What Angles Do Colliding Pendulums Reach Post-Elastic and Inelastic Collisions?

[jg597]

Homework Statement

A small (65g) steel ball bearing is attached to a 50 cm long string as shown (link below) The ball bearing is lifted (with string stretched) to an angle of 55 degrees and released from rest.

a) when the small bearing reaches its lowest point it collides with another steel ball bearing of larger mass (150g) on a string of the same length. if the collision would have been elastic, to what maximum angle would each bearing reach after the collision.

b) if the small ball bearing actually bounces back to a maximum angle of 15 degrees( collision was NOT elastic), to what maximum angle will the larger ball bearing reach?

c) if the bearings behaved as stated in part (b) how much energy is lost during the collision.

http://s615.photobucket.com/albums/tt233/gaby596/
click on the only image there called pendulum problem thanks!

Homework Equations

this are the ones i think should be used
v1=m1-m2/m1+m2*v initial
v2=2m1Vi/(m1+m2)
and the equations for finding delta KE and Delta PE

The Attempt at a Solution

ive tried it several different ways but i keep getting the wrong answer I've seen the right answers from a answer sheet but they show no explanation on how to work it out

 

[rl.bhat]

The ball bearing is lifted (with string stretched) to an angle of 55 degrees and released from rest.
In this position what is the height of the ball from the lowest point of the pendulum?
Using kinematic equation find its velocity.

 

[jg597]

The ball bearing is lifted (with string stretched) to an angle of 55 degrees and released from rest.

ok so i did this, to find the change in height
i can't figure out this equation stuff on this site but i found the change in height to be 21 cm is that right?
so do i use this equation?
Delta KE + Delta PE=0

1/2m(v^2-Vi^2)+mgdeltah=0

 

[rl.bhat]

Yes. You are in right track. Proceed.

 

[jg597]

ok so for that formula i just made I am confused is the initial velocity zero?

 

[rl.bhat]

Yes. The initial velocity is zero.

 

[jg597]

im i supposed to convert grams into kilograms and cm into meters?

 

[rl.bhat]

Yes.

 

[jg597]

since initial is zero can i do this?
Delta PE= KE
(.065)(9.8)(0.21)=1/2(.065)V^2
.13377=.0325(V^2)
.13377/.0325=V^2
4.116=v^2
V=2.02879m/s

is that right?

 

[jg597]

so if I am right can i use this equations to find the before and after now right?
v1=m1-m2/m1+m2*v initial (for before)
v2=2m1Vi/(m1+m2) (for after)

 

[rl.bhat]

Yes. It is right.

 

[jg597]

yay :)

ok so V1=(.065-.15)(2.02879m/s)/(.065+.15) for before the collision and
V2=2(.065)(2.02879m/s)/(.065+.15)

V before= -.80208 m/s
v after= 1.25 m/s

?

 

[jg597]

for part b i need to solve for delta h again since it changed from 55 degrees to 15 degree?
so do the same thing again that i did for the first one
l= length of string
Delta h=l-lcos
Delta h=50cm-50cmcos15degrees
?

 

[rl.bhat]

yay :)

ok so V1=(.065-.15)(2.02879m/s)/(.065+.15) for before the collision and
V2=2(.065)(2.02879m/s)/(.065+.15)

V before= -.80208 m/s
v after= 1.25 m/s

?

Here V1 is the velocity of .065 kg mass after collision and V2 is that of 0,15 kg.
With these velocities find the heights attained by them using V^2 = 2gh formula.

 

[jg597]

Here V1 is the velocity of .065 kg mass after collision and V2 is that of 0,15 kg.

? I am lost now, so are those formula right?

 

[rl.bhat]

With these velocities find the heights attained by them using V^2 = 2gh formula.

 

[jg597]

V^2 = 2gh formula I am not familiar with this formula
can you show me V1 because i got an extremely small number and i don't know if i did it right.

h= -.032823 and h=.079719 for V2

?

 

[rl.bhat]

V^2 = 2gh formula I am not familiar with this formula
can you show me V1 because i got an extremely small number and i don't know if i did it right.

h= -.032823 and h=.079719 for V2

?

Yes. for V1 it is correct. Using Δh formula find the angle. Repeat the same thing for V2.

 

[jg597]

We are still in part a i thought we were answering the other ones never mind ha ha so we are answering if the collision would have been elastic, to what maximum angle would each bearing reach after the collision. But i still have no clue how to solve for the angle, :(

 

[jg597]

So I am i using this formula?
Delta h=l-lcos

im not sure how to arrange this in terms of cos, can you help me?

 

[rl.bhat]

cosθ = (l - delta h)/l

 

[jg597]

for the small one i got 1.00
for the larger mass i got .998406

is this right?

cos=((50-(-.032823))/50

cos=(50-.079719)/50

is that the correct set up?

 

[rl.bhat]

Check the units of length.

 

[jg597]

do you mean put them in meters?

 

[rl.bhat]

Yes.

 

[jg597]

for smaller i got 1.06565
larger i got .840562

 

[rl.bhat]

cos(theta) cannot be more than one. Take delta h positive in both the cases.

 

[jg597]

Ok so i got part a) to this problem answered but for part b) i have no clue on how to start solving for that one can someone point me in the right direction?

 

[jg597]

with this formula i found delta h to be =.017m
delta(h)=l-lcos
l=lenght
and then i plugged that in for mgh=PE
(.065)(9.8)(.017)=.010829
and then i used Ke=1/2m1Vi^2
since this is inelastic the potential equals the kinetic energy right?

 

[thetruckdawg]

so how start part b,d, and c?