²

/(m

Kgs-1

0.020

3.0

4.18

3.59

2.8

50

0.040

.60

2.14

.87

3.49

25

0.060

.62

.45

.53

2.34

6.6

0.080

.19

.30

.24

.53

2.5

0.100

.19

.0

.09

.18

0

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 + at²

U=0, and

There for s=1/2at²

A=2s/t²...4

Substituting 4 in 3 for a

2s/t² = (m/M g ( 2Ms= t².g.(m

This can be written as

t² = 2Ms/g . 1/(m

m, s and g are constants. So a graph of t² against 1/(m should result in a straight line through the origin.