#$&*
course Phy 201
Sept. 18, 7:50
Using a ramp, some dominoes and a ball:Determine the increase in acceleration of a ball down the ramp, per domino added to the stack at one end of the ramp.
--------------------------------------------------------------------------------
There are two ways to find the acceleration of the ball:
You can attempt to time the ball as it rolls down the ramp.
You can observe the horizontal distance traveled by the ball after leaving the end of the ramp, along with the assumption that the ball requires 0.4 seconds to fall to
the floor.
You should do this both ways, and address the question of which way is more accurate.
Give a synopsis of your data below. Your data is what you actually observed, not what you calculated from what you observed:
****
Ramp with 3 dominos:
ramp length: 60 cm
took 2.35 sec for the steel ball to roll down it
after leaving the ramp, continued horizonally about 17 cm
Ramp with 4 dominos:
ramp length: 60 cm
took 1.9 sec for the steel ball to roll down it
after leaving the ramp, continued horizontally about 22 cm
Both:
started from rest
was clearly getting faster
from the table it took 0.35 sec to reach the floor with a vertical drop
#$&*
To find the acceleration using the time required to roll down the ramp from rest you need to answer the following questions:
What was the average velocity of the ball on the ramp?
****
3 dominos:
60 cm / 2.35 sec = 25.53 com / sec
4 dominos:
60 cm / 1.9 cm / sec = 31.58 cm / sec
#$&*
What therefore were the initial and final velocity of the ball on the ramp?
****
3 dominos:
V_o = 0 cm / sec
V_f = 51.06 cm / sec
4 dominos:
V_o = 0 cm / sec
V_f = 63.16 cm / sec
#$&*
What therefore was the average rate at which the velocity of the ball changed while on the ramp?
****
3 dominos:
a_ave = (51.06 cm/sec) / (2.35 sec) = 21.72 cm/sec^2
4 dominos:
a_ave = (63.16 cm/sec) / (1.9 sec) = 33.24 cm/sec^2
#$&*
To find the acceleration using the horizontal distance traveled by the ball after leaving the ramp you need to answer the following:
What was the horizontal distance traveled by the ball while falling?
****
3 dominos:
calculated: V_f * t = 51.06 cm/sec * 0.35 sec = 17.871 cm
physical: 17 cm
4 dominos:
calculated: V_f * t = 63.16 cm/sec * 0.35 sec = 22.106 cm
physical: 22 cm
#$&*
What therefore was the average horizontal velocity of the falling ball?
****
3 dominos:
V_ave = 17.871 cm / 0.35 sec = 51.06 cm/sec
4 dominos:
22.106 cm / 0.35 sec = 63.16 cm/sec
#$&*
Assuming the horizontal velocity to have been unchanging during the fall, what was the ball's speed as it left the ramp?
****
3 dominos:
51.06 cm/sec
4 dominos:
63.16 cm/sec
#$&*
What therefore was the ball's average velocity while rolling down the ramp?
****
3 dominos:
V_ave = 60 cm / 2.35 sec = 25.53 com / sec
4 dominos:
V_ave = 60 cm / 1.9 cm / sec = 31.58 cm / sec
#$&*
How long would it have then taken the ball to roll the length of the ramp?
****
it took...
with 3 dominos: 2.35 sec
and
with 4 dominos: 1.9 sec
#$&*
What was the change in the ball's velocity along the ramp?
****
3 dominos:
51.06 cm/sec
4 dominos: 63.16 cm/sec
#$&*
What was the average rate of change of the ball's velocity with respect to clock time, for the interval between release and reaching the end of the ramp?
****
3 dominos:
a = (51.06 cm/sec) / (2.35 sec) = 21.73cm/sec^2
4 dominos:
a = (63.16 cm/sec) / (1.9 sec) = 33.24 cm/sec^2
#$&*
Which do you think you were able to determine with less percent error, the difference in the times required for the ball to roll down various ramps, or the differences
in the horizontal distance traveled by the ball?
****
the difference in the horizontal distance traveled by the ball
~carbon paper would make it even more obvious than it already is
#$&*
Which method do you think gave you the more accurate result for the change in acceleration per added domino?
****
all of my calculated answers were pretty close to the observable ones
logic, however, tells you that the visual measurements are more reliable and accurate because of human error in anticipation and reaction time
#$&*
University Physics Only
What was the average rate of change of acceleration with respect to ramp slope?
****
#$&*
The ball came off the ramp at about 4 degrees below horizontal. If it required 1.8 seconds to travel down the ramp, what would have been the horizontal and vertical
components of its velocity at the instant it left the ramp? (Note that if theta is the angle of the velocity vector, as measured counterclockwise from the positive x
axis, and the speed is v, then the x and y components of the velocity are v_x = v cos(theta) and v_y = v sin(theta) ).
****
#$&*
The vertical motion of the ball is characterized by acceleration 9.8 m/s^2 = 980 cm/s^2 in the downward direction. The horizontal motion of the ball is characterized
by acceleration zero.
What therefore was the ball's vertical velocity after falling the .9 meters to the floor?
****
#$&*
How long did it take the ball to reach the floor?
****
#$&*
How far did it therefore travel in the horizontal direction during its fall?
****
#$&*
Hard question: If the ball in the preceding traveled 27 cm in the horizontal direction, what must have been its actual speed at the end of the ramp?
****
#$&*
A pendulum whose actual length is 30 cm has a period of about 1.08 seconds. The period is the time required for the pendulum to return to its point of release, after
being released from rest.
If the ball is released at the same instant as the pendulum, and after moving 3 centimeters collides with the pendulum at the instant the pendulum first passes through
its equilibrium position, then what is the acceleration of the ball along the ramp?
****
#$&*
If the ball is to collide with the pendulum as the pendulum passes through its equilibrium point for the second time, at what position along the ramp should the
pendulum be located?
****
#$&*
"
&#Very good responses. Let me know if you have questions. &#