Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Does anyone know how to calculate how many times a ball will bounce if dropped from some height? Information: Restitution = 0.85 Mass = 0.7 kg Drop Height = 10 m Gravity = 9.81 m/s Ground is Horizontal. No air resistance.

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
velocity of separation = e (velocity of approach)
A little more explanation please?
|dw:1347425035783:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

does it make sense?
|dw:1347424692991:dw|
why are you subtracting 0?
Velocity of approach = V' - zero...which is velocity of ground
sorry...Velocity of approach = V - zero...which is velocity of ground
ball is colliding with the ground..and use this velocity of separation = e (velocity of approach)
similarly velocity of separation = V'-0
okay. so e is the math constant.. what am i solving for?
so 1st calculate V, the velocity of ball from which it hits the ground 1st time..
no...e is the coefficient of restitution...and for ur problem e =0.85(given)
Potential Energy completely becomes Kinetic Energy just before collision \[E_{system}=0+mgh=(0.7)(9.81)(10)=68.67\]\[E_{kinetic}=E_{system}\]\[v_{before}=\sqrt{2E_{kinetic} \over m}=14.01[ms^{-1}]\]\[\mu_r = {E_{kin. after} \over E_{kin.before}}\]\[E_{kin. after}=\mu E_{kin. before}=(0.85)(68.67)=58.67[J]\]\[v_{after}=\sqrt{2E_{kinetic} \over m}=12.91[ms^{-1}]\] I have both the velocities. Before and after the bounce.
but see 4th line in ur solution..
The ball will bounce an infinite number of times, but the process will have a finite duration.
velocity of separation = e (velocity of approach) it does not mean that... Energy after collision =e( energy before collision)?
That's one of the formula we were given.. that restitution is energy after divided by energy before. so i just used that to calculate energy after collision and then the velocity.
But the correct formula is for velocity not for energy velocity of separation = e( velocity of approach)
http://scienceworld.wolfram.com/physics/CoefficientofRestitution.html check this
So our professor gave us the wrong equation... awesome.
where are you from?
I'm studying in New Zealand.
ok
anyway...u can solve this problem by using correct formula
Wait, what do I do now that I've got the velocity before and after collision? vel approach =14.01 vel separation = 11.91
yes...If u have done correctly... Better use kinematics relation..and use symbols dont plug in the value now..just use h,g and all..then u'll get better understanding of this problem
|dw:1347427306605:dw|
t' = time of reaching at the peak after 1st collision
does it make sense?
So far yes.
|dw:1347427668995:dw|
good...:)
e, e^2, e^3...will keep decreasing on...
general formula \[v_n=\mu_r^n \sqrt{2gh}\]\[t_n=\mu_r^n \sqrt{2h \over g}\] I need to find n.
yes...now u can tell the answer :)
the ball will stop when v(n)=0...right?
well not really. because I want to find when it stops bouncing right? so v=0 or t=0. ends up just being n=0.. unless i'm failing at algebra
and restitution <1..so tell me the answer?
the ball will stop when v(n)=0
v(n)=0 \[0=\mu_r^n \sqrt{2gh}\] Right?
but now the question is when v(n) will be zero?
yes...are u noticing something? e, e^2, e^3... with each v(n) for n=1,2,3,4....
and e=0.85 which is less than 1
now tell me... e^n will approach zero when n approaches....?
infinity
yes...perfectly right:)
so the ball will bounce infinite times...
well i got that answer like 3 hours ago... but i figured that there would be an actual answer.
ok
actually after each collision h keeps on decreasing...and theoretically h will approach zero after infinite bounces..
Well, thanks for the help! Showed me one of the equations was wrong which let me fix the first bit of the question. I guess I'm happy with infinite bounces for now. But I'll have a talk with my professor next week and see what he says.
but practically ball can't go up to a size of atom after some bounces for large n
as according to ur problem..u r starting with h=10m...later on ...h will reach in nanometer scale...and it's not practical that ball will bounce up to some nm height.
What you CAN work out is not number of bounces (infinite), but total time taken.
@Vincent-Lyon.Fr Out of curiosity how would you go about doing that?

Not the answer you are looking for?

Search for more explanations.

Ask your own question