Recent content by Deadawake

If "m" is big enough to create 9.8 m/s2 acceleration. so a=g . Logically,I think, it could be the case. Actually I can't find a reason why "(g-a)" must be positive when "a" is bigger than g... maybe that's the direction of the accelaration? if it minus this term will be always positive. but we...

mm.. If g=a we get 0 . which means there is no tension in the ropes?

Homework Statement M = M There is another mass "m" attached to mass "M" with a rope There is an acceleration (clockwise) Does Tension2 smaller,bigger or equal to Tension1? Homework Equations F = ma The Attempt at a Solution So this is what I understand just by the logic: In any case -...

Some other question is - What height does the small mass ball reach after the collision? (hf) (Just a reminder - we are talking about vertical standing circle/ring...) So I have the initial speed of the small ball (after the collision) and I can use conservation of energy ½mv02 = mghf...

Correct. this statement is result of conservation of energy and momentum equations. Vbefore collision = √(3Rg) 1) MV0 + 0 = MVf+mvf ⇒ 1.5mV0 = 1.5mVf +mvf ⇒ vf = 1.5⋅√(3Rg) - 1.5Vf 2) V0 + Vf = v0 + vf ⇒ √(3Rg) +Vf = 0 +vf ⇒vf = √(3Rg) +Vf ⇒⇒⇒√(3Rg) +Vf = 1.5⋅√(3Rg)...

What am I missing?

Thank you very much. For some reason I thought it needs to bounce backwards.

You right. So basically the answers tell me that the bigger mass in this case didn't move backwards, It continued forward with third of its velocity?

Homework Statement Please see the attached photo. (down) Hminitial= 1.5R M = 2/3m Perfectly elastic collision What is the velocity of object m immidiatly after the collision? (by m,g,R) Homework Equations Conservation of energy Conservation of momentum The Attempt at a Solution I assumed...

Thanks a lot. Is it logical that after the explosion/splitting the bigger mass has more kinetic energy than the smaller mass? this is what I got here and it also doesn't make sense to me.

Thanks. If it doesn't care about the values inside why it gives me different answers ?

Hi, Homework Statement A free falling object of mass "m" falling from some height, collides the floor in speed of 20 m/s (perfectly elastic collision). In his 1/2 height back up he splits into 2 pieces- ¼m which going downward and ¾m keeping upward. The ¼m reaching the floor after ½ second. 1)...

Thank you . So did I do it right or am I missing something? Can someone explain me why in the moment of the collision (the impact) the conservation law does not apply?

Ops... :0 I'm correcting: Compression with no friction 0.044m or 4.4cm Velocity after collision: -0.2 m/s Fk = -9.81N ½⋅100⋅x2 + 9.81⋅x = ½⋅5.005⋅(-0.2)2 x1,2 = 0.0097m (/1cm), -0.206m (/-20cm) Still I left with the same question

I think so - According to the momentum conservation principle: m1v1+m2v2 = (m1+m2)Vfinal Vf = 5⋅0+200⋅0.05 / 5.05 = 1.98 m/s