Question

From the data:

Experiment 1, when the hanging mass is 100g, what was the velocity through the first photogate?

-281442181 Lab Manual Background ProceduresLab Notes Lab Notes EXPERIMENT 1 99.0 m(subscript 1) airtrack cart starting position 10 cm photogate 1- 50 cm pg2- 150 cm 1. 100 g weight, mass-210, photogate 1- 2.03, 2 3.74 2. 300 g weight, mass 210, pg 1- 2.48, pg 2 4.58 3. 600 g weight, mass 210, pg 1- 2.66, pg 2- 4.90 EXPERIMENT 2 mass of aircarts 99.0+50# 149.0 g 1. hanging weight 300 g, pg 1- 2.35, pg 2- 4.32 2. mass of aircart- 99.0+200-299.0 g, pg 1-2.04, pg 2- 3.75 3, " " " " 99.0+300. 399.0 g, pg 1-1.89, pg 2# 3.48 Save Notes

Answer 1

The force on an aircraft will be equal to the tension force, T.

_$\sum$T = m_hanging g - m_hanging a = m_aircraft a

m_hanging g = (m_aircraft + m_hanging) a

a = m_hanging g / (m_aircraft + m_hanging)

a = (0.1 kg) (9.8 m/s²) / [(0.21 kg) + (0.1 kg)]

a = (0.98 N) / (0.31 kg)

a = 3.16 m/s²

When the hanging mass is 100 g, what was the velocity through a first photogate?

using a formula, we have

a = v² / 2 d $\Leftrightarrow$ v = _$\sqrt{}$ 2 a d

v = _$\sqrt{}$ 2 (3.16 m/s²) (0.0203 m)

v = _$\sqrt{}$ 0.128296 m²/s²

v = 0.358 m/s