Question

(1) An air track that is 2 meters long is raised by 2 cm on one end. What vertical distance does an air cart move if the air cart moves from the 1 meter mark to the 2 meter mark? 1 cm

(2) If the above air cart starts from rest, what would you expect its velocity to be 0.5 meters along the air track? Use the equations of conservation of energy. Assume the air track is frictionless.

(3) Is the result of Question (2) consistent with what you would expect based on the equations of kinematics? Solve for the expected velocity based on the equations of kinematics and compare.

Answer 1

1)

L = length of the air track = 2 m

h = height by which one end is raised = 2 cm = 0.02 m

= angle made by the horizontal with the air track

From the diagram

Sin = h/L

Sin = 0.02/2

= 0.573 deg

In triangle ABC

y = vertical distance moved by cart

l = length of track traveled = 1 m

Sin = y/l

0.02/2 = y/1

y = 0.01 m

y = 1 cm

b)

l = length of track traveled = 0.5 m

y = height dropped

Sin = y/l

0.02/2 = y/0.5

y = 0.005 m

v = speed gained

Using conservation of energy

Kinetic energy gained = Potential energy lost

(0.5) m v² = mgy

v = sqrt(2gy)

v = sqrt(2(9.8)(0.005))

v = 0.313 m/s

c)

consider motion parallel to incline surface

a = acceleration parallel to incline = g Sin = (9.8) (0.02/2) = 0.098 m/s²

x = distance traveled parallel to incline surface = 0.5 m

v_o = initial velocity = 0 m/s

v = final velocity = ?

Using the kinematics equation

v² = v_o² + 2 a x

v² = 0² + 2 (0.098) (0.5)

v = 0.313 m/s