#$&*
course Phy 241
Using a ramp, some dominoes and a ball:Determine the increase in acceleration of a ball down the ramp, per domino added to the stack at one end of the ramp.
________________________________________
There are two ways to find the acceleration of the ball:
• You can attempt to time the ball as it rolls down the ramp.
• You can observe the horizontal distance traveled by the ball after leaving the end of the ramp, along with the assumption that the ball requires 0.4 seconds to fall to the floor.
You should do this both ways, and address the question of which way is more accurate.
Give a synopsis of your data below. Your data is what you actually observed, not what you calculated from what you observed:
**** 60 cm ramp
#dom, total time, time on ramp, how far out
3 dom, 2.5sec, 1.9sec, 17 cm
@& A ball that travels 17 cm in the horizontal direction during a fall that lasts .4 seconds has a contant horizontal velocity of about 43 cm/s.
That is pretty much equal to the ball's speed at the end of the ramp.
If the ball travels 60 cm from rest and ends up at 43 cm/s, its average veocity is about 22 cm/s so it takes a bit less than 3 sec to travel down the ramp. Velocity changes by 43 cm/s in about 3 sec so acceleration would be about 14 cm/s^2.
That's reasonably consistent with your apparently reported 2.5 sec.
It's not completely clear what the two time intervals you report for each ramp correspond to.
You don't explain how you got most of the results you report in this document, but most are not consistent with this analysis.
*@
4 dom, 2.3 sec, 1.7 sec, 19.5 cm
5 dom, 1.9 sec, 1.5 sec, 23.5 cm
#$&*
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 dom = 31.6 cm/s
4 dom= 35.3cm/s
5 dom= 40 cm/s
#$&*
• What therefore were the initial and final velocity of the ball on the ramp?
****
Initial was 0 for all of them so the final velocity is
3 dom= 31.6 cm/s
4dom=35.4 cm/s
5 dom= 40 cm/s
#$&*
@& The average of 31.6 cm/s and 0 cm/s is 15.8 cm/s. So if the final velocity is 31.6 cm/s the average velocity must have been 15.8 cm/s.
However the average velocity you reported is 31.6 cm/s.
*@
• What therefore was the average rate at which the velocity of the ball changed while on the ramp?
****
3 dom =8.3 cm/s^2
4dom = 10.4cm/s^2
5 dom= 13.3 cm/s^2
#$&*
To find the acceleration using the horizontal distance traveled by the ball after leaving the ramp you need to answer the following:
• What was the horizontal distance traveled by the ball while falling?
****
3 dom= 17 cm
4 dom = 19.5 cm
5 dom= 23.5cm
#$&*
• What therefore was the average horizontal velocity of the falling ball?
****
The average horizontal velocity is 4.9 cm/ dominoe
#$&*
• Assuming the horizontal velocity to have been unchanging during the fall, what was the ball's speed as it left the ramp?
****
It would the final velocity
3 dom= 31.6 cm/s
4dom=35.4 cm/s
5 dom= 40 cm/s
#$&*
• What therefore was the ball's average velocity while rolling down the ramp?
****
3 dom =15.8cm/s
4 dom= 17.7 cm/s
5 dom= 20 cm/s
#$&*
• How long would it have then taken the ball to roll the length of the ramp?
Take the length of the ramp which is 60 cm and divide it by the ave velocity
3 dom = 3.79 s
4 dom= 3.39 s
5 dom= 3 s
#$&*
• What was the change in the ball's velocity along the ramp?
****
3 dom= 6.45cm/s^2
4 dom= 7.22 cm/s^2
5 dom= 8.16 cm/s^2
#$&*
• What was the average rate of change of the ball's velocity with respect to clock time, for the interval between release and reaching the end of the ramp?
****
3 dom= 3.23 cm/s^2
4 dom= 3.61 cm/s^2
5 dom= 4.08 cm/s^2
#$&*
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?
****
I feel that the data had a less percent error when we measure where it hit after the ball left the ramp.
#$&*
Which method do you think gave you the more accurate result for the change in acceleration per added domino?
****
the distance the ball traveled after leaving the ramp
#$&*
University Physics Only
What was the average rate of change of acceleration with respect to ramp slope?
****
I did not calculate the slope of the ramp but it I had know the height of the ramp I could have calculated the slope then I would have took the average rate of change of velocity and divided it by the slope of the ramp
#$&*
The ball came off the ramp at about 4 degrees below horizontal. If it required 1.8 seconds to travel down the ramp, what would have been the horizontal and vertical components of its velocity at the instant it left the ramp? (Note that if theta is the angle of the velocity vector, as measured counterclockwise from the positive x axis, and the speed is v, then the x and y components of the velocity are v_x = v cos(theta) and v_y = v sin(theta) ).
****
If it took 1.8 s to travel on a 60 cm long ramp the average velocity is 33.3 cm/s so the final would be 66.6cm/s then you would do 66.6 cm/s *cos(356)= 66.4 cm/s in the horizontal and 66.6 cm/s * sin(356)= -4.65 cm/s in the vertical direction.
#$&*
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?
****
Vf^2= (0)^2+2(9.8m/s^2)(.9m)=sqrt(17.64m^2/s^2)=4.2m/s
#$&*
• How long did it take the ball to reach the floor?
****
dT=.9m/4.2m/s=.214 seconds
#$&*
@& The time to the floor is displacement / average velocity, not displacement / final velocity.*@
• How far did it therefore travel in the horizontal direction during its fall?
****
.214 s*66.4cm/s=14.21cm
#$&*
Hard question: If the ball in the preceding traveled 27 cm in the horizontal direction, what must have been its actual speed at the end of the ramp?
****
60cm+27cm=87cm so the final velocity is 96.66cm/s then divide the final velocity by 1.8 s =53.7 cm/s^2
#$&*
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?
****
I took 1.08s/.25 = .27s then I took 3cm/.27s=11.1 cm/s is the average velocity so the acceleration 41.1 cm/s^2
#$&*
@& You need to divide change in velocity by time interval, not average velocity.*@
• If the ball is to collide with the pendulum as the pendulum passes through its equilibrium point for the second time, at what position along the ramp should the pendulum be located?
****
If 3 cm was ¼ of a cycle then it would ¾ of a cycle so it would need to be placed at 9 cm
#$&*
@& The ball is speeding up, so if the time interval is 3 times as long the distance interval will be more than 3 times as great.*@
@& I've inserted some notes. You're going to need some revisions, and you do need to indicate at least a brief sample of how you did each type of calculation.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@