Q1-5! TY! 7. 8. mig-T=m,a , T=m2a , (82) (8.3 where a is the acceleration of the

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Q1-5! TY!

7. 8. mig-T=m,a , T=m2a , (82) (8.3 where a is the acceleration of the masses m, and m, and T is the tension in the string, By substituting equation (8 3 equation (8.2), we eliminate the tension T, and for the acceleration a, we obtain: (84) m + m m where m-m, + m2 and F = mig . Thus the net fore which acts on the system of two masses, m = m, + 1s F = mig . In this experiment we will determine: i) acceleration a of the system by varying the mass ofth glider while keeping the net force F constant; Tacceleration a of the system by varying the net force while keeping the total mass of the syste m=m1 + m2 constant, by transferring masses fro the glider to the carriage. m2g Fig. 8.2. A free-body diagram. The acceleration of the system will be determin by measuring the instantaneous velocities of the gli at the location of photogate timers using the sa mig procedure as was established during the previous experiments. This acceleration is At (8.5)

Explanation / Answer

1. F is the resultant force on mass m2

now, T = m2*a [ where as is acceleration of the system and T is tensin in the string]

also m1*g - T = m1a

hence

m1*g - m2*a = m1*a

a = m1*g/(m1 + m2) = F/m

so in terms of experimental appratus, F = weight of hanging mass

m = total mass of hanging mass and the glider

2. adding masses from the carriage to the glider means m1 + m2 remains constant and hence we get a graph just between F and m1

3. air track is low friction experimental appratus that helps to measure acceelrations using photogates. hence we are using air track in this experiment

4. FRICITONAL FORCE OF THE PULLEY reduces acceleration of the system but since it is negligible, it can be ignored

5. T = m2*a = m1m2*a/(m1 + m2)

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