Kinematics, Energy, Collisions

[kingwinner]

1)

[The main issue for me here is that I am not sure where the collision will occur. If Odie drops the birdbath right before Garfield touches the it, would the collision occur BELOW 1.5m above the ground? (because the birdbath will be moving AWAY from Garfield? Or will it still occur at exactly 1.5m above the ground as if no dropping of the birdbaths occurs?]

 

2)

[i don't quite understand the situtation here (seems complicated). When Snoppy starts moving, will Charlie Brown also be moving? When will Snoppy come to rest? And is it a momentarily rest or will it stay at rest? At that instant, will Charlie Brown also be at rest?]

 

These problems seem lengthy and complicated to me. Any help is greatly appreciated. :cocktail:

[Jay-qu]

number one - by saying the last instant it is released you can assume that that is the collision before the bath starts moving. First consider the momentum at this point and that will equal combined momentum the instant after collision. Then you can work out the velocity, the rest you can just use normal kinematics (constant acceleration) equations or use energy equations - potential converting to kinetic.

[ronthepon]

The second question is not a peice of cake.

 

In the ideal situation, we can say that the friction with ice is zero. Charlie will not be able to provide any reaction force, and the spring will not be streched at all!

 

I could be mistaken.

[Jay-qu]

I believe you are not mistaken ron, and that is the point of saying that the spring is on the point of stretching, still there must have been some tension in the string otherwise he wouldnt move at all.. An educated guess at the answer would be 2.5 - but at the same time I dont see why they would stop at all on a frictionless surface..

[Tim_Lou]

2. the spring will undergo simple harmonic motion there. "comes to rest" seems quite impossible considering the ice is frictionless. anyway, i'll analyze the whole situation there, from it anything can be calculated.

 

first, find out the center of mass of the system, it will be moving at constant velocity. take this point as the frame of reference. substract and/or add velocity accordingly. assume that the compressing and stretching of the spring have the same constant and the spring obeys the hook's law, you can work out an equation:

 

[math]F_{net}=m_1{{d^2x_1}\over{dt^2}}=-kx_{total}[/math]

 

relate [math]x_{total}[/math] to [math]x_1[/math] using center of mass, and solve the differiential equation. substitute in initial conditions and stuffs to find out constants.

 

relative to the center of mass:

 

[math]m_1x_1=m_2x_2[/math]

[math]x_1+x_2=x_{total}[/math]

 

solve for x_total

 

 

from there, everything can be calculated.

just remember that everything is in reference to the center of mass.

 

edited for typoes and clarifications.

edit: perhaps "comes to rest" is saying that the body's velocity becomes zero at one instant. well, you can just work it out using the above equation. (Relative to the ice, when body 1 moves at a velocity opposite to that of the center of mass while have the same magnitude, it will be at "rest" relative to the ice)

[Tim_Lou]

actually, if "comes to rest" is what i think it is. 2 is quite simple.

 

first, use conservation of momentum, calculate the velocity of the second body when the first "comes to rest".

 

since no energy is lost, the energy difference will be conserved in the spring. equate the energy difference to the energy of the spring and calculate the extension length.

[Jay-qu]

but it cant stay extended on frictionless ice..

[sebbysteiny]

What's wrong with this?

 

Go to the centre of mass frame (ie work out the velocity of the centre of mass and now make all velocities relative to that).

 

m1 v1 + m2 v2 = m(Centre of Mass frame) v(CoMf).

v2 = 0.

m = m1 + m2

v(CoMf) = m1v1 / (m1 + m2).

 

Then, work out the total kinetic energy of Charlie AND snoopy in that frame at time t=0.

 

KE = 1/2 m1 (v1-v[CoMf])^2 + 1/2 m2(v2-v[CoMf])^2.

 

when snoopy comes to rest, total kinetic energy at t=0 is equal to the potential energy stored in the spring. E = 1/2 k l^2 where l is the length of the spring and the length of the spring is the same in all frames of reference including that of the stationary ice.

 

l = Sqr[2 KE / k]

[sebbysteiny]

Oh b*****, I just realised what's wrong.

 

My new answer is identical to Tim_Lou's

[Tim_Lou]

what? is something wrong?

 

 

momentum of the system:

15*12=180

 

final velocity of charlie when snoopy "comes to rest"

25*vf=180

vf=7.2

 

initial kinetic energy:

.5*15*12^2=1080

 

finial kinetic energy:

.5*25*7.2^2=648

 

hence potential energy must be increased by:

1080-648=432=.5*k*x^2

 

x=2.4

 

forget what i say about center of mass... if your not solving for the differiential equation. there is no point to change reference to the center of mass...

[sebbysteiny]

what? is something wrong?

 

My new answer is identical to Tim_Lou's

 

No Tim, having seen the error in my version, my new version is identical to yours.

[Tim_Lou]

oh, i thought some of my logic was wrong... :) glad that we arrived at the same answer.