Physics...?

Question

Two balls, of masses Ma=0.036kg and Mb=0.06kg are suspended on 0.3m strings next to each other. The lighter ball is pulled away to a 60degree angle wiht the vertical and released. WHat is the velocity of the lighter ball before impact? What is the velocity of each ball after the elastic collision? What will be the maximum height of each ball(above the collision point) after the elastic collision?

Answer 1

Ma=0.036kg

Mb=0.06kg

length of string =L=0.3 m

The vertical distance through which lighter ball of0.036 kg is raised, when pulled away=h=L[1-costheta]

h=0.3*[1-cos60 ]=0.3*[1-0.5]=0.150 m

the velocity of the lighter ball before impact=u1 =sq rt 2gh

u1= sq rt 2.94 =1.715 m/s

The velocity of the lighter ball before impact is 1.715 m/s
_______________________________________________

As second ball is at rest, momentum of system before collision pi=0.0617 kgm/s

momentum of system before collision pf=0.036v1+0.06v2

As collision is elastic pf=pi

0.036v1+0.06 v2=0.0617

36v1+60v2 = 61.7____________(1)

As second ball is at rest, kinetic energy of system before collision Ki=0.05292 J

kinetic energy of system after collision Kf=0.018v1^2+0.03v2^2

As collision is elastic, Kf=Ki

0.018v1^2+0.03v2^2=0.05292

18v1^2+30v2^2 =52.92___________(2)

solving the equations(1) and(2)

v1= -0.42875 m/s

velocity of lighter ball after collision v1=0.42875 m/s in the direction it was pulled

maximum height of lighter ball =h1=0.0093789 m

v2=1.28625 m/s

velocity of heavier ball after collision v2=1.28625 m/s in direction lighter ball was moving before collision

maximum height of heavier ball = h2 =0.0844 m

Answer 2

Pre collision conservation of energy:
Initial PE of ball = mgr * (1 - cos theta)
= pre-collision KE of ball = 1/2 mva^2
So speed = va = 2gr (1 - cos theta)

Now the collision:
Conservation of energy
Initial KE = 1/2 ma va^2
= Final KE = 1/2 ma vafinal^2 + 1/2 mb vbfinal^2
Conservation of momentum:
Initial p = ma va
= Final p = ma vafinal + mb vbfinal
Two equations. Two unknowns--vafinal and vbfinal. Do the algebra to find them.

Post collision conservation of energy. Last step is setting the post collision KEs equal to PEs just like in stop 1 to find out how high the balls go after the collision. Have fun!

Answer 3

a) Potential energy Pe gets converted to kinetic energy
KePe=Ke
mgh=0.5mV^2
V=sqrt(2gh)
V=sqrt(2gL(1-cos(60))

b) m1V1+ m2V2=m1u1+ m2u2 since V2=0
m1V1=m1u1+ m2u2 (elastic collision)
Ke(before) =Ke (after)
0.5m1V1^2=0.5m1u1^2+ 0.5m2u2^2 or
m1V1^2=m1u1^2+ m2u2^2
we have a system of equations
m1V1=m1u1+ m2u2
m1V1^2=m1u1^2+ m2u2^2
find u1 and u2

c) hmax= Ke max/(mg)
Ke1=Pe1
Ke2=Pe2
h1= Ke1/(m1g)
h2= Ke2/(m2g)