Question

A group of students performed the same "Newton's Second Law". They
obtained the following results:

m_{1}(kg) |
t_{1}(s) |
v_{1}(m/s) |
t_{2}(s) |
v_{2}(m/s) |
---|---|---|---|---|

0.050 | 1.2000 | 0.2500 | 1.7733 | 0.4378 |

0.100 | 1.2300 | 0.3240 | 1.6288 | 0.6690 |

0.150 | 1.1500 | 0.3820 | 1.4751 | 0.8485 |

0.200 | 1.1100 | 0.4240 | 1.3937 | 0.9558 |

where *m*_{1} is the value of the hanging mass
(including the mass of the hanger), *v*_{1} is the
average velocity and *t*_{1} is the time at which
*v*_{1} is the instantaneous velocity for the first
photogate, and *v*_{2} is the average velocity and
*t*_{2} is the time at which *v*_{2}
is the instantaneous velocity for the second photogate.

(a) Use Excel to construct a spreadsheet to do the following.
(You will not submit this spreadsheet. However, the results will be
needed later in this problem.)

(i) Enter the above data.

(ii) Compute the acceleration, *a*, for each trial.

(iii) Create a graph of the hanging weight
*m*_{1}*g* vs. the acceleration.

(iv) Use the trendline option to draw the best fit line for the
above data and determine the slope and *y*-intercept from
it.

(v) report your results below.

slope | = |

y-intercept |
= |

(b) Use the information you obtained from your graph to determine
the total mass of the system *M* = *m*_{1} +
*m*_{2}.

*M* =

(c) Using the information you obtained in parts (a) and (b),
predict what the value of the acceleration would be if the value of
the hanging mass were increased to *m*_{1} = 0.35
kg.

*a* =

Answer #1

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