Read each question carefully. You must show all of your work in a neat manner. B

Question

Read each question carefully. You must show all of your work in a neat manner. Burt and Ernie are rolling a ball along inclined and level tracks, as was demonstrated in class, while ignoring friction. The distance, x, is measured along the track and is increasing toward the right. The track has four sections labeled A, B, C and Das shown in the figure. The sloped sections, B and D, form the same angle with the horizontal. The track is not drawn to scale. There are measured distances and times for the ball along the different sections of track shown in the figure, e.g. on section A it took the ball 3.00s to traverse 150cm. Calculate the change of velocity that the ball undergoes during the time period when it enters section B until it leaves that section. What is the acceleration of the ball during the time period found in (1)? What is the length of section B? What is the acceleration on section D? Briefly explain your reasoning. How far up section D would the ball roll momentarily came to rest? How much time does it take the ball to go up section D and come to rest?

Explanation / Answer

Q1.

velocity in track A=150 cm/3 sec=50 cm/s

velocity in track c=180 cm/15 s=12 cm/s

so change in velocity=velocity in track C-velocity in track A=12-50 =-38 cm/s

Q2.

acceleration=change in velocity/time

=-38 cm/s /(2 seconds)

=-19 cm/s^2

Q3.initial speed at the start of track B=50 cm/s

final speed at the end of track B=12 cm/s

acceleration=-19 cm/s^2

using the formula:

final speed^2-initial speed^2=2*acceleration*distance

==>length of track B=(12^2-50^2)/(2*(-19))=62 cm

Q4. as acceleration is due to gravity only and the inclination of both track B and track D are same

acceleration on D=acceleration on B=-19 cm/s^2

Q5.

initial speed at the beginning of D=12 cm/s^2

acceleration=-19 cm/s^2

final speed when the ball comes to rest=0 cm/s

so using the formula:
final speed^2-initial speed^2=2*acceleration*distance

==>0^2-12^2=-2*19*distance

==>distance=3.78947 cm

so the ball will roll till 3.78947 cm

Q6. time taken=(final speed-initial speed)/acceleration

=(0-12)/(-19)=0.63157 seconds

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